Chapter 9, Problem 28E

Section 1:

To determine

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

Answer to Problem 28E

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:

$\begin{array}{c}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{183}{231}\\ =0.7922\end{array}$

And,

$\begin{array}{c}{\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{154}{732}\\ =0.2104\end{array}$

Step 1: The null hypothesis $\left(H_0\right)$ and alternate hypothesis $\left(H_1\right)$ needs 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:

$\begin{array}{c}\widehat{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{183+154}{231+732}\\ =\frac{337}{963}\\ =0.35\end{array}$

The test statistic is calculated as:

$\begin{array}{c}{z}_0=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{\widehat{p}\left(1-\widehat{p}\right)}\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}\\ =\frac{0.7922-0.2104}{\sqrt{0.35\times \left(1-0.35\right)}\sqrt{\frac{1}{231}+\frac{1}{732}}}\\ =16.163\end{array}$

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

Section 2:

To determine

To test: The relationship between the two variables.

Answer to Problem 28E

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: The variables ‘Being harassed online’ and ‘Being harassed in person’ are independent.

Alternative hypothesis: The variables ‘Being harassed online’ and ‘Being harassed in person’ are dependent.

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.

Step 5: Press ok two-times.

The desired test statistic value and p-value corresponding to one degree of freedom are obtained as ${\chi }_0^2=261.295$ and 0.000 respectively.

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 level of significance. It can be said that the variables ‘Harassed online’ and ‘Harassed in person’ are dependent on each other.

Section 3:

To determine

To explain: The comparison of results obtained above.

Answer to Problem 28E

Solution: The square of the test statistic $\left(z_0\right)$ obtained in Section 1 is almost equivalent to the test statistic $\left({\chi }_0^2\right)$ obtained in Section 2.

Explanation of Solution

The test statistic in Section 1 and Section 2 are calculated as $z_0=16.163$ and ${\chi }_0^2=261.295$ respectively. Now,

$\begin{array}{c}{\left(z_0\right)}^2={\left(16.163\right)}^2\\ =261.24\\ \approx {\chi }_0^2\end{array}$

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 $\left({\chi }_0^2\right)$.

Section 4:

To determine

To explain: The differences in the results for boys and girls.

Answer to Problem 28E

Solution: The results obtained for boys are equivalent to the results obtained for the girls.

Explanation of Solution

The z- test conducted for testing the differences in two population proportions discarded the null hypothesis that the proportions are equal for both boys and girls when ‘Harassed online’ was treated as the explanatory variable.

The chi-square test conducted for testing the relationship between two variables also discarded the null hypothesis that the two categorical variables ‘Harassed online’ and ‘Harassed in person’ are independent.

Therefore, it can be concluded that there is no difference in the results obtained for both boys and girls.