Question

Is the response rate for questionnaires affected by including some sort of incentive to respond along with the questionnaire? In one experiment, 110 questionnaires with no incentive resulted in 75 being returned, whereas 98 questionnaires that included a chance to win a lottery yielded 66 responses ("'Charities, No; Lotteries, No; Cash, Yes," Public Opinion Quarterly, 1996: 542–562). Does this data suggest that including an incentive increases the likelihood of a response? State and test the relevant hypotheses at significance level . 10 .

Short answer

Do not reject null hypothesis, which means an incentive does not mean that there are going to be more responses.

Step-by-step solution

To find an incentive increases the likelihood of a response

The hypotheses of interest are \({H_0}:{p_1} - {p_2} = 0 versus {H_a}:{p_1} - {p_2} < 0 where {p_1}\) is the true proportion of returned questionnaires that include no incentive, and \({p_2}\) is the true proportion of returned questionnaires that includes an incentive.

The sample size is big enough, thus:

A Large-Sample Test Procedure:

The two proportion z statistic - Consider test in which\({H_0}:{p_1} - {p_2} = 0\). The test statistic value for the large samples is

\(z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {\hat p\hat q \times \left( {\frac{1}{m} + \frac{1}{n}} \right)} }}\)

where

\(\hat p = \frac{m}{{m + n}} \times {\hat p_1} + \frac{n}{{m + n}} \times {\hat p_2}\)

Depending on alternative hypothesis, the P value can be determined as the corresponding area under the standard normal curve. The test should be used when \(m \times {\hat p_1},m \times {\hat q_1},n \times {\hat p_1} ,and n \times {\hat q_1} are atleast 10\)

To State and test the relevant hypotheses at significance level . 10 .

The estimates are

\(\begin{array}{l}{{\hat p}_1} = \frac{{75}}{{110}} = 0.682\\{{\hat p}_2} = \frac{{66}}{{98}} = 0.673\end{array}\)

Notice that in z value the numerator

\({\hat p_1} - {\hat p_2} > 0\)

which indicates that

\(z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {\hat p\hat q \times \left( {\frac{1}{m} + \frac{1}{n}} \right)} }} > 0\)

The test is lower tailed test, and the z value is positive, the P value has to be bigger than 0.5 because \(P(Z < z) > 0.5\)for all z>0.

For any reasonable significance level do not reject null hypothesis because \(P > 0.5 > \alpha \)

Thus, including an incentive does not mean that there are going to be more responses.