solving problems in hypothesis testing

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9.E: Hypothesis Testing with One Sample (Exercises)

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These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

9.1: Introduction

9.2: null and alternative hypotheses.

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.

State the null hypothesis, $H_{0}$, and the alternative hypothesis. $H_{a}$, in terms of the appropriate parameter $(\mu \text{or} p)$.

The mean number of years Americans work before retiring is 34.
At most 60% of Americans vote in presidential elections.
The mean starting salary for San Jose State University graduates is at least $100,000 per year.
Twenty-nine percent of high school seniors get drunk each month.
Fewer than 5% of adults ride the bus to work in Los Angeles.
The mean number of cars a person owns in her lifetime is not more than ten.
About half of Americans prefer to live away from cities, given the choice.
Europeans have a mean paid vacation each year of six weeks.
The chance of developing breast cancer is under 11% for women.
Private universities' mean tuition cost is more than $20,000 per year.
$H_{0}: \mu = 34; H_{a}: \mu \neq 34$
$H_{0}: p \leq 0.60; H_{a}: p > 0.60$
$H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000$
$H_{0}: p = 0.29; H_{a}: p \neq 0.29$
$H_{0}: p = 0.05; H_{a}: p < 0.05$
$H_{0}: \mu \leq 10; H_{a}: \mu > 10$
$H_{0}: p = 0.50; H_{a}: p \neq 0.50$
$H_{0}: \mu = 6; H_{a}: \mu \neq 6$
$H_{0}: p ≥ 0.11; H_{a}: p < 0.11$
$H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000$

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:

$p < 0.30$
$p \leq 0.30$
$p \geq 0.30$
$p > 0.30$

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:

$p = 0.20$
$p > 0.20$
$p < 0.20$
$p \leq 0.20$

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:

$H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5$
$H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5$
$H_{0}: \mu = 4.75, H_{a}: \mu > 4.75$
$H_{0}: \mu = 4.5, H_{a}: \mu > 4.5$

9.3: Outcomes and the Type I and Type II Errors

State the Type I and Type II errors in complete sentences given the following statements.

The mean number of cars a person owns in his or her lifetime is not more than ten.
Private universities mean tuition cost is more than $20,000 per year.
Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.

For statements a-j in Exercise 9.109 , answer the following in complete sentences.

State a consequence of committing a Type I error.
State a consequence of committing a Type II error.

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?

To conclude the drug is safe when in, fact, it is unsafe.
Not to conclude the drug is safe when, in fact, it is safe.
To conclude the drug is safe when, in fact, it is safe.
Not to conclude the drug is unsafe when, in fact, it is unsafe.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.

at least 20%, when in fact, it is less than 20%.
20%, when in fact, it is 20%.
less than 20%, when in fact, it is at least 20%.
less than 20%, when in fact, it is less than 20%.

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours

is more than seven hours.
is at most seven hours.
is at least seven hours.
is less than seven hours.

to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

9.4: Distribution Needed for Hypothesis Testing

$N\left(7.24, \frac{1.93}{\sqrt{22}}\right)$
$N\left(7.24, 1.93\right)$

9.5: Rare Events, the Sample, Decision and Conclusion

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

Is this a test of one mean or proportion?
State the null and alternative hypotheses. $H_{0}$ : ____________________ $H_{a}$ : ____________________
Is this a right-tailed, left-tailed, or two-tailed test?
What symbol represents the random variable for this test?
In words, define the random variable for this test.
$x =$ ________________
$n =$ ________________
$p′ =$ _____________
Calculate $\sigma_{x} =$ __________. Show the formula set-up.
State the distribution to use for the hypothesis test.
Find the $p\text{-value}$.
Reason for the decision:
Conclusion (write out in a complete sentence):

9.6: Additional Information and Full Hypothesis Test Examples

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link] . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

If you are using a Student's $t$ - distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using $\alpha = 0.05$, is the data highly inconsistent with the claim?

$H_{0}: \mu \geq 50,000$
$H_{a}: \mu < 50,000$
Let $\bar{X} =$ the average lifespan of a brand of tires.
normal distribution
$z = -2.315$
$p\text{-value} = 0.0103$
Check student’s solution.
alpha: 0.05
Decision: Reject the null hypothesis.
Reason for decision: The $p\text{-value}$ is less than 0.05.
Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
$(43,537, 49,463)$

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?

$H_{0}: \mu = $1.00$
$H_{a}: \mu \neq $1.00$
Let $\bar{X} =$ the average cost of a daily newspaper.
$z = –0.866$
$p\text{-value} = 0.3865$
$\alpha: 0.01$
Decision: Do not reject the null hypothesis.
Reason for decision: The $p\text{-value}$ is greater than 0.01.
Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
$($0.84, $1.06)$

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let $x =$ the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?

$H_{0}: \mu = 10$
$H_{a}: \mu \neq 10$
Let $\bar{X}$ the mean number of sick days an employee takes per year.
Student’s t -distribution
$t = –1.12$
$p\text{-value} = 0.300$
$\alpha: 0.05$
Reason for decision: The $p\text{-value}$ is greater than 0.05.
Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
$(4.9443, 11.806)$

In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

$H_{0}: p \geq 0.6$
$H_{a}: p < 0.6$
Let $P′ =$ the proportion of students who feel more enriched as a result of taking Elementary Statistics.
normal for a single proportion
$p\text{-value} = 0.1308$
Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.

The “plus-4s” confidence interval is $(0.411, 0.648)$

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Refer to Exercise 9.119 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

$H_{0}: \mu = 4$
$H_{a}: \mu \neq 4$
Let $\bar{X}$ the average I.Q. of a set of brown trout.
two-tailed Student's t-test
$t = 1.95$
$p\text{-value} = 0.076$
Reason for decision: The $p\text{-value}$ is greater than 0.05
Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
$(3.8865,5.9468)$

According to an article in Newsweek , the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?

A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

$H_{a}: p < 0.13$
Let $P′ =$ the proportion of Americans who have seen or sensed angels
–2.688
$p\text{-value} = 0.0036$
Reason for decision: The $p\text{-value}$e is less than 0.05.
Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.

The“plus-4s” confidence interval is (0.0022, 0.0978)

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

Use the “Lap time” data for Lap 4 (see [link] ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.

$H_{0}: \mu \geq 129$
$H_{a}: \mu < 129$
Let $\bar{X} =$ the average time in seconds that Terri finishes Lap 4.
Student's t -distribution
$t = 1.209$
Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
$(128.63, 130.37)$

Use the “Initial Public Offering” data (see [link] ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

The following questions were written by past students. They are excellent problems!

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

$H_{0}: p = 0.60$
$H_{a}: p < 0.60$
Let $P′ =$ the proportion of family members who shed tears at a reunion.
–1.71
Reason for decision: $p\text{-value} < \alpha$
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the $p\text{-value}$ and alpha are quite close, so other tests should be done.
We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. $(0.3829, 0.6171)$. The“plus-4s” confidence interval (see chapter 8) is $(0.3861, 0.6139)$

Note that here the “large-sample” $1 - \text{PropZTest}$ provides the approximate $p\text{-value}$ of 0.0438. Whenever a $p\text{-value}$ based on a normal approximation is close to the level of significance, the exact $p\text{-value}$ based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

"The Problem with Angels," by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.

On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be

In angels, Americans do believe.

Then, it was time to rise to the task,

Ninety-five high school and college students I did ask.

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,

"Do you believe in angels?" Tell me, do!

Hypothesizing at the start,

Totally believing in my heart

That the proportion who said yes

Would be equal on this test.

Lo and behold, seventy-three did arrive,

Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

$H_{0}: \mu \geq 22$
$H_{a}: \mu < 22$
Let $\bar{X} =$ the mean number of bubbles per blow.
–2.667
$p\text{-value} = 0.00486$
Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
$(18.501, 21.499)$

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese:

5 stores @ $2.02
15 stores @ $0.25
3 stores @ $1.29
6 stores @ $0.35
4 stores @ $2.27
7 stores @ $1.50
5 stores @ $1.89
8 stores @ 0.75.

