h_0 hypothesis

Module 9: Hypothesis Testing With One Sample

Null and alternative hypotheses, learning outcomes.

• Describe hypothesis testing in general and in practice

The actual test begins by considering two  hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 : The null hypothesis: It is a statement about the population that either is believed to be true or is used to put forth an argument unless it can be shown to be incorrect beyond a reasonable doubt.

H a : The alternative hypothesis : It is a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make adecision. There are two options for a  decision . They are “reject H 0 ” if the sample information favors the alternative hypothesis or “do not reject H 0 ” or “decline to reject H 0 ” if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in  H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

H 0 : No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ 30

H a : More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.

H 0 : The drug reduces cholesterol by 25%. p = 0.25

H a : The drug does not reduce cholesterol by 25%. p ≠ 0.25

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are:

H 0 : μ = 2.0

H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 66 H a : μ __ 66

• H 0 : μ = 66
• H a : μ ≠ 66

We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:

H 0 : μ ≥ 5

H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 45 H a : μ __ 45

• H 0 : μ ≥ 45
• H a : μ < 45

In an issue of U.S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses.

H 0 : p ≤ 0.066

H a : p > 0.066

On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : p __ 0.40 H a : p __ 0.40

• H 0 : p = 0.40
• H a : p > 0.40

Concept Review

In a  hypothesis test , sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: Evaluate the null hypothesis , typically denoted with H 0 . The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤ or ≥) Always write the alternative hypothesis , typically denoted with H a or H 1 , using less than, greater than, or not equals symbols, i.e., (≠, >, or <). If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties.

Formula Review

H 0 and H a are contradictory.

• OpenStax, Statistics, Null and Alternative Hypotheses. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:58/Introductory_Statistics . License : CC BY: Attribution
• Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
• Simple hypothesis testing | Probability and Statistics | Khan Academy. Authored by : Khan Academy. Located at : https://youtu.be/5D1gV37bKXY . License : All Rights Reserved . License Terms : Standard YouTube License

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• Knowledge Base

Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

• State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a  or H 1 ).
• Collect data in a way designed to test the hypothesis.
• Perform an appropriate statistical test .
• Decide whether to reject or fail to reject your null hypothesis.
• Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

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.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

• H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

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For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

• an estimate of the difference in average height between the two groups.
• a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

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The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

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Introduction to Hypothesis Testing

A statistical hypothesis is an assumption about a population parameter .

For example, we may assume that the mean height of a male in the U.S. is 70 inches.

The assumption about the height is the statistical hypothesis and the true mean height of a male in the U.S. is the population parameter .

A hypothesis test is a formal statistical test we use to reject or fail to reject a statistical hypothesis.

The Two Types of Statistical Hypotheses

To test whether a statistical hypothesis about a population parameter is true, we obtain a random sample from the population and perform a hypothesis test on the sample data.

There are two types of statistical hypotheses:

The null hypothesis , denoted as H 0 , is the hypothesis that the sample data occurs purely from chance.

The alternative hypothesis , denoted as H 1 or H a , is the hypothesis that the sample data is influenced by some non-random cause.

Hypothesis Tests

A hypothesis test consists of five steps:

1. State the hypotheses. 

State the null and alternative hypotheses. These two hypotheses need to be mutually exclusive, so if one is true then the other must be false.

2. Determine a significance level to use for the hypothesis.

Decide on a significance level. Common choices are .01, .05, and .1. 

3. Find the test statistic.

Find the test statistic and the corresponding p-value. Often we are analyzing a population mean or proportion and the general formula to find the test statistic is: (sample statistic – population parameter) / (standard deviation of statistic)

4. Reject or fail to reject the null hypothesis.

Using the test statistic or the p-value, determine if you can reject or fail to reject the null hypothesis based on the significance level.

The p-value  tells us the strength of evidence in support of a null hypothesis. If the p-value is less than the significance level, we reject the null hypothesis.

5. Interpret the results. 

Interpret the results of the hypothesis test in the context of the question being asked. 

The Two Types of Decision Errors

There are two types of decision errors that one can make when doing a hypothesis test:

Type I error: You reject the null hypothesis when it is actually true. The probability of committing a Type I error is equal to the significance level, often called  alpha , and denoted as α.

Type II error: You fail to reject the null hypothesis when it is actually false. The probability of committing a Type II error is called the Power of the test or  Beta , denoted as β.

One-Tailed and Two-Tailed Tests

A statistical hypothesis can be one-tailed or two-tailed.

A one-tailed hypothesis involves making a “greater than” or “less than ” statement.

For example, suppose we assume the mean height of a male in the U.S. is greater than or equal to 70 inches. The null hypothesis would be H0: µ ≥ 70 inches and the alternative hypothesis would be Ha: µ < 70 inches.

A two-tailed hypothesis involves making an “equal to” or “not equal to” statement.

For example, suppose we assume the mean height of a male in the U.S. is equal to 70 inches. The null hypothesis would be H0: µ = 70 inches and the alternative hypothesis would be Ha: µ ≠ 70 inches.

Note: The “equal” sign is always included in the null hypothesis, whether it is =, ≥, or ≤.

Related:   What is a Directional Hypothesis?

Types of Hypothesis Tests

There are many different types of hypothesis tests you can perform depending on the type of data you’re working with and the goal of your analysis.

