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 patients walked on the treadmill every day and after six months the mean walking distance was measured. A control group of patients that did not walk on the treadmill were also measured for mean walking distance. The data for each group is as follows:

Group	Sample Size	Mean	Standard Deviation
Treadmill	48	348	80
Control	46	309	89

Test the claim that the mean distance walked for patients using a treadmill is greater than the mean for the controls. Use α = .05.

a. State whether the test is

i)          a two-sample t-test (independent samples)

ii)        a matched pairs

iii)       a two sample proportion test

b. Write H0 and H1

c. Using Minitab, list

• the test statistic
• the p-value
• your conclusion: reject H0 or do not reject H0. Note: if α is not provided, use a 0.05 significance level

d.   Write a sentence that explains your conclusion in context with the claim. Include the significance level and p-value in this sentence.

e.   Copy and paste the relevant Minitab output into the document. Answers alone are sufficient, you do not need to copy the exercise into the document.

Answer 1

a).here, we will do :-

a two-sample t-test (independent samples)

[ as the data are independent..and we want to compate two means]

b).hypothesis:-

[ i am denoting Treadmill by 1 and control by 2]

our claim is the alternative hypothesis.

c).test statistic = 2.23

p value = 0.014

conclusion:- we reject H0

[p value =0.014 <0.05 (alpha)]

d).here, as the p value is less than 0.05 we reject the null hypothesis.

we conclude that,

there is sufficient evodence to support the claim that the mean distance walked for patients using a treadmill is greater than the mean for the controls at 0.05 level of significance.

e).as instructed i am using minitab to solve the problem.

steps

stat basic statistics 2 sample tselect summmarized data from teh drop down menu in sample 1, type 48 in sample size,348 in sample mean and 80 in standard deviation  in sample 2, type 46 in sample size,309 in sample mean and 89 in standard deviationoptions in confidence level type 95,in hypothesied difference type 0,select the alternative hypothesis as difference hypothesized difference okok.

your minitab output be:-

Two-Sample T-Test and CI

Method

μ₁: mean of Sample 1

µ₂: mean of Sample 2

Difference: μ₁ - µ₂

Equal variances are not assumed for this analysis.

Descriptive Statistics

Sample	N	Mean	StDev	SE Mean
Sample 1	48	348.0	80.0	12
Sample 2	46	309.0	89.0	13

Estimation for Difference

Difference	95% Lower Bound for Difference
39.0	9.9

Test

Null hypothesis	H₀: μ₁ - µ₂ = 0
Alternative hypothesis	H₁: μ₁ - µ₂ > 0

T-Value	DF	P-Value
2.23	90	0.014

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