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)

$H_{0}: \mu \leq 1$
$H_{a}: \mu > 1$
Let $\bar{X} =$ the mean cost in dollars of macaroni and cheese in a certain town.
Student's $t$-distribution
$t = 0.340$
$p\text{-value} = 0.36756$
Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
$(0.8291, 1.241)$

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

THE CHARACTERS (in order of appearance):

HAMLET, Prince of Denmark and student of Statistics
POLONIUS, Hamlet’s tutor
HOROTIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons

(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)

(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

$H_{0}: p = 0.01$
$H_{a}: p > 0.01$
Let $P′ =$ the proportion of errors generated
Normal for a single proportion
Decision: Reject the null hypothesis
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.

The“plus-4s” confidence interval is $(0.004, 0.144)$.

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

$H_{0}: p = 0.50$
$H_{a}: p < 0.50$
Let $P′ =$ the proportion of friends that has a pierced ear.
–1.70
$p\text{-value} = 0.0448$
Reason for decision: The $p\text{-value}$ is less than 0.05. (However, they are very close.)
Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
Confidence Interval: $(0.245, 0.515)$: The “plus-4s” confidence interval is $(0.259, 0.519)$.

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.

$H_{0}: p = 0.40$
$H_{a}: p < 0.40$
Let $P′ =$ the proportion of schoolmates who fear public speaking.
–1.01
$p\text{-value} = 0.1563$
Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
Confidence Interval: $(0.3241, 0.4240)$: The “plus-4s” confidence interval is $(0.3257, 0.4250)$.

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

According to an article in Bloomberg Businessweek , New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.

$H_{0}: p = 0.14$
$H_{a}: p < 0.14$
Let $P′ =$ the proportion of NYC residents that smoke.
–0.2756
$p\text{-value} = 0.3914$
At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
Confidence Interval: $(0.0502, 0.2070)$: The “plus-4s” confidence interval (see chapter 8) is $(0.0676, 0.2297)$.

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

$H_{0}: \mu = 69,110$
$H_{0}: \mu > 69,110$
Let $\bar{X} =$ the mean salary in dollars for California registered nurses.
$t = 1.719$
$p\text{-value}: 0.0466$
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
$($68,757, $73,485)$

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

After conducting the test, your decision and conclusion are

Reject $H_{0}$: There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
Do not reject $H_{0}$: There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
Do not reject $H_{0}$: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
Reject $H_{0}$: There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.

At a 1% level of significance, an appropriate conclusion is:

There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.

At a significance level of $a = 0.05$, what is the correct conclusion?

There is enough evidence to conclude that the mean number of hours is more than 4.75
There is enough evidence to conclude that the mean number of hours is more than 4.5
There is not enough evidence to conclude that the mean number of hours is more than 4.5
There is not enough evidence to conclude that the mean number of hours is more than 4.75

Instructions: For the following ten exercises,

Hypothesis testing: For the following ten exercises, answer each question.

State the null and alternate hypothesis.

State the $p\text{-value}$.

State $\alpha$.

What is your decision?

Write a conclusion.

Answer any other questions asked in the problem.

According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

$H_{0}: p = 0.488$ $H_{a}: p \neq 0.488$
$p\text{-value} = 0.0114$
$\alpha = 0.05$
Reject the null hypothesis.
At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
The survey does not appear to be accurate.

Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using $\alpha = 0.05$, is the AAA proportion accurate?

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the $\alpha = 0.05$ level in Kentucky? Are the results applicable across the country? Why?

$H_{0}: p = 0.517$ $H_{0}: p \neq 0.517$
$p\text{-value} = 0.9203$.
$\alpha = 0.05$.
Do not reject the null hypothesis.
At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use $\alpha = 0.01$ level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the $\alpha = 0.05 level$, can it be concluded that the mean rainfall was below the reported average? What if $\alpha = 0.01$? Assume the amount of summer rainfall follows a normal distribution.

$H_{0}: \mu \geq 11.52$ $H_{a}: \mu < 11.52$
$p\text{-value} = 0.000002$ which is almost 0.
At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
We would make the same conclusion if alpha was 1% because the $p\text{-value}$ is almost 0.

A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the $\alpha = 0.10$ level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals

3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1

At the $\alpha = 0.05$ level can it be concluded that the sample mean is higher than 5.8 visits per year?

$H_{0}: \mu \leq 5.8$ $H_{a}: \mu > 5.8$
$p\text{-value} = 0.9987$
At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:

5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2

At $\alpha = 0.05$ level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

$H_{0}: \mu \geq 150$ $H_{0}: \mu < 150$
$p\text{-value} = 0.0622$
$\alpha = 0.01$
At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
The student academic group’s claim appears to be correct.

9.7: Hypothesis Testing of a Single Mean and Single Proportion

9.4 Full Hypothesis Test Examples

Tests on means, example 9.8.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

H 0 : μ = 16.43 H a : μ < 16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: X ¯ X ¯ = the mean time to swim the 25-yard freestyle.

Distribution for the test: X ¯ X ¯ is normal (population standard deviation is known: σ = 0.8)

X ¯ ~ N ( μ , σ X n ) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) X ¯ ~ N ( 16.43 , 0.8 15 )

μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15.

Calculate the p -value using the normal distribution for a mean:

p -value = P ( x ¯ x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16.

p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16.

μ = 16.43 comes from H 0 . Our assumption is μ = 16.43.

Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p -value:

α = 0.05 p -value = 0.0187 α > p -value

Make a decision: Since α > α > p -value, reject H 0 .

This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds.

Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles.

The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in at least 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

The mean throwing distance of a football for Marco, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Example 9.9

Jasmine has just begun her new job on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of 108 dollars with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least 100 dollars against the alternative that it is less than 100 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jasmine has met this requirement at the significance level of 95%?

H 0 : µ ≤ 100 H a : µ > 100 The null and alternative hypothesis are for the parameter µ because the number of dollars of the contracts is a continuous random variable. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not "too high" a number. This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis.
Test statistic: t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67 t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67
Critical value: t a = 1.753 t a = 1.753 with n-1 degrees of freedom= 15

The test statistic is a Student's t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t t ( t a ) ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jasmine's performance meets company standards.

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, state your conclusion, and identify the Type I errors.

Example 9.10

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, s 2 s 2 . Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of significance in the analysis should be 99%.

Again we will follow the steps in our analysis of this problem.

STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus:

STEP 2 : Decide the level of significance and draw the graph showing the critical value.

This problem has already set the level of significance at 99%. The decision seems an appropriate one and shows the thought process when setting the significance level. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table or the t-table at 0.005 column and infinite degrees of freedom. We draw the graph and mark these points.

STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph.

STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis.

STEP 5 : Reach a Conclusion

Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of significance we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of significance we conclude that the machine is under filling the bottles and is in need of repair”.

Try It 9.10

A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2).

Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 .

Hypothesis Test for Proportions

Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p .

The population parameter for the binomial is p . The estimated value (point estimate) for p is p′ where p′ = x/n , x is the number of successes in the sample and n is the sample size.

When you perform a hypothesis test of a population proportion p , you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five ( np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = np μ = np and σ = npq σ = npq . Remember that q = 1 – p q = 1 – p . There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p .

Again, we begin with the standardizing formula modified because this is the distribution of a binomial.

Substituting p 0 p 0 , the hypothesized value of p , we have:

This is the test statistic for testing hypothesized values of p , where the null and alternative hypotheses take one of the following forms:

The decision rule stated above applies here also: if the calculated value of Z c shows that the sample proportion is "too many" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is "too many" is pre-determined by the analyst depending on the level of significance required in the test.

Example 9.11

The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50% . They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance.

STEP 1 : Set the null and alternative hypothesis.

