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91Ó°ÊÓ

State the conclusion of the test based on this p-value in terms of "Reject \(H_{0} "\) or "Do not reject \(H_{0} "\), if we use a \(5 \%\) significance level. p-value \(=0.0007\)

Short Answer

Expert verified
Given that the p-value of 0.0007 is less than the significance level of 5% or 0.05, we reject the null hypothesis \(H_{0}\).

Step by step solution

01

Understand Terminologies

In hypothesis testing, the null hypothesis (\(H_{0}\)) represents a statement of no effect or no difference. The p-value is the probability, under the null hypothesis about the unknown distribution, of randomly drawing a value equal to or more extreme than the one obtained by testing. The significance level, often denoted by \(α\), is a threshold pre-determined at which if the calculated p-value in the test is less than this level, we reject the null hypothesis in favor of the alternative hypothesis.
02

Comparing the P-Value with the Significance Level

In this case, the given p-value is 0.0007, and the significance level is 5% or 0.05. We determine whether to reject the null hypothesis \(H_{0}\) based on whether the p-value is less than or equal to the significance level.
03

Formulate the Conclusion

Seeing that 0.0007 is less than 0.05, we would reject the null hypothesis \(H_{0}\). However, it is important to remember that rejecting the null hypothesis does not prove the alternative hypothesis; it merely provides evidence to suggest the alternative hypothesis may be true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

null hypothesis
The null hypothesis, often represented by the symbol \(H_0\), is a fundamental concept in statistical hypothesis testing. It is essentially a statement of no effect or no change, implying that any observed difference or pattern in data is due to chance. In a variety of studies, the null hypothesis typically proposes that there is no relationship between two phenomena or that a treatment has no effect.

Consider the null hypothesis as the default position that a researcher seeks to test against. This concept helps in understanding whether there is enough statistical evidence to infer that a particular condition or effect genuinely exists. For instance, if a pharmaceutical company claims that a new drug is effective against a disease, the null hypothesis might state that the drug has no effect compared to a placebo.
  • The null hypothesis is always tested indirectly by testing the opposite, known as the alternative hypothesis.
  • Rejecting the null hypothesis in hypothesis testing suggests that the alternative hypothesis may be true.
p-value
A p-value is a crucial statistical measure used in hypothesis testing to interpret the results of an experiment or study. It helps us determine the strength of evidence against the null hypothesis. Simply put, the p-value tells us the probability of obtaining results at least as extreme as those observed, given that the null hypothesis is true.

In hypothesis testing, a smaller p-value indicates stronger evidence against the null hypothesis, suggesting that the observed data is unlikely under the null hypothesis. Thus, statisticians and scientists set thresholds known as significance levels to decide whether to reject the null hypothesis.
  • If the p-value is less than or equal to the significance level, it indicates significant results, prompting rejection of the null hypothesis.
  • P-values are objective, offering a standardized method to gauge evidence from data.
significance level
The significance level, denoted by \( \alpha \), is a predetermined threshold in hypothesis testing that defines the probability of rejecting the null hypothesis when it is true. Typically set at values like 0.05 or 0.01, the significance level reflects the risk one is willing to take to reject the null hypothesis incorrectly.

Choosing a significance level involves balancing the potential for Type I errors (incorrectly rejecting a true null hypothesis) and Type II errors (failing to reject a false null hypothesis). A typical significance level of 0.05 implies a 5% risk of committing a Type I error.
  • Researchers choose a significance level before conducting the experiment to guide decision-making.
  • A smaller \( \alpha \) reduces the chance of a Type I error but increases the chance of a Type II error.
  • Interpreting the test results involves comparing the p-value to the significance level. If the p-value is below \( \alpha \), the null hypothesis is rejected.

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Most popular questions from this chapter

Approval Rating for Congress In a Gallup poll \(^{51}\) conducted in December 2015 , a random sample of \(n=824\) American adults were asked "Do you approve or disapprove of the way Congress is handling its job?" The proportion who said they approve is \(\hat{p}=0.13,\) and a \(95 \%\) confidence interval for Congressional job approval is 0.107 to 0.153 . If we use a 5\% significance level, what is the conclusion if we are: (a) Testing to see if there is evidence that the job approval rating is different than \(14 \%\). (This happens to be the average sample approval rating from the six months prior to this poll.) (b) Testing to see if there is evidence that the job approval rating is different than \(9 \%\). (This happens to be the lowest sample Congressional approval rating Gallup ever recorded through the time of the poll.)

Hypotheses for a statistical test are given, followed by several possible confidence intervals for different samples. In each case, use the confidence interval to state a conclusion of the test for that sample and give the significance level used. Hypotheses: \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2} .\) In addition, in each case for which the results are significant, state which group ( 1 or 2 ) has the larger mean. (a) \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : 0.12 to 0.54 (b) \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : -2.1 to 5.4 (c) \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : -10.8 to -3.7

Give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) Sample data: \(\hat{p}=28 / 40=0.70\) with \(n=40\)

A confidence interval for a sample is given, followed by several hypotheses to test using that sample. In each case, use the confidence interval to give a conclusion of the test (if possible) and also state the significance level you are using. A \(95 \%\) confidence interval for \(p: 0.48\) to 0.57 (a) \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) (b) \(H_{0}: p=0.75\) vs \(H_{a}: p \neq 0.75\) (c) \(H_{0}: p=0.4\) vs \(H_{a}: p \neq 0.4\)

Introductory statistics students fill out a survey on the first day of class. One of the questions asked is "How many hours of exercise do you typically get each week?" Responses for a sample of 50 students are introduced in Example 3.25 on page 244 and stored in the file ExerciseHours. The summary statistics are shown in the computer output below. The mean hours of exercise for the combined sample of 50 students is 10.6 hours per week and the standard deviation is 8.04 . We are interested in whether these sample data provide evidence that the mean number of hours of exercise per week is different between male and female statistics students. $$\begin{array}{lllrrrr} \text { Variable } & \text { Gender } & \text { N } & \text { Mean } & \text { StDev } & \text { Minimum } & \text{ Maximum } \\\\\text { Exercise } & \mathrm{F} & 30 & 9.40 & 7.41 & 0.00 & 34.00 \\\& \mathrm{M} & 20 & 12.40 & 8.80 & 2.00 & 30.00\end{array}$$ Discuss whether or not the methods described below would be appropriate ways to generate randomization samples that are consistent with \(H_{0}: \mu_{F}=\mu_{M}\) vs \(H_{a}: \mu_{F} \neq \mu_{M} .\) Explain your reasoning in each case. (a) Randomly label 30 of the actual exercise values with "F" for the female group and the remaining 20 exercise values with "M" for the males. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\). (b) Add 1.2 to every female exercise value to give a new mean of 10.6 and subtract 1.8 from each male exercise value to move their mean to 10.6 (and match the females). Sample 30 values (with replacement) from the shifted female values and 20 values (with replacement) from the shifted male values. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\) (c) Combine all 50 sample values into one set of data having a mean amount of 10.6 hours. Select 30 values (with replacement) to represent a sample of female exercise hours and 20 values (also with replacement) for a sample of male exercise values. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\) .

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