The following tutorials provide an explanation of the most common types of hypothesis tests:

Introduction to the One Sample t-test Introduction to the Two Sample t-test Introduction to the Paired Samples t-test Introduction to the One Proportion Z-Test Introduction to the Two Proportion Z-Test

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Choosing $H_0$ and $H_a$ in hypothesis testing

There seems to be some ambiguity or contradiction in how to correctly choose the null and alternative hypotheses, both online and in my instructor's notes. I'm trying to figure out if this stems merely from my lack of understanding or if there actually is a disagreement in the scientific community at large. I've seen the following two ideas on choosing $H_0$ and $H_a$

The null hypothesis is the status quo, the state of things already accepted and/or shown to be true by previous data. We assume it to be true and need convincing evidence to reject it. The alternative hypothesis is the one being proposed based on data from the experiment in question, and is assumed to be false unless the data supporting it can convincingly show otherwise.

The null hypothesis is always the one that includes the equality, and the alternative hypothesis is the complement to it. It doesn't matter whether the equality is the status quo or is being claimed by the researcher, it is always H0.

An example I made up myself for demonstrative purposes, I'm not looking for an actual solution. Only interested in the hypotheses:

A researcher believes that children in economically disadvantages areas are more likely to be raised in single-parent homes. He surveys 1000 children from such an area and finds that 317 of them are raised in a single-parent home. Can we conclude with 95% confidence that 30% or more of the children in economically disadvantages areas are raised in single-parent homes?

What would be the H0 and Ha in this case and why? My professor provided the correct answer (for an equivalent question but with different numbers) to be

H0 : p >= 0.3; Ha : p < 0.3

With the rationale that H0 must include the equality, which in this case is greater or equal to 30% . Her solution then failed to reject the null hypothesis and concluded that the researcher's claim is therefore correct. To me this seems like assuming the claim to be true to begin with and giving it the benefit of the doubt, which is the opposite of what I thought was the correct approach.

A professor in this related question Difference between "at least" and "more than" in hypothesis testing? seemingly took the same approach.

I wish I could talk to my professor about this, but unfortunately there's a significant language barrier.

• hypothesis-testing

3 Answers 3

Both ideas of the null and alternative hypothesis are true. The null hypothesis must always include an equals sign, whether it be $\geq\text{, } \leq\text{, or just}=$. Usually, however, it's just $=$. The alternative hypothesis is what we wish to show.

The null hypothesis in this case is that the proportion of children in economically disadvantaged areas raised in single-parent homes is $30$%.

The alternative hypothesis is that the proportion of children in economically disadvantaged areas raised in single-parent homes is greater than $30$%.

More formally

$$H_0 : p=0.3$$

$$H_a : p \gt 0.3$$

There are two ways you can test this hypothesis if you so wish. Letting $X$ be the number of children raised in single-parent homes, you can use normal approximation to the binomial:

$$P(X\geq317)=1-P(X\lt317)=1-\Phi\left(\frac{316.5-300}{\sqrt{1000\cdot0.3\cdot0.7}}\right)$$

where I used a continuity correction

In R statistical software

You could also, using software, find the exact probability using the standard binomial distribution:

$$P(X\geq317)=\sum_{k=317}^{1000} {1000 \choose k}\cdot0.3^k\cdot0.7^{1000-k}$$

Since $n$ is large, the normal approximation does very well.

At $\alpha=0.05$ we fail to reject the null hypothesis.

• 1 $\begingroup$ You choice is also the one I would have gone with, since for a continuous distribution >= 30% and >30% are the same thing. However, both my professor, and the professor in the linked related question, picked H0 : p = 0.3; H1 : p < 0.3 . That's what I'm trying to understand. $\endgroup$ –  Egor Commented Apr 9, 2018 at 17:42

Your null hypothesis is $H_0:p=0.3$

The alternative hypothesis is $H_1:p>0.3$

You need to calculate $$p(X\geq317)$$ using $X\sim Bin(1000,0.3)$

Can you finish?

Just to clarify:

• The null hypothesis always has an equal sign and never an inequality symbol
• In this particular example we conclude that $317$ is not in the critical region.

We conclude that in accepting the null hypothesis there is insufficient evidence that the probability is more than $30$%

• $\begingroup$ The question is purely hypothetical, this isn't an actual homework assignment. I'm trying to understand the overall strategy of picking the null and alternative hypotheses. If you check the related question, you'll see that a professor would have picked H1 : p < 0.3 instead. Why? $\endgroup$ –  Egor Commented Apr 9, 2018 at 17:35
• $\begingroup$ It wouldn't make sense to choose this alternative hypothesis, since 317 is greater than the mean, we should be looking in the upper part of the distribution, not the lower $\endgroup$ –  David Quinn Commented Apr 9, 2018 at 17:38
• $\begingroup$ please see my additional comments to my previous answer. $\endgroup$ –  David Quinn Commented Apr 9, 2018 at 18:08
• $\begingroup$ The way my professor explained it in class, we have two hypotheses: p >= 0.3 (since the question states "30% or more") and p < 0.3 as the complement. We must pick the one that has an equality as the null hypothesis, and the other one as the alternative. $\endgroup$ –  Egor Commented Apr 9, 2018 at 18:13
• $\begingroup$ The null hypothesis always states a particular value, in this case $0.3$. The alternative hypothesis is always either $<$ or $>$ or $\neq$ $\endgroup$ –  David Quinn Commented Apr 9, 2018 at 18:16

You always have to choose $H_a$ so that the sample’s estimation fulfills $H_a$.

The reason is that otherwise the rejection rule will always vote for $H_0$ as in the incorrect choice of your professor.

In your case you want to test a probability against $0.3$, the sample’s estimation was $0.37$, hence $H_a\colon p>0.3$ as $0.37>0.3$. And it does in no way matter where the equal-sign occurs as long as you’re dealing with continuous random variables.