H 0 : p = 0.50 H a : p ≠ 0.50

The words "is the same or different from" tell you this is a two-tailed test. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of borrowers is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time borrowers differs from 50% when, in fact, the proportion does differ from 50%. (You fail to reject the null hypothesis when the null hypothesis is false.)

STEP 2 : Decide the level of significance and draw the graph showing the critical value

The level of significance has been set by the problem at the 5% level. Because this is two-tailed test one-half of the alpha value will be in the upper tail and one-half in the lower tail as shown on the graph. The critical value for the normal distribution at the 95% level of confidence is 1.96. This can easily be found on the student’s t-table at the very bottom at infinite degrees of freedom remembering that at infinity the t-distribution is the normal distribution. Of course the value can also be found on the normal table but you have go looking for one-half of 95 (0.475) inside the body of the table and then read out to the sides and top for the number of standard deviations.

STEP 3 : Calculate the sample parameters and critical value of the test statistic.

The test statistic is a normal distribution, Z, for testing proportions and is:

For this case, the sample of 100 found 53 of these loans were smaller than those of other borrowers. The sample proportion, p′ = 53/100= 0.53 The test question, therefore, is : “Is 0.53 significantly different from .50?” Putting these values into the formula for the test statistic we find that 0.53 is only 0.60 standard deviations away from .50. This is barely off of the mean of the standard normal distribution of zero. There is virtually no difference from the sample proportion and the hypothesized proportion in terms of standard deviations.

STEP 4 : Compare the test statistic and the critical value.

The calculated value is well within the critical values of ± 1.96 standard deviations and thus we cannot reject the null hypothesis. To reject the null hypothesis we need significant evident of difference between the hypothesized value and the sample value. In this case the sample value is very nearly the same as the hypothesized value measured in terms of standard deviations.

STEP 5 : Reach a conclusion

The formal conclusion would be “At a 5% level of significance we cannot reject the null hypothesis that 50% of first-time borrowers take out smaller loans than other borrowers.” Notice the length to which the conclusion goes to include all of the conditions that are attached to the conclusion. Statisticians, for all the criticism they receive, are careful to be very specific even when this seems trivial. Statisticians cannot say more than they know, and the data constrain the conclusion to be within the metes and bounds of the data.

Try It 9.11

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

Example 9.12

Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones.

Here is an abbreviate version of the system to solve hypothesis tests applied to a test on a proportions.

Try It 9.12

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

Example 9.13

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98; 1.02; .95; .95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05.

Let’s follow a four-step process to answer this statistical question.

H 0 : μ ≤ 1
H a : μ > 1
Plan : We are testing a sample mean without a known population standard deviation with less than 30 observations. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
Do the calculations and draw the graph .
State the Conclusions : We cannot accept the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

Try It 9.13

The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows:

205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205

Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal.

Example 9.14

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

H 0 : p ≤ 0.00034
H a : p > 0.00034

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np' = 420,019(0.00034) = 142.8, nq' = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p' = 0.00034. Thus we will be able to generalize our results to the population.

Try It 9.14

In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

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Hypothesis Testing

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A hypothesis test is a statistical inference method used to test the significance of a proposed (hypothesized) relation between population statistics (parameters) and their corresponding sample estimators . In other words, hypothesis tests are used to determine if there is enough evidence in a sample to prove a hypothesis true for the entire population.

The test considers two hypotheses: the null hypothesis , which is a statement meant to be tested, usually something like "there is no effect" with the intention of proving this false, and the alternate hypothesis , which is the statement meant to stand after the test is performed. The two hypotheses must be mutually exclusive ; moreover, in most applications, the two are complementary (one being the negation of the other). The test works by comparing the $p$-value to the level of significance (a chosen target). If the $p$-value is less than or equal to the level of significance, then the null hypothesis is rejected.

When analyzing data, only samples of a certain size might be manageable as efficient computations. In some situations the error terms follow a continuous or infinite distribution, hence the use of samples to suggest accuracy of the chosen test statistics. The method of hypothesis testing gives an advantage over guessing what distribution or which parameters the data follows.

Definitions and Methodology

Hypothesis test and confidence intervals.

In statistical inference, properties (parameters) of a population are analyzed by sampling data sets. Given assumptions on the distribution, i.e. a statistical model of the data, certain hypotheses can be deduced from the known behavior of the model. These hypotheses must be tested against sampled data from the population.

The null hypothesis $($denoted $H_0)$ is a statement that is assumed to be true. If the null hypothesis is rejected, then there is enough evidence (statistical significance) to accept the alternate hypothesis $($denoted $H_1).$ Before doing any test for significance, both hypotheses must be clearly stated and non-conflictive, i.e. mutually exclusive, statements. Rejecting the null hypothesis, given that it is true, is called a type I error and it is denoted $\alpha$, which is also its probability of occurrence. Failing to reject the null hypothesis, given that it is false, is called a type II error and it is denoted $\beta$, which is also its probability of occurrence. Also, $\alpha$ is known as the significance level , and $1-\beta$ is known as the power of the test. $H_0$ $\textbf{is true}$$\hspace{15mm}$ $H_0$ $\textbf{is false}$ $\textbf{Reject}$ $H_0$$\hspace{10mm}$ Type I error Correct Decision $\textbf{Reject}$ $H_1$ Correct Decision Type II error The test statistic is the standardized value following the sampled data under the assumption that the null hypothesis is true, and a chosen particular test. These tests depend on the statistic to be studied and the assumed distribution it follows, e.g. the population mean following a normal distribution. The $p$-value is the probability of observing an extreme test statistic in the direction of the alternate hypothesis, given that the null hypothesis is true. The critical value is the value of the assumed distribution of the test statistic such that the probability of making a type I error is small.

Methodologies: Given an estimator $\hat \theta$ of a population statistic $\theta$, following a probability distribution $P(T)$, computed from a sample $\mathcal{S},$ and given a significance level $\alpha$ and test statistic $t^*,$ define $H_0$ and $H_1;$ compute the test statistic $t^*.$ $p$-value Approach (most prevalent): Find the $p$-value using $t^*$ (right-tailed). If the $p$-value is at most $\alpha,$ reject $H_0$. Otherwise, reject $H_1$. Critical Value Approach: Find the critical value solving the equation $P(T\geq t_\alpha)=\alpha$ (right-tailed). If $t^*>t_\alpha$, reject $H_0$. Otherwise, reject $H_1$. Note: Failing to reject $H_0$ only means inability to accept $H_1$, and it does not mean to accept $H_0$.

Assume a normally distributed population has recorded cholesterol levels with various statistics computed. From a sample of 100 subjects in the population, the sample mean was 214.12 mg/dL (milligrams per deciliter), with a sample standard deviation of 45.71 mg/dL. Perform a hypothesis test, with significance level 0.05, to test if there is enough evidence to conclude that the population mean is larger than 200 mg/dL. Hypothesis Test We will perform a hypothesis test using the $p$-value approach with significance level $\alpha=0.05:$ Define $H_0$: $\mu=200$. Define $H_1$: $\mu>200$. Since our values are normally distributed, the test statistic is $z^*=\frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}}=\frac{214.12 - 200}{\frac{45.71}{\sqrt{100}}}\approx 3.09$. Using a standard normal distribution, we find that our $p$-value is approximately $0.001$. Since the $p$-value is at most $\alpha=0.05,$ we reject $H_0$. Therefore, we can conclude that the test shows sufficient evidence to support the claim that $\mu$ is larger than $200$ mg/dL.

If the sample size was smaller, the normal and $t$-distributions behave differently. Also, the question itself must be managed by a double-tail test instead.

