Question

Compare gain scores (Test 1 & 2) between group 1 and the group 2 protocols with independent t-tests. Significance for the study will be considered as p≤0.05 (noting that p-values near the cutoff will be discussed relative to clinical relevance). I've highlighted the applicable gain scores below.

This will be two separate t-tests, correct? One for Test 1 and one for Test 2. However, do I then compare the two t-tests to each other??

I also have negative gain scores, this is possible, correct??

Currently lost on the next step or how to perform the t-tests.

Group 1 Group 2 Test 1 Difference/Gain Score Test 2 Difference/Gain Sco Test 1 Difference/Gain Score Test 2 Difference/Gain Score 9 19.685 2 7.62 3 22.86 4 -2.54 0 10.16 5 5.08 6 21.273 0 -5.08 2 8.255 2 7.62 0 3.81 4 2.54 5 11.43 9 17.78 6 -2.54 2 10.16 4 5.08 2 0 5 10.16 0 5.08 0 0 0 13.97 4 1.27 2 -1.27 2 9.208 4 7.62 0 0 6 -22.86 0 16.51 4 -12.7

Answer 1

The question is asking you to compare Gain score for Test 1 and Test 2.

This means you need to perform indeoendent t test to check the below Hypotheses. There is no need to perform 2 separate t tests

Null Hypotheses, H_o : mean gain score of test 1 = mean gain score of test 2

H_o :  ₁ = ₂

Alternate Hypotheses,  H_a : mean gain score of test 1 mean gain score of test 2

H_a : ₁₂

Given Data set:

T1	T2
9	19.685
3	22.86
0	10.16
6	21.273
2	8.255
0	3.81
5	11.43
6	-2.54
4	5.08
5	10.16
0	0
4	1.27
2	9.208
0	0
0	16.51
2	7.62
4	-2.54
5	5.08
0	-5.08
2	7.62
4	2.54
9	17.78
2	10.16
2	0
0	5.08
0	13.97
2	-1.27
4	7.62
6	-22.86
4	-12.7

we are given p 0.05. So,

Alpha() = 0.05

Formula to calculate t-statistic is:

Let's go one by one:

For treatment 1:

Total Observations, N1 = 30

Mean, X1 = 3.066

Variance, s₁² = 6.754

For Treatment 2,

Total Observations, N2 = 30

Mean, X2 = 5.672

Variance, s₂² = 95.611

Putting all the value isn above formula, we get

t = -1.41

For our case, N1 = N2 = 30

So, degrees of freedom, (N₁ - 1) + (N₂ - 1) = 58

From p-value calculator, for 58 degrees of freedom,

p = 0.078

Now, p > 0.05, so we have to reject the alternate Hypotheses.

Null Hypotheses is accepted

Note: Since you mentioned, "noting that p-values near the cutoff will be discussed relative to clinical relevance".

So, 0.078 is very close to 0.05. You can consider it as per your requirement.