• $\begingroup$ Why should we choose the alternative according to the observede data? $\endgroup$ –  Thomas Commented Jun 20, 2022 at 6:36

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Hypothesis testing. Why center the sampling distribution on H0?

A p-value is the probability to obtain a statistic that is at least as extreme as the one observed in the sample data when assuming that the null-hypothesis ($H_0$) is true.

Graphically this corresponds to the area defined by the sample statistic under the sampling distribution which one would obtain when assuming $H_0$:

However, because the shape of this assumed distribution is actually based on the sample data, centering it on $\mu_0$ seems like an odd choice to me. If one would instead use the sampling distribution of the statistic, i.e. center the distribution on the sample statistic, then hypothesis testing would correspond to estimating the probability of $\mu_0$ given the samples.

In that case the p-value is the probability to obtain a statistic at least as extreme as $\mu_0$ given the data instead of the above definition.

Additionally, such an interpretation has the advantage of relating well to the concept of confidence intervals: A hypothesis test with significance level $\alpha$ would be equivalent to checking whether $\mu_0$ falls within the $(1-\alpha)$ confidence interval of the sampling distribution.

I thus feel that centering the distribution on $\mu_0$ could be an unneccessary complication. Are there any important justifications for this step which I did not consider?

• hypothesis-testing
• confidence-interval

• 2 $\begingroup$ Please tell us what the sampling distribution will be if you do not assume $H_0$. (Answer: you cannot, except in textbook examples where the alternative hypothesis specifies a unique distribution.) $\endgroup$ –  whuber ♦ Commented Jun 24, 2016 at 14:15
• $\begingroup$ I am not sure if I understand the request correctly but in the above example it would be the sampling distribution of the mean. I have now added a figure to the question which shows this distribution together with a 95% confidence interval/area which should also help illustrating the relationship to confidence intervals. $\endgroup$ –  kuru Commented Jun 24, 2016 at 15:22
• 2 $\begingroup$ You have no way to know the sampling distribution of the mean. To know that, you need to know the true mean: but that's precisely the quantity you are trying to test! Your logic is completely circular. $\endgroup$ –  whuber ♦ Commented Jun 24, 2016 at 15:26
• 1 $\begingroup$ I understood that was your meaning. In general, until you know--or assume--the true parameters of the distribution, you cannot know the distribution of any property of the sample. (In fact, if you could deduce the distribution of any sample property without assuming knowledge of the parameters, that would be proof that it gives you no information about the parameters!) $\endgroup$ –  whuber ♦ Commented Jun 24, 2016 at 15:52
• 1 $\begingroup$ I cannot, because it appears you are not using terms like "mean," "estimated," or even "H0" in their usual statistical senses. I am at a complete loss to understand even what your question is. The only thing that is clear is that it is predicated on a misunderstanding of null hypothesis testing, but your responses to my comments haven't provided any useful indications of what that misunderstanding might be. $\endgroup$ –  whuber ♦ Commented Jun 24, 2016 at 16:26

3 Answers 3

Suppose $\boldsymbol X = (X_1, X_2, \ldots, X_n)$ is a sample drawn from a normal distribution with unknown mean $\mu$ and known variance $\sigma^2$. The sample mean $\bar X$ is therefore normal with mean $\mu$ and variance $\sigma^2/n$. On this much, I think there can be no possibility of disagreement.

Now, you propose that our test statistic is $$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim \operatorname{Normal}(0,1).$$ Right? BUT THIS IS NOT A STATISTIC . Why? Because $\mu$ is an unknown parameter . A statistic is a function of the sample that does not depend on any unknown parameters. Therefore, an assumption must be made about $\mu$ in order for $Z$ to be a statistic. One such assumption is to write $$H_0 : \mu = \mu_0, \quad \text{vs.} \quad H_1 : \mu \ne \mu_0,$$ under which $$Z \mid H_0 = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}} \sim \operatorname{Normal}(0,1),$$ which is a statistic.

By contrast, you propose to use $\mu = \bar X$ itself. In that case, $Z = 0$ identically, and it is not even a random variable, let alone normally distributed. There is nothing to test.

• 1 $\begingroup$ Thank you. This is very straightforward and now I really wonder how I could have missed that before. All that would be left now as an excuse for the second presented case is to rely on the confidence interval calculation. However, because there the margin of error is explicitly added/subtracted from the mean or point estimate, the use of that estimate becomes a step that would need to be justified. $\endgroup$ –  kuru Commented Jun 24, 2016 at 21:12

However, because the shape of this assumed distribution is actually based on the sample data, centering it on H0 seems like an odd choice to me.

This is actually not true. The shape of this assumed distribution comes from accepting $H_0$ as true. Sample is not directly involved in that, other than by some assumptions. Using the sample directly, is not enough. You need also the null hypothesis to hold.

If one would instead use the sampling distribution of the statistic, i.e. center the distribution on the sample statistic, then hypothesis testing would correspond to estimating the probability of H0 given the samples.

The question is: how do you estimate a probability of something which you assume is true. In our case if you assume $H_0$ as true, is futile to try to estimate the probability that $H_0$ is true.

I thus feel that centering the distribution on H0 is an unneccessary complication.

You don't have two distributions there, there is only one, the one assumed to be your ground truth, aka the one which comes with $H_0$. There is however a sampling distribution derived from sample, but this is not involved in the hypotheses you use.

