Question

Exercise: 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 47 patients walked on a treadmill for six minutes every day. After six months, the mean distance walked in six minutes was 341 meters, with a standard deviation of 79 meters. For a control group of 39 patients who did not walk on a treadmill, the mean distance was 305 meters with a standard deviation of 82 meters. Can you conclude that the mean distance walked for patients using a treadmill differs from the mean for the controls? Let μ1 denote the mean distance walked for patients who used a treadmill. Use the a=0.05 level of significance and the TI-84 Plus calculator.

find p value

Answer 1

s.no	distance travelled on threadmill
1	341metres
2	305 metres

mean $\mu$ =341+305/2

= 646/2

=323

Absolute deviation :

X	|X-U|
341	18
305	18
	$\sum_{}^{}$ |X-U| = 36

Mean devition:

mean deviation = $\sum_{}^{}$ |x-u| /N

= 36 / 2

=18

SO, the average mean distance walked on the threadmill is 18

p value = 0.0417

when the p value is less than 0.05 it represents that there is difference between the two distances.