Chapter 24, Problem 25E

(a)

To determine

To explain is there a significant difference in viewers’ abilities to remember brands advertised in shows with violent versus neutral content.

Answer to Problem 25E

Yes, there is a sufficient evidence that viewers’ abilities to remember brands advertised in shows with violent versus neutral content.

Explanation of Solution

It is given in the question that:

  $\begin{array}{l}{\overline{x}}_1=3.02\\ {\overline{x}}_2=4.65\\ n_1=101\\ n_2=103\\ s_1=1.61\\ s_2=1.62\end{array}$

Let us define the hypotheses for testing:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_1:{\mu }_1\ne {\mu }_2\end{array}$

Thus, the value of test statistics is as:

  $\begin{array}{l}t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{3.02-4.65}{\sqrt{\frac{{1.61}^2}{101}+\frac{{1.62}^2}{103}}}\\ =-7.21\end{array}$

And the value of degree of freedom is then:

  $\begin{array}{l}df=\frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})}{\frac{{(\raisebox{1ex}{$s_1^2$}\!\left/ \!\raisebox{-1ex}{$n_1$}\right.)}^2}{n_1-1}+\frac{{(\raisebox{1ex}{$s_2^2$}\!\left/ \!\raisebox{-1ex}{$n_2$}\right.)}^2}{n_2-1}}\\ =201>180\end{array}$

Thus, the P-value will be as:

  $P<0.01$

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

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

Thus, we conclude that there is a sufficient evidence that viewers’ abilities to remember brands advertised in shows with violent versus neutral content.

(b)

To determine

To find a $95\%$ confidence interval for the difference in mean number of brand names remembered between the groups watching shows with sexual content with those watching neutral content and interpret your interval.

Answer to Problem 25E

We are $95\%$ confidence that the mean number of brand names remembered in the group watching shows with sexual content in between $1.4519\text{ and 2}\text{.4081}$ lower than the mean number of brand names remembered in the group watching shows with neutral content.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=2.72\\ {\overline{x}}_2=4.65\\ n_1=106\\ n_2=103\\ s_1=1.85\\ s_2=1.62\\ \alpha =0.05\end{array}$

Thus, the degree of freedom will be as:

  $\begin{array}{l}df=\frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})}{\frac{{(\raisebox{1ex}{$s_1^2$}\!\left/ \!\raisebox{-1ex}{$n_1$}\right.)}^2}{n_1-1}+\frac{{(\raisebox{1ex}{$s_2^2$}\!\left/ \!\raisebox{-1ex}{$n_2$}\right.)}^2}{n_2-1}}\\ =204>180\end{array}$

Thus, the t -value will be:

  $t=1.973$

Therefore, the confidence interval will be:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(2.72-4.65)-1.973\times \sqrt{\frac{{1.85}^2}{106}+\frac{{1.62}^2}{103}}\\ =-2.4081\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(2.72-4.65)+1.973\times \sqrt{\frac{{1.85}^2}{106}+\frac{{1.62}^2}{103}}\\ =-1.4519\end{array}$

Thus, we conclude that we are $95\%$ confidence that the mean number of brand names remembered in the group watching shows with sexual content in between $1.4519\text{ and 2}\text{.4081}$ lower than the mean number of brand names remembered in the group watching shows with neutral content.