91Ó°ÊÓ

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 \(m \times {\hat p_1},m \times {\hat q_1},n \times {\hat p_1} ,and n \times {\hat q_1} are atleast 10\)

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.