Question

A telephone sales solicitor, trying to decide between two alternative sales pitches, randomly alternated between them during a day of calls. Using approach A, 20% of 100 calls led to requests for the mailing of additional product information. For approach B in another 100 calls, only 14 % led to requests for the product information mailing. At the 0.05 significance level, can we conclude that the difference in results was due to chance? Construct the 95% confidence level for the difference between population proportions (π1 - π2). Identify and interpret the p-value for the test.

n1 100 n2 100 p 1 20% p 2 14%

Answer 1

The hypotheses are,

H0:

H1:

Sample proportions are,

p1 = 0.2 and p2 = 0.14

Pooled sample proportion, p = (n1 * p1 + n2 * p2) / (n1 + n2)

= (100 * 0.2 + 100 * 0.14) / (100 + 100) = 0.17

Standard error (SE) of sampling distribution difference between two proportions =

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }

= sqrt{ 0.17 * ( 1 - 0.17 ) * [ (1/100) + (1/100) ] }

= 0.0531225

Z value for 95% confidence interval is 1.96

Sample difference in proportions = p1 - p2 = 0.2 - 0.14 = 0.06

95% confidence interval for difference in proportions = (0.06 - 1.96 * 0.0531225, 0.06 + 1.96 * 0.0531225)

= (-0.0441201, 0.1641201)

Test statistic, z = (p1 - p2) / SE = (0.2 - 0.14) / 0.0531225 = 1.13

P(Z > 1.13) = 0.1292

For two-tail tests, P-value = 2 * 0.1292 = 0.2584

Since, p-value is greater than the 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no significant evidence of difference between population proportions.