Assume a population's cholesterol levels are recorded and various statistics are computed. From a sample of 25 subjects, the sample mean was 214.12 mg/dL (milligrams per deciliter), with a sample standard deviation of 45.71 mg/dL. Perform a hypothesis test, with significance level 0.05, to test if there is enough evidence to conclude that the population mean is not equal to 200 mg/dL. Hypothesis Test We will perform a hypothesis test using the $p$-value approach with significance level $\alpha=0.05$ and the $t$-distribution with 24 degrees of freedom: Define $H_0$: $\mu=200$. Define $H_1$: $\mu\neq 200$. Using the $t$-distribution, the test statistic is $t^*=\frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}}=\frac{214.12 - 200}{\frac{45.71}{\sqrt{25}}}\approx 1.54$. Using a $t$-distribution with 24 degrees of freedom, we find that our $p$-value is approximately $2(0.068)=0.136$. We have multiplied by two since this is a two-tailed argument, i.e. the mean can be smaller than or larger than. Since the $p$-value is larger than $\alpha=0.05,$ we fail to reject $H_0$. Therefore, the test does not show sufficient evidence to support the claim that $\mu$ is not equal to $200$ mg/dL.

The complement of the rejection on a two-tailed hypothesis test (with significance level $\alpha$) for a population parameter $\theta$ is equivalent to finding a confidence interval $($with confidence level $1-\alpha)$ for the population parameter $\theta$. If the assumption on the parameter $\theta$ falls inside the confidence interval, then the test has failed to reject the null hypothesis $($with $p$-value greater than $\alpha).$ Otherwise, if $\theta$ does not fall in the confidence interval, then the null hypothesis is rejected in favor of the alternate $($with $p$-value at most $\alpha).$

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1.2 - the 7 step process of statistical hypothesis testing.

We will cover the seven steps one by one.

Step 1: State the Null Hypothesis

The null hypothesis can be thought of as the opposite of the "guess" the researchers made. In the example presented in the previous section, the biologist "guesses" plant height will be different for the various fertilizers. So the null hypothesis would be that there will be no difference among the groups of plants. Specifically, in more statistical language the null for an ANOVA is that the means are the same. We state the null hypothesis as:

$H_0 \colon \mu_1 = \mu_2 = ⋯ = \mu_T$

for T levels of an experimental treatment.

Step 2: State the Alternative Hypothesis

$H_A \colon \text{ treatment level means not all equal}$

The alternative hypothesis is stated in this way so that if the null is rejected, there are many alternative possibilities.

For example, $\mu_1\ne \mu_2 = ⋯ = \mu_T$ is one possibility, as is $\mu_1=\mu_2\ne\mu_3= ⋯ =\mu_T$. Many people make the mistake of stating the alternative hypothesis as $\mu_1\ne\mu_2\ne⋯\ne\mu_T$ which says that every mean differs from every other mean. This is a possibility, but only one of many possibilities. A simple way of thinking about this is that at least one mean is different from all others. To cover all alternative outcomes, we resort to a verbal statement of "not all equal" and then follow up with mean comparisons to find out where differences among means exist. In our example, a possible outcome would be that fertilizer 1 results in plants that are exceptionally tall, but fertilizers 2, 3, and the control group may not differ from one another.

Step 3: Set $\alpha$

If we look at what can happen in a hypothesis test, we can construct the following contingency table:

You should be familiar with Type I and Type II errors from your introductory courses. It is important to note that we want to set $\alpha$ before the experiment ( a-priori ) because the Type I error is the more grievous error to make. The typical value of $\alpha$ is 0.05, establishing a 95% confidence level. For this course, we will assume $\alpha$ =0.05, unless stated otherwise.

Step 4: Collect Data

Remember the importance of recognizing whether data is collected through an experimental design or observational study.

Step 5: Calculate a test statistic

For categorical treatment level means, we use an F- statistic, named after R.A. Fisher. We will explore the mechanics of computing the F- statistic beginning in Lesson 2. The F- value we get from the data is labeled $F_{\text{calculated}}$.

Step 6: Construct Acceptance / Rejection regions

As with all other test statistics, a threshold (critical) value of F is established. This F- value can be obtained from statistical tables or software and is referred to as $F_{\text{critical}}$ or $F_\alpha$. As a reminder, this critical value is the minimum value of the test statistic (in this case $F_{\text{calculated}}$) for us to reject the null.

The F- distribution, $F_\alpha$, and the location of acceptance/rejection regions are shown in the graph below:

Step 7: Based on Steps 5 and 6, draw a conclusion about $H_0$

If $F_{\text{calculated}}$ is larger than $F_\alpha$, then you are in the rejection region and you can reject the null hypothesis with $\left(1-\alpha \right)$ level of confidence.

Note that modern statistical software condenses Steps 6 and 7 by providing a p -value. The p -value here is the probability of getting an $F_{\text{calculated}}$ even greater than what you observe assuming the null hypothesis is true. If by chance, the $F_{\text{calculated}} = F_\alpha$, then the p -value would be exactly equal to $\alpha$. With larger $F_{\text{calculated}}$ values, we move further into the rejection region and the p- value becomes less than $\alpha$. So, the decision rule is as follows:

If the p- value obtained from the ANOVA is less than $\alpha$, then reject $H_0$ in favor of $H_A$.

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A Beginner’s Guide to Hypothesis Testing in Business

Business professionals performing hypothesis testing

30 Mar 2021

Becoming a more data-driven decision-maker can bring several benefits to your organization, enabling you to identify new opportunities to pursue and threats to abate. Rather than allowing subjective thinking to guide your business strategy, backing your decisions with data can empower your company to become more innovative and, ultimately, profitable.

If you’re new to data-driven decision-making, you might be wondering how data translates into business strategy. The answer lies in generating a hypothesis and verifying or rejecting it based on what various forms of data tell you.

Below is a look at hypothesis testing and the role it plays in helping businesses become more data-driven.

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What Is Hypothesis Testing?

To understand what hypothesis testing is, it’s important first to understand what a hypothesis is.

A hypothesis or hypothesis statement seeks to explain why something has happened, or what might happen, under certain conditions. It can also be used to understand how different variables relate to each other. Hypotheses are often written as if-then statements; for example, “If this happens, then this will happen.”

Hypothesis testing , then, is a statistical means of testing an assumption stated in a hypothesis. While the specific methodology leveraged depends on the nature of the hypothesis and data available, hypothesis testing typically uses sample data to extrapolate insights about a larger population.

Hypothesis Testing in Business

When it comes to data-driven decision-making, there’s a certain amount of risk that can mislead a professional. This could be due to flawed thinking or observations, incomplete or inaccurate data , or the presence of unknown variables. The danger in this is that, if major strategic decisions are made based on flawed insights, it can lead to wasted resources, missed opportunities, and catastrophic outcomes.

The real value of hypothesis testing in business is that it allows professionals to test their theories and assumptions before putting them into action. This essentially allows an organization to verify its analysis is correct before committing resources to implement a broader strategy.

As one example, consider a company that wishes to launch a new marketing campaign to revitalize sales during a slow period. Doing so could be an incredibly expensive endeavor, depending on the campaign’s size and complexity. The company, therefore, may wish to test the campaign on a smaller scale to understand how it will perform.

In this example, the hypothesis that’s being tested would fall along the lines of: “If the company launches a new marketing campaign, then it will translate into an increase in sales.” It may even be possible to quantify how much of a lift in sales the company expects to see from the effort. Pending the results of the pilot campaign, the business would then know whether it makes sense to roll it out more broadly.

Related: 9 Fundamental Data Science Skills for Business Professionals

Key Considerations for Hypothesis Testing

1. alternative hypothesis and null hypothesis.

In hypothesis testing, the hypothesis that’s being tested is known as the alternative hypothesis . Often, it’s expressed as a correlation or statistical relationship between variables. The null hypothesis , on the other hand, is a statement that’s meant to show there’s no statistical relationship between the variables being tested. It’s typically the exact opposite of whatever is stated in the alternative hypothesis.

For example, consider a company’s leadership team that historically and reliably sees $12 million in monthly revenue. They want to understand if reducing the price of their services will attract more customers and, in turn, increase revenue.