I good exercise would be to try to replicate the same logic with an asymmetric distribution. Take chi-square distribution like in chi square independence test. Are you able to reproduce it? I think the answer is no.

• $\begingroup$ " This is actually not true. The shape of this assumed distribution comes from accepting H0 as true. Sample is not directly involved in that, other than by some assumptions. " But in the case of the one sample t-test presented above, the test statistic includes the SEM and the sample mean and is thus dependent on the sample data. $$t = \frac{\bar{x}-\mu_0}{\frac{s}{{\sqrt{n}}}}$$ Furthermore the degrees of freedom which determine the height of the tails depend on the sample size. $\endgroup$ –  kuru Commented Jun 24, 2016 at 11:57
• 1 $\begingroup$ My formulation was misleading. I was trying to say that you can use any information you have, also the sample itself, but it is not enough. To evaluate p-values and to have a distribution you need to assume also the null hypothesis. I reformulate in the post also. $\endgroup$ –  rapaio Commented Jun 24, 2016 at 21:02
• 1 $\begingroup$ ... Take for example your formula for $t$, it uses $\mu_0$ which I suppose it the value from the null hypothesis $H_0 : \mu = \mu_0$ $\endgroup$ –  rapaio Commented Jun 24, 2016 at 21:14

From what I gather, you are arguing that it makes more sense to 'flip' $H_0$ and $H_1$.

I find it helpful to think of hypothesis testing as a proof by contradiction. We assume $H_0$ to be true, then show that evidence indicates such an assumption is flawed, thus justifying the rejection of $H_0$ in favor of $H_1$.

This works because when we assume $H_0$ and center our distribution there, we can determine how likely/unlikely our observation is. For example, if $H_0: \mu = 0$ vs. $H_1: \mu \neq 0$ and we determine from our testing that there is a less that 5% chance that the true mean $\mu$ actually equals 0, we can reject $H_0$ with 95% confidence.

The reverse is not necessarily true. Say we do an experiment and determine that there is actually a 30% chance that the null hypothesis still holds. We cannot reject the null, but we also do not accept it . This situation does not show that $H_0$ (the null) is true, but that we do not have the evidence to show that it is false.

Now imagine if we flipped this situation. Say we assume $H_1$ and find that given our results, the likelihood of $H_0$ is 5% or less, what does that mean? Sure we can reject the null, be can we necessarily accept $H_1$? It is hard to justify accepting the thing we assumed to be true in the beginning.

Showing that $H_0$ is false is not the result we are after; we want to argue in favor of $H_1$. By doing the test in the way you describe, we are showing that we do not have evidence to say that $H_1$ is false, which is subtly different than arguing $H_1$ is true.

• $\begingroup$ Since the hypothesis test doesn't allow us to completely eliminate uncertainty I would't see it as a proof . Perhaps I did not make my point clear enough but I am essentially asking for a logical rather than a semantic reason to shift the sampling distribution to $H_0$. $\endgroup$ –  kuru Commented Jun 24, 2016 at 15:39
• $\begingroup$ And in general, H1 is pretty vague (mu != 0), making likelihood calculations problematic. Though I suppose that is often a good incentive for people to go Bayesian. :) $\endgroup$ –  Hao Ye Commented Jun 24, 2016 at 19:31

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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 \(H_A\).

In 1989, the average age of a divorced man was 36.2 years of age, according to data obtained from the U.S. Department of Health and Human Services. A sociologist wants to investigate if the average age of a divorced man is now different. She randomly selects 32 divorced men and finds that the mean age is 32.5 years with a sample standard deviation s = 9.6. At a 95% level of confidence, we wish to investigate if the average age of a divorced man is now different than it was in 1989. 24. State the Null and Alternate Hypothesis for this study. A) H0: µ = 33.9 H1: µ 33.9 B) H0: µ = 36.2 H1: µ 36.2 C) H0: µ >= 33.9 H1: µ <33.9 D) H0: µ >= 36.2 H1: µ < 36.2

On solving the provided question, P = 0.037 reject H o,P-value < 0.05 ( level of significance )

If there is no statistical significance between two variables, the null hypothesis is that there isn't any. The researcher or experimenter generally aims to refute or undermine the theory. An alternate explanation is that the two variables have a statistically significant association.

based on P value -

P = 0.037 reject H o

P-value < 0.05 ( level of signifance )

we are 95% y confidence that the average age of divorced man is how different than it was in 1989.

we would have concluded that the average age is not different wher, it really was different.

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Related Questions

7. A customer ordered 20 kg of fertilizer that contains 15% nitrogen. To fulfill the customer's order, how much of the stock 30% nitrogen fertilizer must be mixed with the 10% nitrogen fertilizer?

5 kg of 30% nitrogen an d 15 kg of 10% nitrogen.

Step-by-step explanation:

Which figure has four sides that are the same length

1) 2x+7x= 2) 8b-5b= 3) 5m+2m-4m= 4) a+a+a+a+a= 5) 2x+3x+6x-4x= 6) 2a+2b+a+3b= 7) 2x+7x-4= 8) 2m-2b-m+4b= 9) 10a+2b+2c+3a-b+4c= 10) 5+2 b-4+7 b= 11) 9 p+3 q-2 p-3 q= 12) a+2 b+a-b-10=

Answer: 1: 9x

9: 13a+3b+6c

12: 2a+1b-10

Find the solution of the inequality 5 > r – 3.

