2 An article reported the results of 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 44.00 patients walked on a treadmill for 6 minutes every day. After 6 months, the mean distance walked in 6 minutes was 348 m, with a standard deviation of 80 m. For a control group of 42.00 patients who did not walk on a treadmill, the mean distance was 309 m with a standard deviation of 89 m. Can you conclude that the mean distance walked for patients using a treadmill is greater than the mean for the controls? Use the a = 0.05 level of significance. Find the P-value and state a conclusion. (Round the final answer to four decimal places.) The P-value is 0.0172 x We cannot conclude that the mean distance walked for patients using a treadmill is greater than the mean for the controls.

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.2: Expected Value And Variance Of Continuous Random Variables
Problem 10E
Question
2
An article reported the results of 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 44.00 patients walked on a
treadmill for 6 minutes every day. After 6 months, the mean distance walked in 6 minutes was 348 m, with a standard deviation of 80
m. For a control group of 42.00 patients who did not walk on a treadmill, the mean distance was 309 m with a standard deviation of 89
m. Can you conclude that the mean distance walked for patients using a treadmill is greater than the mean for the controls? Use the a
= 0.05 level of significance. Find the P-value and state a conclusion. (Round the final answer to four decimal places.)
The P-value is 0.0172 x
We cannot conclude that the mean distance walked for patients using a treadmill is greater than the mean for the controls.
Transcribed Image Text:2 An article reported the results of 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 44.00 patients walked on a treadmill for 6 minutes every day. After 6 months, the mean distance walked in 6 minutes was 348 m, with a standard deviation of 80 m. For a control group of 42.00 patients who did not walk on a treadmill, the mean distance was 309 m with a standard deviation of 89 m. Can you conclude that the mean distance walked for patients using a treadmill is greater than the mean for the controls? Use the a = 0.05 level of significance. Find the P-value and state a conclusion. (Round the final answer to four decimal places.) The P-value is 0.0172 x We cannot conclude that the mean distance walked for patients using a treadmill is greater than the mean for the controls.
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