Question

Medical researchers conducted a study to determine whether treadmill exercise could improve the walking ability of patients suffering from claudication, which is pain caused by insufficient blood flow to the muscles of the legs. a sample of 51 patients walked on a treadmill for six minutes every day. after six months, the mean distance walked in six minutes was 370 meters, with a standard deviation of 83 meters. for a control group of 56 patients who did not walk on a treadmill, the mean distance was 376 meters with a standard deviation of 99 meters. can you conclude that the mean distance walked for patients using a treadmill is less than the mean for the controls?

let µ₁ denote the mean distance walked for patients who used a treadmill. use the α = 0.1 level of significance and the ti-84 plus calculator.

Answer 1

To determine if treadmill exercise affects the mean distance walked by patients with claudication, a two-sample t-test is used with a significance level of 0.1. The null hypothesis is that the treadmill exercises make no difference or improve walking distance, while the alternative hypothesis is it reduces walking distance. The TI-84 Plus calculator can assist in calculating the test statistic and p-value.

Step-by-step explanation:

Conducting a Hypothesis Test for Mean Distance Walked

To determine if the mean distance walked for patients using a treadmill (µ₁) is less than the mean distance walked for controls, we conduct a hypothesis test. Given the details, this is a two-sample t-test comparing the means of two independent groups. We will use the following hypotheses:

• Null hypothesis (H0): µ₁ ≥ µ₂ (The treadmill exercise does not reduce the mean distance walked.)
• Alternative hypothesis (Ha): µ₁ < µ₂ (The treadmill exercise reduces the mean distance walked.)

We use an α = 0.1 significance level for this test. The TI-84 Plus calculator can perform this test using its two-sample t-test function by entering the given means, standard deviations, and sample sizes for both groups.

To interpret the output, compare the p-value to the significance level (α = 0.1). If the p-value is less than 0.1, we reject the null hypothesis; otherwise, we fail to reject it. A conclusion that rejects the null hypothesis would suggest that treadmill exercise does reduce walking distance, contrary to the claims.

For the second part of the question about the matched-pairs t-test:

The null hypothesis is H0: µd = 0 (no change in time to run a mile after the exercise program). The alternative hypothesis is Ha: µd < 0 (a decrease in time to run a mile after the exercise program).

Using the provided data (x = -5, sa = 6, n = 30), the test statistic is calculated using the formula t = x / (sa / √n). The degrees of freedom would be n - 1, which is 29. A decision on the effectiveness would depend on whether the calculated t-value falls into the critical region defined by α = 0.05. If the test statistic exceeds the critical value, we reject the null hypothesis and conclude the exercise program is effective.