5 > r – 3

add 3 to both sides

move variable to the left and flip the sign

Evaluate the line integral, where C is the given plane curve. 1649 (x®y + sin(x)) dy, C is the arc of the parabola y = x2 from (0, 0) to (1, x2)

Line integral will be ∫c(x²y + sinx )dy = 1/3π²×π³ + 2π

C is the arc of parabola y = x² from (0,0) to (π,π²)

Let x be the parameter ; since the parabola is given as function of x

the parametric equations are

x=x,y=x², for 0≤x≤π

By using concept

∫c(x²y + sinx)dy = [tex]\int\limits^0_x {(x^2(x^2) + sinx} \, ).2xdx[/tex]

by applying the limits of integration we get ,

= 2 [ π².π³/6 + [ -πcosπ + sinπ - (0 + sin0)] ]

by calculating,

= 2 ( π².π³/6 + π )

= 1/3 π².π³ + 2π

∫c(x²y + sinx )dy = 1/3π²×π³ + 2π

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Suppose you have two Single Factor ANOVA experiments with the same degrees of freedom, if the resulting F test statistics are: Experiment 1, F = 2.68 Experiment 2, F = 20.15 Which statement is TRUE in regards to comparing Experiment 1 and Experiment 2? A.Procedure 1 has a smaller test statistic and therefore will result in stronger evidence in favor of the alternative hypothesis. B.Procedure 2 has a larger test statistic and therefore will result in stronger evidence in favor of the alternative hypothesis. C.We should reject the null for both tests. D.Impossible to know without more information.

TRUE - Procedure 2 has a larger test statistic and therefore will result in stronger evidence in favor of the alternative hypothesis.

Degrees of freedom are the maximum number of logically independent, or potentially different, values in the sample of data. One is subtracted from the number of elements in the data sample to determine the degree of freedom.

There are degrees of flexibility in the third column. Degrees of freedom (df1) between treatments equals k-1. df2 = N - k is the degree of freedom for the error. The total degree of freedom is N-1; additionally, (k-1) plus (N-k) equals N-1.

Fisher's Least Significant Difference (LSD) and the Bonferroni Method stand out among the techniques. These two use the t-test as their foundation. According to Fisher's LSD, run an F-test first, then if you find evidence to reject the null hypothesis, run regular t tests on all possible pairs of means. Similar but only requiring that you choose beforehand is the Bonferroni technique.

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what is 24divided by 2

24/2 = 12 bc 12*2=24

Can someone please help me answer the second part of this question

The compositions of the two given functions are:

g(f(x)) =   √(2*x + 29) - 2

f(g(x)) = √(2*x  + 25)

Here we have the two functions :

f(x) = √(2x + 29)

g(x) = x - 2

We want to find the compositions :

So we just need to evaluate f(x) on g(x), we will get:

f(g(x)) = √(2*g(x) + 29)

Now we can replace g(x) there:

f(g(x)) = √(2*(x - 2) + 29)

f(g(x)) = √(2*x - 2*2 + 29)

The other composition is:

g(f(x)) = f(x) - 2

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Math geometry, GIVEN

The SSS congruence theorem is used to prove that triangles MAH and THA are congruent to each other. See proof below.

"SSS" stands for side-side-side congruence theorem , which says that given the criteria that two triangles both have three corresponding sides that are equal or congruent to each other, both triangles are congruent.

The proof below shows how we can apply the SSS to prove that triangles MAH and THA are congruent to each other.

Statement                                 Reason                                            

1. MA ≅ TH                               1. Given

2. MH ≅ AT                              2. Given

3. AH ≅ HA                              3. Reflexive property of congruence

4. ΔMAH ≅ ΔTHA                   4. SSS congruence theorem

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Triangle Congruence: Question 3 In the proof, what is the reason for (5) ? Given: V W║Z Y W X=Y X Prove: ΔVWX ΔXYZ Select one: SSA H SAS ASA SSS

In the proof, The reason for (5) ΔVWX ≅ ΔXYZ is ASA

Two triangles are congruent when their three angles and three sides are equal to each other. The two triangles need to be of the same size and shape to be congruent. Both the triangles under consideration should superimpose on each other.

There are 4 property we use to congruent two triangles ( ASA , RHS , SSS and SAS)

According to the question,

Given = V W║Z Y

We have to prove :  ΔVWX ≅ ΔXYZ

Proof =>

It is given that V W║Z Y ,

Using parallel line property ,

(1) ∠W = ∠Y (interior alternate angles )

(2) ∠V = ∠Z ( Interior angle property )

(3) W X=Y X (given)

Therefore , By using ASA property

ΔVWX ≅ ΔXYZ

Hence , In the proof, the reason for (5) is ASA

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The averages wait time to see an E.R. doctor is said to be 150 minutes. You think the wait time is actually less. You take a random sample of 30 people and find their average wait is 148 minutes with a standard deviation of 5 minutes. Assume the distribution is normal. Using a significance level of 0.05 and p test, what would you recommend to the winery? Write down the hypothesis and show all steps for testing the hypothesis.

The average wait time to see an E.R. doctor is significantly less than 150 minutes, and we can recommend to the winery that the wait time is actually less than what is reported.

To test the hypothesis that the average wait time to see an E.R. doctor is less than 150 minutes, we can follow these steps:

State the null hypothesis : The null hypothesis is the assumption that there is no difference between the observed result and what we expect to see. In this case, the null hypothesis is that the average wait time to see an E.R. doctor is 150 minutes.

State the alternative hypothesis: The alternative hypothesis is the opposite of the null hypothesis. In this case, the alternative hypothesis is that the average wait time to see an E.R. doctor is less than 150 minutes.

