A postal researcher wanted to test the theory that a higher response rate is achieved when a postal questionnaire is sent out with a personalized covering letter (A) than when the covering letter is impersonal (B).
Two random samples of 100 people were selected. When the questionnaires were dispatched, the first sample-received letter A and the second received letter B. The response rates to letter A were 70% and to letter B, 55%.
a) Do these results provide evidence in support of the theory?
b) Explain the reasoning behind the test.
Call the response rate for each sample pA and pB respectively and the number in each sample nA and nB.
Please utilize and see the Student's t-distribution below. In using the t-table data, assume the number of data  corresponds to large distribution of (Infinity) in the table [see next pages, the t-table data].
Also please notice that this is a case of 2 independent samples from 2 populations. We cannot assume that the variance of the 2 populations are the same [not known], so we should utilize t-distribution formula and include both variances.
Also, remember that for proportions the equations are similar to the mean differences except proportions are replaced for means.
The final formula for student's t-test of 2 independent groups with unequal population variances [2 independent samples from 2 populations]
(X1-bar − X2-bar) − (U (X1-bar - X2bar) )
S (X1-bar - X2-bar)
Sample means differences − the 2 population means difference
Standard error [SE.]of the difference between the sample means
[Also, this is called standard deviation of the
difference between the sample means]
In t-test for proportion of these 2 samples, we use the same formula, but with proportions, i.e.,
S.E (PA− PB )
S.E (PA− PB )
Where your S.E (PA− PB ) is =
Please proceed by setting your HO & H1 and carry the 2 proportion t-test as mentioned above.
See attachment for detailed description.
A Complete, Neat and Step-by-step Solution is provided in the attached Excel file.