Chapter 9, Problem 27E

(a)

To determine

To test: The hypothesis for the significant difference between the two proportions.

Answer to Problem 27E

Solution: There is an adequate evidence to conclude that the two population proportions $p_1\text{and}{p}_2$ are different from each other.

Explanation of Solution

Calculation: Let $p_1$ and $p_2$ be the proportion of success in the first and second population which are estimated by the sample proportions ${\widehat{p}}_1$ and ${\widehat{p}}_2$, treating ‘Harassed Online’ as the explanatory variable.

The sample proportions, ${\widehat{p}}_1$ and ${\widehat{p}}_2$ are calculated as:

$\begin{array}{c}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{321}{361}\\ =0.8892\end{array}$

And,

$\begin{array}{c}{\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{200}{641}\\ =0.3120\end{array}$

Step 1: The null hypothesis $\left(H_0\right)$ and alternate hypothesis $\left(H_1\right)$ need to be defined. The two-tailed hypothesis is defined as follows:

$\begin{array}{c}{H}_0:p_1=p_2\\ H_1:p_1\ne p_2\end{array}$

Step 2: The level of significance is denoted by $\alpha =0.05$.

Step 3: Calculate the pooled estimate of $p$, that is $\widehat{p}$. The test statistic $z_0$ will be calculated and the critical values $z_{-\frac{\alpha }{2}}$ and $z_{\frac{\alpha }{2}}$, will be computed using the table provided in the book. The null hypothesis will be not be discarded if $z_{-\frac{\alpha }{2}}<z_0<z_{\frac{\alpha }{2}}$.

The pooled estimate is calculated as:

$\begin{array}{c}\widehat{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{321+200}{361+641}\\ =0.52\end{array}$

The test statistic is calculated as:

$\begin{array}{c}{z}_0=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{\widehat{p}\left(1-\widehat{p}\right)}\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}\\ =\frac{0.8892-0.3120}{\left(\sqrt{0.52\times \left(1-0.52\right)}\right)\left(\sqrt{\frac{1}{361}+\frac{1}{641}}\right)}\\ =17.56\end{array}$

The critical values for a two-tailed test for a given level of significance can be computed form the Table A published in the book.

The value corresponding to the left tail and right tail is denoted by $z_{-0.025}$ and $z_{0.025}$ respectively and the values are:

$z_{-0.025}=-1.96\text{ and }{z}_{0.025}=1.96$

Therefore, $z_0>z_{0.025}$.

Step 5: There is an adequate indication at 5% significance level to discard the null hypothesis.

Conclusion: Therefore, it can be concluded that the two population proportions are not equal.

(b)

To determine

To test: The relationship between the provided variables.

Answer to Problem 27E

Solution: There is an adequate indication to conclude that there is association between the two variables.

Explanation of Solution

Calculation: The hypothesis for the given problem has been defined in the following manner.

Null Hypothesis: There is no significant association between the being harassed online and in person.

Alternative Hypothesis: There is a significant association between being harassed online and in person.

The following procedure has to be followed in Minitab to obtain the test statistic value:

Step 1: Input the data of ‘Harassed in person’ and ‘Harassed online’, in column C1 and C2 respectively, in terms of ‘1’ and ‘2’, where 1 and 2 denotes ‘Yes’ and ‘No’ respectively, corresponding to both the variables. Input the frequencies for both the variables in column C3.

Step 2: Go to $\text{Stat}\to \text{Tables}\to \text{Cross Tabulation and Chi-Square}\text{.}$

Step 3: Drag the variable C1 and C2 in the space provided for ‘Rows’ and ‘Columns’ respectively.

Step 4: Click on Chi-Square and select Chi-Square Test and finally click on OK twice.

The desired test statistic value and p-value corresponding to one degree of freedom are obtained as ${\chi }_0^2=308.23$ and 0.0000.

Conclusion: Therefore, there is an adequate indication at $\alpha =0.05$ level of significance to discard the null hypothesis because the p-value is less than the significance level. Hence, it can be said that the variables ‘Harassed online’ and ‘Harassed in person’ are dependent on each other.

(c)

To determine

To explain: The comparison of results obtained using chi-square test and z test.

Answer to Problem 27E

Solution: The square of the test statistic $\left(z_0\right)$ obtained in part (a) is almost equivalent to the test statistic $\left({\chi }_0^2\right)$ obtained in part (b). Thus, both the tests provide same results.

Explanation of Solution

The test statistic obtained in above parts are calculated as $z_0=17.56$ and ${\chi }_0^2=308.23$ respectively. Now,

$\begin{array}{c}{\left(z_0\right)}^2={\left(17.56\right)}^2\\ =308.35\\ \approx {\chi }_0^2\end{array}$

Therefore, it can be decided that the square of the test statistic for testing the difference of two proportions using the normal model $\left(z_0\right)$ is approximately equal to the test statistic for testing the homogeneity of proportions using Chi-square test $\left({\chi }_0^2\right)$.

(d)

To determine

To explain: The reason for less number of girls reporting.

Answer to Problem 27E

Solution: The number of girls reporting harassment is low because girls are reluctant to share the details of such incidents.

Explanation of Solution

The number of girls who report cases of harassment remains on a lower side because such events ruin their reputation in the society. Due to some societal pressure, all the girls may not share the information.