Calculate the test statistic : The test statistic is a measure of the difference between the observed result and the expected result. In this case, the test statistic is the difference between the observed average wait time of 148 minutes and the expected average wait time of 150 minutes, divided by the standard deviation of 5 minutes.

Determine the critical value: The critical value is the value that determines whether the test statistic is significant or not. To determine the critical value, we need to determine the p-value of the test statistic using a significance level of 0.05.

Compare the test statistic to the critical value: If the test statistic is greater than the critical value, then the null hypothesis can be rejected. If the test statistic is less than the critical value, then the null hypothesis cannot be rejected.

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Question 1 of 10 For the x-values 1,2,3, and so on, the y-values of a function form a geometric sequence that increases in value. What type of function is it? A. Increasing linear B. Decreasing linear C. Exponential growth D. Exponential decay

The type of function is it is C . Exponential growth .

For the x-values 1,2,3, and so on, the y -values of a function form a geometric sequence that increases in value .

If  for values of x values of y increases and forms a geometric sequence then it would be an exponential growth function because geometric sequence is an exponential function and since, it is increasing hence, an exponential growth.

Option B is incorrect because  because y is not decreasing

Option C and D are incorrect because geometric sequence can never be linear since, it gives common ratio .

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What is the role of galaxies, planets, satellites, comets, and asteroids in shaping the universe?​

We can say that the role of galaxies, planets, satellites, comets and asteroids , in the conformation of the universe , is based on being elements to structure it.

In physics, it is mentioned that the universe is all physical existence that exists, this goes from our planet to what is found abroad.

Within the universe , as an entity, how to find different elements that structure it, for example, we have:

¡ Hope this helped !

none of the answer choices is correct. the sample consisted of all veterinarians who treat horses with nsaids. the sample consisted of all veterinarians on the list and therefore equaled the target population. the sample consisted of all veterinarians who returned the questionnaire.

equaled the target population. the sample consisted of all veterinarians who returned the questionnaire is None of the answer choices is correct.

The sample in this case consists of all veterinarians who treat horses with Non-Steroidal Anti-Inflammatory Drugs (NSAIDs). The target population for this study would be all veterinarians who treat horses with NSAIDs. Therefore, since the sample is made up of all veterinarians in the target population, the sample equals the target population. None of the answer choices provided is correct in this scenario.

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When two variables are not correlated at all, the correlation coefficient would be _______. a. -1 b. 0 c. 1 d. -2 e. 0.5

Pls choose me as brainliest!

Simplify for all questions.

-----------------------------------------

root of 63x to the power of 11 yz to the power of 10

The value of the equation is [tex]A = (\sqrt{63x}^{(11yz)^1^0}[/tex]

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation . The "=" sign and terms on both sides must always be present when writing an equation .

Given data ,

Let the equation be represented as A

Now , the value of A is given by

Root of 63x to the power of 11 yz to the power of 10

The value of root of 63x = [tex]\sqrt{63x}[/tex]

Now , the value of root of 63x to the power of 11 yz is

The value of root of 63x to the power of 11 yz = [tex](\sqrt{63x}^{(11yz)[/tex]

The value of root of 63x to the power of 11 yz to the power of 10 is

Value of root of 63x to the power of 11 yz to the power of 10 is [tex](\sqrt{63x}^{(11yz)^1^0}[/tex]

Therefore , the value of A is [tex](\sqrt{63x}^{(11yz)^1^0}[/tex]

Hence , The value of the equation is [tex]A = (\sqrt{63x}^{(11yz)^1^0}[/tex]

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The Variance Inflationary Factor (VIF) measures the: a. correlation of the X variables with the Y variable. b. correlation of the X variables with each other. c. standard deviation of the slope. d. contribution of each X variable with the Y variable after all other X variables are included in the model. e. both a and b.

The X variables have an overall correlation of 0. VIF, or Variance Inflationary Factor , measures the

Regression analysis's level of multicollinearity is gauged by a variance inflation factor, or VIF. When several independent variables in some kind of a model with multiple regression correlate with one another, this is known as multicollinearity . The regression findings may be significantly impacted by this.

Greater correlation between the variable and other variables is indicated by a higher value. With values of Ten or more being considered very high, values greater than 4 or 5 are occasionally considered to be moderate to high.

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Imaginary Numbers: Simplify [tex] \sqrt{ - 80} [/tex] ​

=√ –( 16 × 5

= √ –(4² × 5)

this is what I think I hope I helped

researchers wanted to check if carpeted rooms in hospitals contained more bacteria than uncarpeted rooms. to determine the amount of bacteria in a room, researchers pumped the air from the room over a petri dish for eight carpeted and eight uncarpeted rooms. colonies of bacteria were allowed to form in the 16 petri dishes. the results are presented in the table. (measured as bacteria per cubic foot) carpeted: 11.8, 10.8, 8.2, 10.1, 7.1, 14.6, 13.0, 14.0 uncarpeted: 12.1, 12.0, 8.3, 11.1, 3.8, 10.1, 7.2, 13.7 do carpeted rooms have more bacteria than uncarpeted rooms at a

No , the carpeted rooms does not have more bacteria than uncarpeted rooms .

mean = ∑x / n

x1 = 11.8+10.8+7.1+14.6+8.2+10.1+13.0+14.0 / 8 = 11.2

x2 =12.1+12.0+3.8+10.1+8.3+11.1+7.2+13.7 /8 = 9.78

H0 : μ 1 =μ 2

H1 : μ 1  ≠μ 2

​Either the alternative hypothesis or the null hypothesis applies to the claim. The equality must be present in the null hypothesis. The alternative hypothesis declares the opposite of the null hypothesis if the claim is the null hypothesis.

