In order to study the effect of the front cover on the response rate of mail sur

Question

In order to study the effect of the front cover on the response rate of mail surveys, a social scientist sent the same survey in 400 addresses. In 205 of them the cover was plain, but in the other ones the front cover had a skydiver. 100 of the surveys with plain cover were returned and 100 of the surveys with a skydiver cover were returned. (a) Obtain a 100(1 - alpha)% confidence interval for the difference in the response rates, where alpha = 1%. Does the confidence interval include 0? (b) Test the hypothesis that there is no difference in the response rates at alpha = 1% significance level. (c) Test the hypothesis that a non-plain front cover increases response rate at alpha = 1% significance level. (d) Repeat (a), (b), (c) with alpha = 10%.

Explanation / Answer

total address = 400

Plain cover = 205; Skydiver cover = 195

response rate pplain = p1 = (100)/205 = 0.4878

response rate pskydiver = p2 = ( 100)/ 195 = 0.5128

(a) Obtain a 99% confidence interval = (p2 - p1) +- Z/2 * sqrt [  p1 (1-p1)/n1 + p2 (1-p2)/n2]

= (0.5128 - 0.4878) +- 2.575 * sqrt [ 0.5128 * 0.4872 /195 + 0.4878 *0.5122/205]

= 0.025 +- 2.575 * 0.05 = (-0.10375, 0.15375)

yes, the confidence interval include 0

(b) Null Hypothesis : H0 : There is no differrence in the response rate p1 = p2

Alternative Hypothesis : Ha : There is significant difference in the response rate . p1 p2

Test Statistic:

pooled estimate = (100 + 100)/ 400 = 0.5

Standard Error of estimate SEo = sqrt [ p* (1-p*) * (1/n1 + 1/n2 )] = sqrt [ 0.5 * 0.5 * (1/195 + 1/205)] =  0.05

Test Statistic

Z = ( p2 -p1)/ SE0 = (0.5128 - 0.4878)/ 0.05 = 0.5

and Zcrtical = 2.575

so Z < Zcritical so we cannot reject the null hypothesis.

(c) Here there is nothing new but it will change the value of critical Z = 1.645 still it is greater than Z value of difference = 0.5 . so we can not conclude that non- plain front cover response rate increases.

(d)

Obtain a 90% confidence interval = (p2 - p1) +- Z/2 * sqrt [  p1 (1-p1)/n1 + p2 (1-p2)/n2]

= (0.5128 - 0.4878) +- 1.645 * sqrt [ 0.5128 * 0.4872 /195 + 0.4878 *0.5122/205]

= 0.025 +- 1.645 * 0.05 = (-0.0.5725, 0.10725)

yes, the confidence interval include 0

(b) Null Hypothesis : H0 : There is no differrence in the response rate p1 = p2

Alternative Hypothesis : Ha : There is significant difference in the response rate . p1 p2

Test Statistic:

pooled estimate = (100 + 100)/ 400 = 0.5

Standard Error of estimate SEo = sqrt [ p* (1-p*) * (1/n1 + 1/n2 )] = sqrt [ 0.5 * 0.5 * (1/195 + 1/205)] =  0.05

Test Statistic

Z = ( p2 -p1)/ SE0 = (0.5128 - 0.4878)/ 0.05 = 0.5

and Zcrtical =1.645

so Z < Zcritical so we cannot reject the null hypothesis.

Here there is nothing new in this case but it will change the value of critical Z = 1.28 still it is greater than Z value of difference = 0.5 . so we can not conclude that non- plain front cover response rate increases.

so even decreasing the confidence interva we will not be able to tell that a non- plain front cover increases resonse rate.

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