Question

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, \(\bar{x}_{F}-\bar{x}_{M}\) .

Short answer

Method A: Not appropriate as it ignores the actual genders of the students. Method B: Not appropriate as it artificially standardizes the means. Method C: Appropriate because it maintains the original structure of the data while respecting the presumption of the null hypothesis.

Step-by-step solution

Understanding Data and Formulating Hypotheses

Evaluate the data provided for the exercise hours of female and male students. The null hypothesis is that the mean hours of exercise for both genders are equal (\(H_0: \mu_{F}=\mu_{M}\)). The alternative hypothesis states that they are not equal (\(H_a: \mu_{F} \neq \mu_{M}\)).

Evaluating Method A

Assess the appropriateness of Method A, which suggests randomly labelling 30 of the exercise values as female and remaining 20 as male. This method is not appropriate because it ignores the actual genders of the students. We must maintain the structure of the original gender assignments when generating randomization samples under the null hypothesis.

Evaluating Method B

Assess the appropriateness of Method B, which involves adjusting the mean hours for both genders to match the overall average, then re-sampling. This method is not appropriate as it artificially standardizes the means, which is not what we expect under the null hypothesis. Our goal under the null hypothesis is to adjust the labels of the observations, not the means of the groups.

Evaluating Method C

Evaluate the appropriateness of Method C, which combines all exercise hours, randomly select 30 for females, and 20 for males. This method is appropriate because under the null hypothesis, we expect no difference in exercise time due to gender. By amalgamating the data into one group and randomly assigning observations, we maintain the original structure of the data while respecting the presumption of the null hypothesis.