On calculating we get the test statistic t as,

The null hypothesis is rejected if the test statistic's value falls within the rejection zone.

0.956<1.771,

therefore , fail to reject H .

Therefore , there is no support for the null hypothesis , so we accept the alternate hypothesis as true.

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Evaluate f(x) = 3x - 6: a) x = -4 b) f(x) = 36.

f(x) = 3x -6

f(-4) = 3(-4) - 6

f(-4) = -12 - 6

f(-4) = -18

36 = 3x - 6  Add 6 to both sides

36 + 6 = 3x - 6 + 6

42 = 3x  Divide both side by 3

[tex]\frac{42}{3}[/tex] = [tex]\frac{3x}{3}[/tex]

In Exercises 1, 2, 3, 4, 5, and 6, show that v is an eigenvector of A and find the corresponding eigenvalue. A= [ -1 16 0 ] V = [1 -2]

Av = λv , so we can conclude that v is an eigenvector of A and the corresponding eigenvalue is λ = -2.

A*v = [-1 16 0] * [1 -2] = [-1 -32 0]

λv = [-2 -4 0]

Since Av = λv, we can conclude that v is an eigenvector of A and the corresponding eigenvalue is λ = -2.

1. First, we multiply the matrix A with vector v to find the product Av.

2. A*v = [-1 16 0] * [1 -2] = [-1 -32 0]

3. Then, we multiply the eigenvalue λ with vector v which gives the result λv.

4. λv = [-2 -4 0]

5. We compare Av and λv. If they are equal , then v is an eigenvector of A and the corresponding eigenvalue is λ.

6. Av = λv, so we can conclude that v is an eigenvector of A and the corresponding eigenvalue is λ = -2.

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the sugar sweet company delivers sugar to its customers. let be the total cost to transport the sugar (in dollars). let be the amount of sugar transported (in tons). the company can transport up to tons of sugar. suppose that gives as a function of . identify the correct description of the values in both the domain and range of the function. then, for each, choose the most appropriate set of values.

The domain and the range of a function are the possible input and output values of the function.

The function is given as:

C = 130S + 3500

This is the possible S values of the function.

S cannot be less than 0, because it represents a physical quantity (i.e. tons of sugar)

The value of S can be any value greater than 0

Hence, the domain of the function is [0, ∞)

This is the possible C values of the function.

C = 130(0) + 3500

The above means that:

C cannot be less than 3500

The value of C can be any value greater than 3500

Hence, the range of the function is [3500, ∞)

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GIVING BRAINLIEST & 30 POINTS! (please do the math and don't use other sources I know answering this question means nothing to you other than points but just think about me and others who might get the question wrong if you other sources.) Ty! −np − 5 = 4(c − 2) Which of the following solves for n? n = the quantity 4 times c minus 3 all over negative p n = the quantity 4 times c minus 13 all over negative p n = the quantity 4 times c plus 3 all over p n = the quantity 4 times c plus 13 all over p

n = the quantity 4 times c minus 3 all over negative p

Explanation:

Interpretation of answers to equation

n = the quantity 4 times c minus 3 all over negative p = [tex]\frac{4c-3}{-p}[/tex]

n = the quantity 4 times c minus 13 all over negative p = [tex]\frac{4c-13}{-p}[/tex]

n = the quantity 4 times c plus 3 all over p = [tex]\frac{4c+3}{p}[/tex]

n = the quantity 4 times c plus 13 all over p = [tex]\frac{4c+13}{p}[/tex]

Step-by-step:

[tex]-np-5=4(c-2)[/tex]

[tex]-np-5=4c-4(2)[/tex]

[tex]-np-5=4c-8[/tex]

[tex]-np-5+5=4c-8+5[/tex]

[tex]-np=4c-3[/tex]

Divide both side by -p

[tex]n=\frac{4c-3}{-p}[/tex]

The answer is the option A, which is: A.) n ≥ − the quantity 4 times c minus 3 all over p.

You have −np − 5 ≤ 4(c − 2)

When you solve for n, you obtain:

-np-5 ≤ 4(c-2)

Therefore, you have:

n≤-(4c-3/p)

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There are 4 consecutive even integers that add up to 100. What is the least of the 4 integers?

Let's say the 4 consecutive integers are a, b, c, and d

a=b-2  ==> Consecutive even numbers have a difference of 2: Ex. (4-2=2)

a+b+c+d=100

Solve for a:

a=b-2  ==> add 2 on both sides

a+(a+2)+c+d=100  ==> substitute a+2 for b

a+2=c-2  ==> b=c-2 and a+2=b, so a+2=c-2

a+2+2=c-2+2  ==> add 2 on both sides

a+(a+2)+(a+4)+d=100 ==> substitute a+4 for c

a+4=d-2  ==> c=d-2 and a+4=c, so a+4=d-2

a+4+2=d-2+2  ==> add 2 on both sides

a+(a+2)+(a+4)+(a+6)=100  ==> substitute a+6 for d

a+a+2+a+4+a+6=100  ==> solve for a

a+a+a+a+2+4+6=100

4a+12=100 ==> simplify

4a+12-12=100-12 ==> isolate a

4a=88  ==> simplify

Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of a. (Enter your answers as a comma-separated list.) 5 f(x) a = 2 1 + x

The first four nonzero terms of the series for f(x) from the taylor series is F(x)=5/3-5/9(x-2)+5/27(x-2)²-5/81(x-2)³+....