In this case, the alternative hypothesis may take the form of a statement such as: “If we reduce the price of our flagship service by five percent, then we’ll see an increase in sales and realize revenues greater than $12 million in the next month.”

The null hypothesis, on the other hand, would indicate that revenues wouldn’t increase from the base of $12 million, or might even decrease.

Check out the video below about the difference between an alternative and a null hypothesis, and subscribe to our YouTube channel for more explainer content.

2. Significance Level and P-Value

Statistically speaking, if you were to run the same scenario 100 times, you’d likely receive somewhat different results each time. If you were to plot these results in a distribution plot, you’d see the most likely outcome is at the tallest point in the graph, with less likely outcomes falling to the right and left of that point.

With this in mind, imagine you’ve completed your hypothesis test and have your results, which indicate there may be a correlation between the variables you were testing. To understand your results' significance, you’ll need to identify a p-value for the test, which helps note how confident you are in the test results.

In statistics, the p-value depicts the probability that, assuming the null hypothesis is correct, you might still observe results that are at least as extreme as the results of your hypothesis test. The smaller the p-value, the more likely the alternative hypothesis is correct, and the greater the significance of your results.

3. One-Sided vs. Two-Sided Testing

When it’s time to test your hypothesis, it’s important to leverage the correct testing method. The two most common hypothesis testing methods are one-sided and two-sided tests , or one-tailed and two-tailed tests, respectively.

Typically, you’d leverage a one-sided test when you have a strong conviction about the direction of change you expect to see due to your hypothesis test. You’d leverage a two-sided test when you’re less confident in the direction of change.

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4. Sampling

To perform hypothesis testing in the first place, you need to collect a sample of data to be analyzed. Depending on the question you’re seeking to answer or investigate, you might collect samples through surveys, observational studies, or experiments.

A survey involves asking a series of questions to a random population sample and recording self-reported responses.

Observational studies involve a researcher observing a sample population and collecting data as it occurs naturally, without intervention.

Finally, an experiment involves dividing a sample into multiple groups, one of which acts as the control group. For each non-control group, the variable being studied is manipulated to determine how the data collected differs from that of the control group.

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Learn How to Perform Hypothesis Testing

Hypothesis testing is a complex process involving different moving pieces that can allow an organization to effectively leverage its data and inform strategic decisions.

If you’re interested in better understanding hypothesis testing and the role it can play within your organization, one option is to complete a course that focuses on the process. Doing so can lay the statistical and analytical foundation you need to succeed.

Do you want to learn more about hypothesis testing? Explore Business Analytics —one of our online business essentials courses —and download our Beginner’s Guide to Data & Analytics .

About the Author

Hypothesis Testing: A Comprehensive Guide to Scientific Decision-Making

Master hypothesis testing to bolster your scientific decision-making skills with our in-depth guide. Elevate your research now!

In scientific research and experimentation, one needs a structured framework for answering questions, confirming results, and making decisions.

This framework, known as hypothesis testing , plays a pivotal role both in research and in various industries like healthcare, finance, and technology.

This guide will delve into the principles and processes of hypothesis testing, offering readers a holistic understanding of this fundamental aspect of scientific decision-making.

Definition of Hypothesis Testing

Hypothesis testing is a method used in statistics to decide whether a statement about a population parameter is likely to be true based on sample data.

The process involves making an initial assumption, observing data, then determining how compatible the data is with the assumption. It's a core part of many o nline certificate programs and widely used in fields requiring data analysis.

Importance of Hypothesis Testing in Research and Industry

The value of hypothesis testing goes beyond science and research. Businesses use it for making crucial decisions, such as whether a new product will succeed in the market, or if a change in strategy will lead to increased profit margins.

Similarly, in the healthcare sector, hypothesis testing helps determine if a new medication is more effective than the current standard treatment. This broad applicability underlies the significance of a problem-solving course that includes hypothesis testing.

Understanding the Basics of Hypothesis Testing

Before embarking on the journey of hypothesis testing, it's crucial to understand its fundamental elements - the Null and Alternate Hypotheses.

Explanation of Null Hypothesis

Definition Null Hypothesis

The Null Hypothesis, symbolized as H0, is a statement we test for possible rejection under the assumption that it is true. In most cases, it anticipates no effect, no difference, or no relationship between variables.

How to Formulate a Null Hypothesis

Formulating a null hypothesis requires identifying your research question, specifying your outcome variable, and expressing a statement of no effect or difference. For instance, you may hypothesize, "There is no significant difference between the performance of students who have breakfast and those who don't."

Practical Examples of Null Hypotheses

Consider a beverage company aiming to reduce its plastic use by 20% within a year. The null hypothesis might state that "There has been no decrease in the company's plastic use."

Explanation of Alternate Hypothesis

Definition Alternate Hypothesis

The alternative hypothesis, symbolized as H1, is the statement we accept when there's sufficient evidence against the null hypothesis. It anticipates an effect, a difference, or a relationship between variables.

How to Construct an Alternate Hypothesis

In constructing an alternate hypothesis, we simply state the counter of the null hypothesis. Following the above example, the alternate hypothesis would be, "There's a significant difference between the performance of students who have breakfast and those who don't."

Case Examples of Alternate Hypotheses

Referring to the beverage company example, the alternate hypothesis would state, "There has been a decrease in the company's plastic use."

Differentiating between Null and Alternate Hypotheses

While both form the crux of hypothesis testing, their roles differ significantly. The null hypothesis is the claim we test for possible rejection, while the alternate hypothesis is accepted when there's evidence against the null. However, neither proof nor disproof of either hypothesis is definitive since all statistical tests are susceptible to errors.

Understanding Errors in Hypothesis Testing

A critical aspect of hypothesis testing is the recognition and management of two types of errors: Type I and Type II errors. Understanding these errors is paramount for interpreting the results accurately and making informed decisions.

Type I Error: False Positive

A Type I error occurs when the null hypothesis is wrongly rejected when it is actually true. This is akin to a false alarm, where, for instance, a test indicates a drug is effective against a disease when it actually isn't. The probability of committing a Type I error is denoted by alpha (α), often set at 0.05 or 5%, indicating a 5% risk of rejecting the null hypothesis incorrectly.

Type II Error: False Negative

Conversely, a Type II error happens when the null hypothesis is not rejected when it is false. This can be compared to a missed detection, such as failing to identify the effectiveness of a beneficial drug. The probability of a Type II error is denoted by beta (β), and researchers strive to minimize this risk to ensure that genuine effects are detected.

Balancing the Risks: Power of the Test

The power of a statistical test is the probability that it correctly rejects a false null hypothesis, essentially avoiding a Type II error. High-powered tests are more reliable for detecting true effects. The power is influenced by the sample size, effect size, significance level, and variability within the data. Optimizing these factors can reduce the chances of both Type I and Type II errors, leading to more trustworthy conclusions.

Steps in Hypothesis Testing

Hypothesis testing involves a series of structured steps to guide researchers and professionals through the decision-making process:

Formulate Hypotheses : Clearly define the null and alternative hypotheses based on the research question or problem statement.

Choose a Significance Level (α) : Decide on the alpha level, which determines the threshold for rejecting the null hypothesis.

Select the Appropriate Test : Based on the data type and study design, choose a statistical test that aligns with the research objectives.

Collect and Analyze Data : Gather the necessary data and perform the statistical test to calculate the test statistic and p-value.

Make a Decision : Compare the p-value to the significance level. If the p-value is less than α, reject the null hypothesis in favor of the alternative. Otherwise, do not reject the null hypothesis.

Hypothesis testing is a cornerstone of scientific inquiry, providing a rigorous framework for evaluating theories, exploring relationships, and making decisions based on empirical evidence.

Whether in academia, healthcare, finance, or technology, the principles of hypothesis testing enable practitioners to draw conclusions with a defined level of confidence, navigate uncertainties, and contribute to advancements in their fields. By understanding its fundamentals, errors, and steps, professionals can apply hypothesis testing to enhance decision-making processes and achieve more reliable outcomes.

