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# Statistics - Hypothesis test exercise

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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.

Hint:
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 &#8722; X2-bar) &#8722; (U (X1-bar - X2bar) )
S (X1-bar - X2-bar)

Sample means differences &#8722; 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.,

PA&#8722; PB
S.E (PA&#8722; PB )

PA&#8722; PB
S.E (PA&#8722; PB )

Where your S.E (PA&#8722; PB ) is =

Please proceed by setting your HO & H1 and carry the 2 proportion t-test as mentioned above.

See attachment for detailed description.