Given that,

We have to get the first four nonzero terms of the series for f(x), centering at the specified value of a, use the definition of a Taylor series. (Separate your responses into a list using commas.)

f(x)=5/1+x and a=2

We know that,

f(2)=5/1+2=5/3

f'(x)=-5/(1+x)²         f'(2)=-5/(1+2)²=-5/9                  ( differentiating )

f''(x)=10/(1+x)³        f''(2)=10/(1+2)³=10/27

f'''(x)=-30/(1+x)⁴     f'''(2)=-30/(1+2)⁴=-30/81=-10/27

Using the taylor series

F(x)=f(a)-(x-a)f'(a)+(x-a)²/2!f''(a)-(x-a)³/3!f'''(a)+....

F(x)=f(2)-(x-2)f'(2)+(x-2)²/2f''(2)-(x-2)³/3!f'''(2)+....

F(x)=5/3-(x-2)(5/9)+(x-2)²/2(10/27)-(x-2)³/6(-10/27)+....

F(x)=5/3-5/9(x-2)+5/27(x-2)²-5/81(x-2)³+....

Therefore, The first four nonzero terms of the series for f(x) from the taylor series is F(x)=5/3-5/9(x-2)+5/27(x-2)²-5/81(x-2)³+....

To learn more about series visit: https://brainly.com/question/15415793

If a sample mean is 32, which of the following is most likely the range of possible values that best describes an estimate for the population mean? A. (28,36) B. (34,42) C. (32, 42) D. (30,38)​

If a sample mean is 32, the range of possible values that best describes an estimate for the population mean is most likely (30, 38).

The sample mean is a statistic that is used to estimate the population mean. In general, the sample mean is a good estimate of the population mean, but it is not always perfectly accurate. There is always some degree of uncertainty or variability associated with any estimate, and this is especially true when the sample size is small.

One way to quantify this uncertainty is to use a confidence interval, which is a range of values that is likely to contain the population mean with a certain level of confidence. For example, a 95% confidence interval is a range of values that is likely to contain the population mean with 95% confidence.

In this case, if a sample mean is 32, a 95% confidence interval for the population mean would likely be a range of values centered around 32 and extending approximately 2 standard errors in either direction. For a sample mean of 32, a range of values extending 2 standard errors in either direction would be approximately (30, 38). Therefore, the range of possible values that best describes an estimate for the population mean is most likely (30, 38).

If a sample mean is 32, (30, 38) is most likely the range of possible values that best describes an estimate for the population mean. the correct option is option D.

Range serves as a statistical measure of dispersion in mathematics, or how widely spaced a particular data collection is from smallest to biggest. The range in a piece of data is the distinction between the highest and lowest number. The confidence interval of 95% for the population mean in this scenario, assuming the sample mean is 32, is likely to include a range of values centred around 32.

Extending around 2 standard errors in each direction. A range of results spanning two standard deviations in either way would result in around (30, 38) for a sample mean of 32. Thus, the range of potential values that most accurately captures an estimate of the population mean is (30, 38).

Therefore, the correct option is option D.

To know more about range , here:

https://brainly.com/question/30821383

show two examples of the associative property

Solve the equation. 3^2x= 6,561 x=3 x=4 x=8 x= 16

Step-by-step explanation

then think 3 raise to power what is 6561

the answer is 8

three cancel three

divide both sides by 2

In a board game, a certain number of points is awarded to a player upon rolling a six sided die (labeled 1 to 6) according to the function f(x)=4x+4, where xx is the value rolled on the die. Find and interpret the given function values and determine an appropriate domain for the function.

The numeric values of the function are given f(1) = 8, meaning when a 1 is rolled on the die, the player is awarded 8 points. This interpretation makes sense in the context of the problem.

The function in this problem is defined as follows:

f(x) = 4x+4

The variables are given as x is the number rolled and y is the score.

The numbers that can be rolled are:

1, 2, 3, 4, 5 and 6.

Hence the domain is

{x ∈ N | 1 ≤ x ≤ 4}.

The numeric values are obtained replacing the variable x by the input, as follows:

f(1) = 4 + 4 =8

f(10) = 4(10) + 4= 44-> does not make sense, as a 10 cannot be rolled .

This interpretation does not make sense in the context.

Learn more about the numeric values at brainly.com/question/28367050

9.1 Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 : The null hypothesis: It is a statement of no difference between the variables—they are not related. This can often be considered the status quo and as a result if you cannot accept the null it requires some action.

H a : The alternative hypothesis: It is a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 . This is usually what the researcher is trying to prove.

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are "reject H 0 " if the sample information favors the alternative hypothesis or "do not reject H 0 " or "decline to reject H 0 " if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

Example 9.1

H 0 : No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ .30 H a : More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.

Example 9.2

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are: H 0 : μ = 2.0 H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 66
• H a : μ __ 66

Example 9.3

We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are: H 0 : μ ≥ 5 H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 45
• H a : μ __ 45

Example 9.4

In an issue of U. S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066

On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : p __ 0.40
• H a : p __ 0.40

Collaborative Exercise

Bring to class a newspaper, some news magazines, and some Internet articles . In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

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• Authors: Barbara Illowsky, Susan Dean
• Publisher/website: OpenStax
• Book title: Introductory Statistics 2e
• Publication date: Dec 13, 2023
• Location: Houston, Texas
• Book URL: https://openstax.org/books/introductory-statistics-2e/pages/1-introduction
• Section URL: https://openstax.org/books/introductory-statistics-2e/pages/9-1-null-and-alternative-hypotheses

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