Chapter 20, Problem 20.28E

(a)

To determine

To explain is there good evidence that one approach is more likely to be effective than the other and state the null and alternative hypotheses, obtain test statistics and P-value and conclude in context.

Answer to Problem 20.28E

Yes, there is good evidence that one approach is more likely to be effective than the other and hypotheses are:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

And,

  $\begin{array}{l}z=-2.8163\\ p=0.0048\end{array}$

Explanation of Solution

In the question, it is given that they found that $\text{412 of the 549}$ respondents chosen to ignore online harassment thought it was an effective way and out of the $\text{368 the 305}$ respondents thought that chosen to respond is effective method. Thus, we want to know is there good evidence that one approach is more likely to be effective than the other. Then the null and alternative hypotheses is defined as:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The proportions are as:

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{412}{549}\\ =0.750455\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{305}{368}\\ =0.828804\end{array}$

Thus, to do the mechanics of a hypothesis test for equality of population proportions we will use the calculator $\text{TI}-89$ . Then select $6:2-$ PropZTest from the STAT Tests menu. Enter the observed counts and sample sizes. Indicate what kind of test you want: one-tail upper tail, lower tail, or two-tail. And specify whether results should simply be calculated or displayed with the area corresponding to the P-value of the test shaded. Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}z=-2.8163\\ p=0.0048\end{array}$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we have sufficient evidence to conclude that one approach is more likely to be effective than the other.

(b)

To determine

To explain how large is the difference and obtain and interpret in context a $\text{95\%}$ confidence interval for the difference between the proportions of American adults who find ignoring online harassment effective and those who find responding to it effective.

Answer to Problem 20.28E

We are $\text{95\%}$ confident that the difference between the proportions of American adults who find ignoring online harassment effective and those who find responding to it effective is between $(-0.13288,-0.023819)$ .

Explanation of Solution

In the question, it is given that they found that $\text{412 of the 549}$ respondents chosen to ignore online harassment thought it was an effective way and out of the $\text{368 the 305}$ respondents thought that chosen to respond is effective method. Thus, we want to know is there good evidence that one approach is more likely to be effective than the other. The proportions are as:

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{412}{549}\\ =0.750455\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{305}{368}\\ =0.828804\end{array}$

And the differences are:

  $\begin{array}{l}{\widehat{p}}_1-{\widehat{p}}_2=0.750455-0.828804\\ =-0.078349\end{array}$

Now, a $\text{95\%}$ confidence interval for the difference between the proportions of American adults who find ignoring online harassment effective and those who find responding to it effective is calculated as:

  $\begin{array}{l}SE=\sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1+2}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2+2}}\\ =\sqrt{\frac{0.750455(1-0.750455)}{549}+\frac{0.828804(1-0.828804)}{368}}\\ =0.02782\\ z_{\alpha /2}=z_{0.025}=1.96\\ \text{Confidence interval}=(0.750455-0.828804)\pm 1.96\times 0.02782\\ =-0.078349\pm 0.05453\\ =(-0.13288,-0.023819)\end{array}$

Thus, we are $\text{95\%}$ confident that the difference between the proportions of American adults who find ignoring online harassment effective and those who find responding to it effective is between $(-0.13288,-0.023819)$ .

(c)

To determine

To explain how effective is that best method and obtain and interpret in context a $\text{95\%}$ confidence interval for the proportions of American adults who find the best method effective.

Answer to Problem 20.28E

We are $\text{95\%}$ confident that the proportions of American adults who find the best method effective is between $(0.790318,0.86729)$ .

Explanation of Solution

In the question, it is given that they found that $\text{412 of the 549}$ respondents chosen to ignore online harassment thought it was an effective way and out of the $\text{368 the 305}$ respondents thought that chosen to respond is effective method. Thus, we want to know is there good evidence that one approach is more likely to be effective than the other. Here the best way is to respond by part (a) so, the proportion is as:

  $\begin{array}{l}{\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{305}{368}\\ =0.828804\end{array}$

And the $\text{95\%}$ confidence interval for the proportions of American adults who find the best method effective is calculated as:

  $\begin{array}{l}SE=\sqrt{\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2+2}}\\ =\sqrt{\frac{0.828804(1-0.828804)}{368}}\\ =0.019636\\ z_{\alpha /2}=z_{0.025}=1.96\\ \text{Confidence interval}=0.828804\pm 1.96\times 0.019636\\ =0.828804\pm 0.038486\\ =(0.790318,0.86729)\end{array}$

Thus, we are $\text{95\%}$ confident that the proportions of American adults who find the best method effective is between $(0.790318,0.86729)$ .