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After exercise, massage is often used to relieve pain, and a recent study 33 shows that it also may relieve inflammation and help muscles heal. In the study, 11 male participants who had just strenuously exercised had 10 minutes of massage on one quadricep and no treatment on the other, with treatment randomly assigned. After 2.5 hours, muscle biopsies were taken and production of the inflammatory cytokine interleukin-6 was measured relative to the resting level. The differences (control minus massage) are given in Table 4.11 . $$ \begin{array}{lllllllllll} 0.6 & 4.7 & 3.8 & 0.4 & 1.5 & -1.2 & 2.8 & -0.4 & 1.4 & 3.5 & -2.8 \end{array} $$ (a) Is this an experiment or an observational study? Why is it not double blind? (b) What is the sample mean difference in inflammation between no massage and massage? (c) We want to test to see if the population mean difference \(\mu_{D}\) is greater than zero, meaning muscle with no treatment has more inflammation than muscle that has been massaged. State the null and alternative hypotheses. (d) Use Statkey or other technology to find the p-value from a randomization distribution. (e) Are the results significant at a \(5 \%\) level? At a \(1 \%\) level? State the conclusion of the test if we assume a \(5 \%\) significance level (as the authors of the study did).

Short Answer

Expert verified
The study is an experiment, not double-blind. After calculating, you will get the sample mean difference. Hypotheses should be set where the null hypothesis assumes no difference in inflammation level between treatments, and the alternative assumes greater inflammation in untreated muscles. P-value is found using software and compared to significance levels: if it's less than 0.05 or 0.01, null hypothesis is rejected supporting the claim massage helps reduce inflammation.

Step by step solution

01

Identifying the Type of Study and Its Constraints

This is an experiment. In this study, treatments (massage or no massage) were randomly assigned to subjects (participants). The researchers actively manipulated the conditions, hence it's not observational. It's not double-blind because the person performing the massage therapy was aware of the conditions of the study.
02

Calculating the Sample Mean Difference

The sample mean difference can be calculated by adding all the differences and dividing the sum by the number of data points, which is 11 in this case.
03

Formulating the Hypotheses

The null hypothesis (\(H_0\)) is that the population mean difference (\(\mu_D\)) equals zero, meaning that massage treatment has no effect on inflammation level. The alternative hypothesis (\(H_a\)) is that the population mean difference is greater than zero, which means the untreated muscle has more inflammation than the massaged muscle.
04

Finding the P-Value

Using Statkey or similar software, input the data and select one-sided test to find the p-value. This will simulate the randomization distribution under the null hypothesis and calculate the p-value, which is the probability of observing a mean as extreme as, or more than, the actual sample mean.
05

Determining the Significance of Results

Compare the p-value to significance levels. If the p-value is less than 0.05, reject the null hypothesis at the 5% significance level and conclude that the massage significantly reduces inflammation. If the p-value is less than 0.01, also reject the null hypothesis at 1% significance level, further strengthening the conclusion. If the p-values are greater, do not reject the null hypothesis.

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Most popular questions from this chapter

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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