Through this exploration of hypothesis testing, it becomes clear that the method is not just a statistical tool but a comprehensive approach to answering complex questions across various domains. As researchers and industry professionals continue to harness its power, the potential for innovation and discovery remains boundless.

What is the fundamental concept and importance of hypothesis testing in scientific decision-making?

Understanding hypothesis testing.

Hypothesis testing is a cornerstone of scientific inquiry. It involves making an assumption, the hypothesis, about a population parameter. Scientists test these assumptions through experimentation and observation.

The Essence of Hypotheses

At its core, a hypothesis is a predictive statement. It usually pertains to an outcome or a relationship between variables. The hypothesis asserts a specific effect, direction, or magnitude will emerge under certain conditions.

Types of Hypotheses

There are two primary hypotheses in testing: null and alternative. The null hypothesis ( H0 ) suggests no effect or relationship exists. It represents a default position, waiting for evidence to challenge it. The alternative hypothesis ( H1 ) posits there is an effect or relationship. It states the specific condition the researcher believes is true.

Role of Evidence

Evidence plays a critical role. Researchers collect data through controlled methods. They aim to either support or refute the hypothesis. This data must be empirical and measurable, ensuring objectivity.

Decision-Making with P-Values

The p-value is a crucial concept in hypothesis testing. It is the probability of observing a test statistic as extreme as the one observed, given the null hypothesis is true. A low p-value indicates the observed data is unlikely under the null hypothesis. This typically leads to rejection of the null in favor of the alternative.

The Importance of Hypothesis Testing

Provides structure to research

Ensures consistency in methods

Allows quantification of evidence

Facilitates replication of studies

Shields from personal biases

Hypothesis testing helps map the unknown territory of scientific phenomena. It allows researchers to make informed decisions grounded in statistical evidence. This rational approach to understanding ensures that conclusions drawn from scientific work are reliable and valid.

The process also shapes the scientific method itself. It demands rigorous standards for evidence and reproducibility. Hypothesis testing thus builds a foundation on which scientific knowledge advances. It underpins the integrity of scientific disciplines. It challenges scientists to prove, disprove, and refine their understanding of the world.

Hypothesis testing is fundamental to the scientific decision-making process. It turns subjective questions into objective inquiries. It drives the pursuit of knowledge through empirical evidence. With hypothesis testing, science moves from conjecture to proven or disproven theories. It is this disciplined approach that adds credibility to scientific findings. Without it, distinguishing between chance results and true discoveries becomes impossible.

Understanding Hypothesis Testing Hypothesis testing is a cornerstone of scientific inquiry. It involves making an assumption, the hypothesis, about a population parameter. Scientists test these assumptions through experimentation and observation. The Essence of Hypotheses At its core, a hypothesis is a predictive statement. It usually pertains to an outcome or a relationship between variables. The hypothesis asserts a specific effect, direction, or magnitude will emerge under certain conditions. Types of Hypotheses There are two primary hypotheses in testing: null and alternative. The null hypothesis ( H0 ) suggests no effect or relationship exists. It represents a default position, waiting for evidence to challenge it. The alternative hypothesis ( H1 ) posits there is an effect or relationship. It states the specific condition the researcher believes is true. Role of Evidence Evidence plays a critical role. Researchers collect data through controlled methods. They aim to either support or refute the hypothesis. This data must be empirical and measurable, ensuring objectivity. Decision-Making with P-Values The p-value is a crucial concept in hypothesis testing. It is the probability of observing a test statistic as extreme as the one observed, given the null hypothesis is true. A low p-value indicates the observed data is unlikely under the null hypothesis. This typically leads to rejection of the null in favor of the alternative. The Importance of Hypothesis Testing Hypothesis testing helps map the unknown territory of scientific phenomena. It allows researchers to make informed decisions grounded in statistical evidence. This rational approach to understanding ensures that conclusions drawn from scientific work are reliable and valid. The process also shapes the scientific method itself. It demands rigorous standards for evidence and reproducibility. Hypothesis testing thus builds a foundation on which scientific knowledge advances. It underpins the integrity of scientific disciplines. It challenges scientists to prove, disprove, and refine their understanding of the world. Hypothesis testing is fundamental to the scientific decision-making process. It turns subjective questions into objective inquiries. It drives the pursuit of knowledge through empirical evidence. With hypothesis testing, science moves from conjecture to proven or disproven theories. It is this disciplined approach that adds credibility to scientific findings. Without it, distinguishing between chance results and true discoveries becomes impossible.

How do Type I and Type II errors relate to hypothesis testing and what are their implications on the results?

Understanding type i and type ii errors.

When delving into hypothesis testing, the concepts of Type I and Type II errors often emerge as critical elements. These errors play a paramount role in the interpretation of results. They convey the instances where our conclusions could be incorrect.

What Are Type I and Type II Errors?

Type I error occurs when we wrongly reject a true null hypothesis. We call this a false positive. It implies that the evidence suggests an effect or difference exists when it does not. In statistical terms, this is the 'alpha' (α), which defines the likelihood of a Type I error.

Type II error , in contrast, happens when we fail to reject a false null hypothesis. This error, termed a false negative, means that one overlooks an actual effect or difference. It's quantified by 'beta' (β), which gives the probability of a Type II error occurring.

Implications of Type I and Type II Errors

The implications of these errors reach far into hypothesis testing and the trustworthiness of results.

Confidence Levels : High risks of Type I errors lower confidence in findings. To mitigate this, researchers set a low alpha level, commonly 0.05. It shows a willingness to accept a 5% chance of a false positive.

Power of the Test : The risk of Type II errors correlates with the power of the test—the probability of correctly detecting an effect when it exists. A high beta value means a higher chance of missing an actual effect due to low test power.

Sample Size : Larger samples reduce both Type I and Type II error risks. They offer more accurate estimates and a clearer distinction between the null and alternative hypotheses.

Consequences : Type I errors might lead to unwarranted actions based on false positives. Type II errors could result in missed opportunities due to unrecognized truths.

Balancing Errors in Hypothesis Testing

Researchers must balance Type I and Type II errors in hypothesis testing. The balance depends on the context and potential consequences of each error.

Safety in Medicine : In drug testing, Type I errors can lead to harmful side effects if a drug isn't actually safe. Minimizing Type I errors is crucial here.

Effectiveness in Treatment : Conversely, Type II errors in medicine may miss a treatment effect. Ensuring sufficient power to detect treatment efficacy is essential.

Type I and Type II errors remind us of the limitations in hypothesis testing. No test is infallible. Decisions on alpha and beta levels depend on the stakes of potential errors.

Understanding and addressing these errors are vital. They enhance credibility in conclusions drawn from statistical testing. Proper balance ensures valuable and trustworthy research outcomes.

Understanding Type I and Type II Errors When delving into hypothesis testing, the concepts of Type I and Type II errors often emerge as critical elements. These errors play a paramount role in the interpretation of results. They convey the instances where our conclusions could be incorrect. What Are Type I and Type II Errors? Type I error occurs when we wrongly reject a true null hypothesis. We call this a false positive. It implies that the evidence suggests an effect or difference exists when it does not. In statistical terms, this is the alpha (α), which defines the likelihood of a Type I error. Type II error , in contrast, happens when we fail to reject a false null hypothesis. This error, termed a false negative, means that one overlooks an actual effect or difference. Its quantified by beta (β), which gives the probability of a Type II error occurring. Implications of Type I and Type II Errors The implications of these errors reach far into hypothesis testing and the trustworthiness of results. Balancing Errors in Hypothesis Testing Researchers must balance Type I and Type II errors in hypothesis testing. The balance depends on the context and potential consequences of each error. Type I and Type II errors remind us of the limitations in hypothesis testing. No test is infallible. Decisions on alpha and beta levels depend on the stakes of potential errors. Understanding and addressing these errors are vital. They enhance credibility in conclusions drawn from statistical testing. Proper balance ensures valuable and trustworthy research outcomes.

