Question

A team of researchers used the Survey of Consumer Attitudes to investigate whether incentives would improve response rates on telephone surveys. A national sample of 735 households were selected for the survey. 368 households were randomly assigned to receive a monetary incentive and of these 286 responded to the survey. The other 367 households were assigned to only receive an advance letter and of these 245 responded to the survey. Identify the observational units in this study. Describe in words the parameters of interest in this study assigning appropriate symbols. State the null and alternative hypotheses for testing whether incentives improve the response rates on telephone surveys using above symbols. Compute the conditional proportions to evaluate if monetary incentives improve the response rate.  Compute the relative risk of responding to a telephone survey, comparing those households who received a monetary reward to those who only received an advance letter. If the above hypothesis test turns out to have a very small p-value, which one of the following is the best conclusion to be taken Conclude that there is no difference in survey response rates between households who received a monetary reward to those received an advance letter. Conclude that there is a difference in survey response rates between households who received a monetary reward to those received an advance letter. Conclude that there is a higher survey response rate from households who received a monetary reward compared to those received an advance letter. None of the above.

Answer 1

1) p1 = 286/368 = 0.777

    p2 = 245/367 = 0.668

The Pooled sample proportion (P) = (p1 * n1 + p2 * n2)/(n1 + n2)

                                                       = (0.777 * 368 + 0.668 * 367)/(368 + 367) = 0.723

SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))

      = sqrt(0.723 * (1 - 0.723) * (1/368 + 1/367))

      = 0.033

The test statstic z = (p1 - p2)/SE

                             = (0.777 - 0.668)/0.033 = 3.30

P-value = P(Z > 3.30)

             = 1 - P(Z < 3.30)

             = 1 - 0.9995

             = 0.0005

2) As the P-value is less than 0.05, so the null hypothesis is rejected.

Option - a) We have very strong evidence against the null hypothesis.

3) The test statstic z = (p1 - p2)/SE

                             = (0.777 - 0.668)/0.033 = 3.30

4) At 5% significance level the critical value is z^* = 1.96

As the test statistic value is greater than the critical value (3.3 > 1.96), so the null hypothesis is rejected.

So its TRUE.