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The null and alternative hypotheses for a test are given as well as some information about the actual sample(s) and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. In addition, indicate whether the test is a left-tail test, a right-tail test, or a twotailed test. Hypotheses: \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) Sample: \(\bar{x}_{1}=2.7\) and \(\bar{x}_{2}=2.1\) Randomization statistic \(=\bar{x}_{1}-\bar{x}_{2}\)

Short Answer

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The randomization distribution would be centered at 0, and the test would be a two-tailed test.

Step by step solution

01

Interpreting the Hypotheses

The null hypothesis (H0) asserts that the means of population 1 and 2 are equal, denoted by \(H_{0}: \mu_{1}=\mu_{2}\). The alternative hypothesis (Ha), on the other hand, contends that the means are not equal, denoted by \(H_{a}: \mu_{1} \neq \mu_{2}\). Given that Ha involves a 'not equals to' condition, it suggests a two-tailed test.
02

Understanding the Sample Means

The given sample means are \(\bar{x}_{1}=2.7\) and \(\bar{x}_{2}=2.1\). Thus, the mean of sample 1 is greater than that of sample 2. This information will be used in the next step.
03

Calculating Randomization Statistic and Predicting the Center

The randomization statistic is given by \(=\bar{x}_{1}-\bar{x}_{2}\), which translates to \(=2.7-2.1=0.6\). However, under the null hypothesis (H0), the two population means are assumed to be equal so the expected difference between them would be zero. Consequently, the center of the randomization distribution should be at 0.

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Most popular questions from this chapter

The same sample statistic is used to test a hypothesis, using different sample sizes. In each case, use StatKey or other technology to find the p-value and indicate whether the results are significant at a \(5 \%\) level. Which sample size provides the strongest evidence for the alternative hypothesis? Testing \(H_{0}: p=0.5\) vs \(H_{a}: p>0.5\) using \(\hat{p}=0.55\) with each of the following sample sizes: (a) \(\hat{p}=55 / 100=0.55\) (b) \(\hat{p}=275 / 500=0.55\) (c) \(\hat{p}=550 / 1000=0.55\)

Influencing Voters When getting voters to support a candidate in an election, is there a difference between a recorded phone call from the candidate or a flyer about the candidate sent through the mail? A sample of 500 voters is randomly divided into two groups of 250 each, with one group getting the phone call and one group getting the flyer. The voters are then contacted to see if they plan to vote for the candidate in question. We wish to see if there is evidence that the proportions of support are different between the two methods of campaigning. (a) Define the relevant parameter(s) and state the null and alternative hypotheses. (b) Possible sample results are shown in Table 4.3 . Compute the two sample proportions: \(\hat{p}_{c},\) the proportion of voters getting the phone call who say they will vote for the candidate, and \(\hat{p}_{f},\) the proportion of voters getting the flyer who say they will vote for the candidate. Is there a difference in the sample proportions? (c) A different set of possible sample results are shown in Table 4.4. Compute the same two sample proportions for this table. (d) Which of the two samples seems to offer stronger evidence of a difference in effectiveness between the two campaign methods? Explain your reasoning. $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Will Vote } \\ \text { Sample A } \end{array} & \text { for Candidate } & \begin{array}{l} \text { Will Not Vote } \\ \text { for Candidate } \end{array} \\ \hline \text { Phone call } & 152 & 98 \\ \text { Flyer } & 145 & 105 \\ \hline \end{array} $$ $$ \begin{array}{lcc} \text { Sample B } & \begin{array}{c} \text { Will Vote } \\ \text { for Candidate } \end{array} & \begin{array}{c} \text { Will Not Vote } \\ \text { for Candidate } \end{array} \\ \hline \text { Phone call } & 188 & 62 \\ \text { Flyer } & 120 & 130 \\ \hline \end{array} $$

A statistics instructor would like to ask "clicker" questions that about \(80 \%\) of her students in a large lecture class will get correct. A higher proportion would be too easy and a lower proportion might discourage students. Suppose that she tries a sample of questions and receives 76 correct answers and 24 incorrect answers among 100 responses. The hypotheses of interest are \(H_{0}: p=0.80\) vs \(H_{a}: p \neq 0.80 .\) Discuss whether or not the methods described below would be appropriate ways to generate randomization samples in this setting. Explain your reasoning in each case. (a) Sample 100 answers (with replacement) from the original student responses. Count the number of correct responses. (b) Sample 100 answers (with replacement) from a set consisting of 8 correct responses and 2 incorrect responses. Count the number of correct mses.

Weight Loss Program Suppose that a weight loss company advertises that people using its program lose an average of 8 pounds the first month, and that the Federal Trade Commission (the main government agency responsible for truth in advertising) is gathering evidence to see if this advertising claim is accurate. If the FTC finds evidence that the average is less than 8 pounds, the agency will file a lawsuit against the company for false advertising. (a) What are the null and alternative hypotheses the FTC should use? (b) Suppose that the FTC gathers information from a very large random sample of patrons and finds that the average weight loss during the first month in the program is \(\bar{x}=7.9\) pounds with a p-value for this result of \(0.006 .\) What is the conclusion of the test? Are the results statistically significant? (c) Do you think the results of the test are practically significant? In other words, do you think patrons of the weight loss program will care that the average is 7.9 pounds lost rather than 8.0 pounds lost? Discuss the difference between practical significance and statistical significance in this context.

Eating Breakfast Cereal and Conceiving Boys Newscientist.com ran the headline "Breakfast Cereals Boost Chances of Conceiving Boys," based on an article which found that women who eat breakfast cereal before becoming pregnant are significantly more likely to conceive boys. \({ }^{42}\) The study used a significance level of \(\alpha=0.01\). The researchers kept track of 133 foods and, for each food, tested whether there was a difference in the proportion conceiving boys between women who ate the food and women who didn't. Of all the foods, only breakfast cereal showed a significant difference. (a) If none of the 133 foods actually have an effect on the gender of a conceived child, how many (if any) of the individual tests would you expect to show a significant result just by random chance? Explain. (Hint: Pay attention to the significance level.) (b) Do you think the researchers made a Type I error? Why or why not? (c) Even if you could somehow ascertain that the researchers did not make a Type I error, that is, women who eat breakfast cereals are actually more likely to give birth to boys, should you believe the headline "Breakfast Cereals Boost Chances of Conceiving Boys"? Why or why not?

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