Can you explain the critical role of the p-value in hypothesis testing and its influence on accepting or rejecting the null hypothesis?

Understanding the p-value.

Researchers often turn to hypothesis testing to understand data. They make an initial assumption called the null hypothesis . This hypothesis suggests no effect or no difference exists. To challenge this, they use an alternative hypothesis.

The Null Hypothesis and P-value

In hypothesis testing, the p-value helps measure the strength of the results against the null hypothesis. It calculates the probability of observing data as extreme as the test results, assuming the null hypothesis is true. A low p-value indicates that the observed data would be very unlikely if the null hypothesis were true.

Significance Threshold

Scientists usually set a significance level before testing. Often, this level is 0.05 . It marks the cut-off for determining statistical significance.

If the p-value is below 0.05, the result is statistically significant.

This means the test provides enough evidence to reject the null hypothesis.

What Does Rejecting the Null Hypothesis Mean?

Rejecting the null does not prove the alternative hypothesis. It merely suggests that the data are not consistent with the null. Researchers can be more confident that an effect or difference might exist.

Misinterpretations of the P-value

A common mistake is seeing the p-value as the odds that the null hypothesis is true or false. It is not. It only assesses how compatible the data are with the null hypothesis.

Influencing Factors

Several factors influence the p-value. This includes the size of the effect and the sample size. Larger samples may detect smaller differences and result in smaller p-values.

The p-value is critical in deciding whether to accept or reject the null hypothesis. It quantifies how surprising the data are, assuming the null is true. A small p-value can lead to rejecting the null, paving the way for new scientific insights. However, it is crucial to use this tool wisely, with an understanding of its limitations and context.

Understanding the P-value Researchers often turn to hypothesis testing to understand data. They make an initial assumption called the null hypothesis . This hypothesis suggests no effect or no difference exists. To challenge this, they use an alternative hypothesis. The Null Hypothesis and P-value In hypothesis testing, the p-value helps measure the strength of the results against the null hypothesis. It calculates the probability of observing data as extreme as the test results, assuming the null hypothesis is true. A low p-value indicates that the observed data would be very unlikely if the null hypothesis were true. Significance Threshold Scientists usually set a significance level before testing. Often, this level is 0.05 . It marks the cut-off for determining statistical significance. What Does Rejecting the Null Hypothesis Mean? Rejecting the null does not prove the alternative hypothesis. It merely suggests that the data are not consistent with the null. Researchers can be more confident that an effect or difference might exist. Misinterpretations of the P-value A common mistake is seeing the p-value as the odds that the null hypothesis is true or false. It is not. It only assesses how compatible the data are with the null hypothesis. Influencing Factors Several factors influence the p-value. This includes the size of the effect and the sample size. Larger samples may detect smaller differences and result in smaller p-values. The p-value is critical in deciding whether to accept or reject the null hypothesis. It quantifies how surprising the data are, assuming the null is true. A small p-value can lead to rejecting the null, paving the way for new scientific insights. However, it is crucial to use this tool wisely, with an understanding of its limitations and context.

He is a content producer who specializes in blog content. He has a master's degree in business administration and he lives in the Netherlands.

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Unlocking Da Vinci's Problem Solving Skills

A close-up of a pile of papers on a table, with various sheets of paper of various sizes and colors scattered around. A white letter 'O' is seen on a black background in the upper left corner of the image. In the lower right corner, a woman is seen wearing a white turtleneck and a black jacket. In the middle of the image, a close-up of a book with a bookmark is visible. Lastly, a screenshot of a black and white photo of a woman is seen in the upper right corner. The papers, letter, woman, book, and photo all appear to be on the same table, creating an interesting image that is suitable for use in an image caption dataset.

Developing Problem Solving Skills Since 1960s WSEIAC Report

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The First Step in Critical Thinking & Problem Solving

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What are Problem Solving Skills?

Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as $H_{0}$. Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as $H_{1}$ or $H_{a}$. For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, $\alpha$ or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$. $\overline{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation and n is the size of the sample.
t = $\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$. s is the sample standard deviation.
$\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}$. $O_{i}$ is the observed value and $E_{i}$ is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

One sample: z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$.
Two samples: z = $\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}$.

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

One sample: t = $\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$.
Two samples: t = $\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}$.

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

$H_{0}$: The population parameter is ≤ some value

$H_{1}$: The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

$H_{0}$: The population parameter is ≥ some value

$H_{1}$: The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

$H_{0}$: the population parameter = some value

$H_{1}$: the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
Step 2: Set up the alternative hypothesis.
Step 3: Choose the correct significance level, $\alpha$, and find the critical value.
Step 4: Calculate the correct test statistic (z, t or $\chi$) and p-value.
Step 5: Compare the test statistic with the critical value or compare the p-value with $\alpha$ to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as $H_{0}$: $\mu$ = 100.

Step 2: The alternative hypothesis is given by $H_{1}$: $\mu$ > 100.

Step 3: As this is a one-tailed test, $\alpha$ = 100% - 95% = 5%. This can be used to determine the critical value.

1 - $\alpha$ = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$.

$\mu$ = 100, $\overline{x}$ = 112.5, n = 30, $\sigma$ = 15

z = $\frac{112.5-100}{\frac{15}{\sqrt{30}}}$ = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Probability and Statistics
Data Handling

Important Notes on Hypothesis Testing

Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
It involves the setting up of a null hypothesis and an alternate hypothesis.
There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. $H_{0}$: $\mu$ = 90, $H_{1}$: $\mu$ > 90 $\overline{x}$ = 110, $\mu$ = 90, n = 5, s = 18. $\alpha$ = 0.05 Using the t-distribution table, the critical value is 2.132 t = $\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$ t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. $H_{0}$: $\mu$ = 80, $H_{1}$: $\mu$ ≠ 80 $\overline{x}$ = 88, $\mu$ = 80, n = 36, $\sigma$ = 10. $\alpha$ = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$ z = $\frac{88-80}{\frac{10}{\sqrt{36}}}$ = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. $H_{0}$: $\mu$ = 90, $H_{1}$: $\mu$ < 90 $\overline{x}$ = 110, $\mu$ = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = $\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$ t = $\frac{82-90}{\frac{18}{\sqrt{6}}}$ t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

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FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$ and for two samples is z = $\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}$.

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide $\alpha$ by 2. This is done as there are two rejection regions in the curve.

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9. Hypothesis Testing

9.1 Hypothesis Testing Problem Solving Steps

$P(\mbox{test statistic} \mid H_{0})$

Interpretation : Summarize results in a sentence and/or present a graphic or table.

A generic test statistic may be defined by :

$\mbox{test value} = \frac{\mbox{(observed value)} - \mbox{(expected }H_{0} \mbox{ value)}}{\mbox{standard error}}.$

