Question

	Group A	Group B
Subject 1	40	30
Subject 2	42	45
Subject 3	30	38
Subject 4	37	32
Subject 5	23	28

A small hypothetical study explored the response to two treatments where each subject in the study received both treatments (e.g. in a cross-over design). The data are given below 1) Analyze the information using a paired t-test (e.g. with GraphPad) a) What is the observed value of the test statistic from this test? b) What can you conclude about difference in mean response to the two treatments? Why?

Answer 1

A small hypothetical study explored the response to two treatments where each subject in the study received both treatments.The data are given below,

	Group A	Group B
Subject 1	40	30
Subject 2	42	45
Subject 3	30	38
Subject 4	37	32
Subject 5	23	28

Now, we analyze the information using a paired t-test .

Null and Alternative Hypotheses:

vs

where, Population mean for Group A and   Population mean for Group B

define, Zi = Xi - Yi   where, Xi = ith observation for Group A and   Yi  = ith observation for Group B

and

and

and

    

For this test our appropriate test statistic given by,

Here, n=number of observation=5 So we get under

T is called " paird-t-statistic "

We perform this test in statistical software Minitab.

Steps For Minitab:-

1) Open Mibitab -> Click on " Stat " -> Click on " Basic Statistics " -> Click on " Paired t "

2) Check each sample is in a columns and select " Group A " as sample 1 and select " Group B " as sample 2

3) Click on " Options " -> Enter " Confidence level " 95% -> Enter " Hypothesized Difference " 0.0 -> Check " Alternative Hypothesis " Difference Hypothesized Difference "

4) Click on OK -> And you get the Result.

Output From Minitaab:-

Paired T-Test and CI: Group A, Group B

Paired T for Group A - Group B

N Mean StDev SE Mean
Group A 5 34.40 7.83 3.50
Group B 5 34.60 6.91 3.09
Difference 5 -0.20 7.46 3.34

95% CI for mean difference: (-9.47, 9.07)
T-Test of mean difference = 0 (vs ≠ 0): T-Value = -0.06 P-Value = 0.955

a) So from the minitab output we get,

Value of the test statistic = T-Value = -0.06

b) Rejection Rule based on p-value of the test: We reject the null hypothesis at level if, [We take ]

P-value

Here, from the minitab output we get,  p-value = 0.955

So we get , p-value = 0.955 >

Result: We accept the null hypothesis.

Conclusion: At level we conclude that there is no difference in mean response to the two treatments. And 95% confidence interval for mean difference is (-9.47, 9.07) [ from minitab outout ]