Introduction to Applied Statistics for Psychology Students Copyright © 2022 by Gordon E. Sarty is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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SOLVING PROBLEMS INVOLVING TEST OF HYPOTHESIS ON POPULATION PROPORTION
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9.E: Hypothesis Testing with One Sample (Exercises)
9.6: Additional Information and Full Hypothesis Test Examples. For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in . Please feel free to make copies of the solution sheets. ... Solve this problem and have some fun. Q 9.6.17
Hypothesis Testing
Table of contents. Step 1: State your null and alternate hypothesis. Step 2: Collect data. Step 3: Perform a statistical test. Step 4: Decide whether to reject or fail to reject your null hypothesis. Step 5: Present your findings. Other interesting articles. Frequently asked questions about hypothesis testing.
9.4 Full Hypothesis Test Examples
Set up the Hypothesis Test: Since the problem is about a mean, this is a test of a single population mean. H 0: μ = 16.43 H a: μ < 16.43. For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed. Determine the distribution needed: Random variable: X ¯ X ¯ = the mean time to swim the 25-yard ...
6a.2
Below these are summarized into six such steps to conducting a test of a hypothesis. Set up the hypotheses and check conditions: Each hypothesis test includes two hypotheses about the population. One is the null hypothesis, notated as H 0, which is a statement of a particular parameter value. This hypothesis is assumed to be true until there is ...
S.3.3 Hypothesis Testing Examples
If the biologist set her significance level $\alpha$ at 0.05 and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic t* were less than -1.6939 (determined using statistical software or a t-table):s-3-3. Since the biologist's test statistic, t* = -4.60, is less than -1.6939, the biologist rejects the null hypothesis.
Significance tests (hypothesis testing)
Unit test. Significance tests give us a formal process for using sample data to evaluate the likelihood of some claim about a population value. Learn how to conduct significance tests and calculate p-values to see how likely a sample result is to occur by random chance. You'll also see how we use p-values to make conclusions about hypotheses.
Hypothesis Testing
A hypothesis test is a statistical inference method used to test the significance of a proposed (hypothesized) relation between population statistics (parameters) and their corresponding sample estimators. In other words, hypothesis tests are used to determine if there is enough evidence in a sample to prove a hypothesis true for the entire population. The test considers two hypotheses: the ...
1.2
Step 1: State the Null Hypothesis. The null hypothesis can be thought of as the opposite of the "guess" the researchers made. In the example presented in the previous section, the biologist "guesses" plant height will be different for the various fertilizers. So the null hypothesis would be that there will be no difference among the groups of ...
Introduction to Hypothesis Testing with Examples
In binary hypothesis testing problems, we'll often be presented with two choices which we call hypotheses, and we'll have to decide whether to pick one or the other. The hypotheses are represented by H₀ and H₁ and are called null and alternate hypotheses respectively. In hypothesis testing, we either reject or accept the null hypothesis.
A Beginner's Guide to Hypothesis Testing in Business
3. One-Sided vs. Two-Sided Testing. When it's time to test your hypothesis, it's important to leverage the correct testing method. The two most common hypothesis testing methods are one-sided and two-sided tests, or one-tailed and two-tailed tests, respectively. Typically, you'd leverage a one-sided test when you have a strong conviction ...
Simple hypothesis testing (practice)
Simple hypothesis testing. Niels has a Magic 8 -Ball, which is a toy used for fortune-telling or seeking advice. To consult the ball, you ask the ball a question and shake it. One of 5 different possible answers then appears at random in the ball. Niels sensed that the ball answers " Ask again later " too frequently.
Hypothesis Testing
Step 2: State the Alternate Hypothesis. The claim is that the students have above average IQ scores, so: H 1: μ > 100. The fact that we are looking for scores "greater than" a certain point means that this is a one-tailed test. Step 3: Draw a picture to help you visualize the problem. Step 4: State the alpha level.
Hypothesis Testing
This statistics video tutorial explains how to solve hypothesis testing problems with proportions. It explains how to calculate the sample proportion and th...
Hypothesis Testing Explained (How I Wish It Was Explained to Me)
The curse of hypothesis testing is that we will never know if we are dealing with a True or a False Positive (Negative). All we can do is fill the confusion matrix with probabilities that are acceptable given our application. To be able to do that, we must start from a hypothesis. Step 1. Defining the hypothesis
Hypothesis Testing: A Comprehensive Guide to Scientific Decision-Making
This broad applicability underlies the significance of a problem-solving course that includes hypothesis testing. Understanding the Basics of Hypothesis Testing. Before embarking on the journey of hypothesis testing, it's crucial to understand its fundamental elements - the Null and Alternate Hypotheses. Explanation of Null Hypothesis
Hypothesis Testing
The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough ...
Hypothesis Testing In Real Life. Real world problems solved with Math
Hypothesis Testing. Hypothesis Tests, or Statistical Hypothesis Testing, is a technique used to compare two datasets, or a sample from a dataset. ... HMMs are probabilistic models used to solve real life problems ranging from weather forecasting to finding the next word in a sentence.
9.1 Hypothesis Testing Problem Solving Steps
9.1 Hypothesis Testing Problem Solving Steps Now that we have some background on setting up hypotheses and finding critical regions, we introduce the steps needed for every hypothesis testing procedure. Hypothesis testing is based directly on sampling theory and the probabilities that the sampling theory gives. Here are the steps we will follow :
Hypothesis Testing Solved Examples(Questions and Solutions)
View Solution to Question 1. Question 2. A professor wants to know if her introductory statistics class has a good grasp of basic math. Six students are chosen at random from the class and given a math proficiency test. The professor wants the class to be able to score above 70 on the test. The six students get the following scores:62, 92, 75 ...
PDF Quarter 4 Module 8: Solving Problems Involving Test of Hypothesis on
Solving Problems Involving Test of Hypothesis on the Population Mean Hypothesis testing is a method of testing a claim or hypothesis about a parameter in a population using data measured in a sample. In this method, we test some hypotheses by determining the likelihood that a sample statistic could have been

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9.E: Hypothesis Testing with One Sample (Exercises)

9.1: Introduction

9.3: Outcomes and the Type I and Type II Errors

9.4: Distribution Needed for Hypothesis Testing

9.5: Rare Events, the Sample, Decision and Conclusion

9.6: Additional Information and Full Hypothesis Test Examples

9.7: Hypothesis Testing of a Single Mean and Single Proportion

9.4 Full Hypothesis Test Examples

Example 9.9

Example 9.10

Try It 9.10

Hypothesis Test for Proportions

Example 9.11

Try It 9.11

Example 9.12

Try It 9.12

Example 9.13

Try It 9.13

Example 9.14

Try It 9.14

Hypothesis Testing

Definitions and Methodology

User Preferences

Keyboard Shortcuts

Step 1: State the Null Hypothesis

Step 2: State the Alternative Hypothesis

Step 3: Set \(\alpha\)

Step 4: Collect Data

Step 5: Calculate a test statistic

Step 6: Construct Acceptance / Rejection regions

Step 7: Based on Steps 5 and 6, draw a conclusion about \(H_0\)

Business Insights

A Beginner’s Guide to Hypothesis Testing in Business

What Is Hypothesis Testing?

Hypothesis Testing in Business

Key Considerations for Hypothesis Testing

2. Significance Level and P-Value

3. One-Sided vs. Two-Sided Testing

4. Sampling

Learn How to Perform Hypothesis Testing

About the Author

Hypothesis Testing: A Comprehensive Guide to Scientific Decision-Making

Definition of Hypothesis Testing

Importance of Hypothesis Testing in Research and Industry

Understanding the Basics of Hypothesis Testing

Explanation of Null Hypothesis

Explanation of Alternate Hypothesis

Differentiating between Null and Alternate Hypotheses

Understanding Errors in Hypothesis Testing

Type I Error: False Positive

Type II Error: False Negative

Balancing the Risks: Power of the Test

Steps in Hypothesis Testing

What is the fundamental concept and importance of hypothesis testing in scientific decision-making?

The Essence of Hypotheses

Types of Hypotheses

Role of Evidence

Decision-Making with P-Values

The Importance of Hypothesis Testing

How do Type I and Type II errors relate to hypothesis testing and what are their implications on the results?

What Are Type I and Type II Errors?

Implications of Type I and Type II Errors

Balancing Errors in Hypothesis Testing

Can you explain the critical role of the p-value in hypothesis testing and its influence on accepting or rejecting the null hypothesis?

The Null Hypothesis and P-value

Significance Threshold

What Does Rejecting the Null Hypothesis Mean?

Misinterpretations of the P-value

Influencing Factors

Unlocking Da Vinci's Problem Solving Skills

Developing Problem Solving Skills Since 1960s WSEIAC Report

The First Step in Critical Thinking & Problem Solving

What are Problem Solving Skills?

Hypothesis Testing

What is Hypothesis Testing in Statistics?

Hypothesis Testing Definition

Null Hypothesis

Alternative Hypothesis

Hypothesis Testing P Value