Chapter 21, Problem 21.25E

(a)

To determine

To state: The hypotheses in terms of the population means for men $\left({\mu }_M\right)$ and women $\left({\mu }_F\right)$.

Answer to Problem 21.25E

The null hypothesis is ${\mu }_F={\mu }_M$.

The alternative hypothesis is ${\mu }_F>{\mu }_M$.

Explanation of Solution

Given info:

The data shows the average number of words spoken per day by men and women.

Explanation:

State the test hypotheses.

Null hypothesis:

$H_0:{\mu }_F={\mu }_M$

Alternative hypothesis:

$H_a:{\mu }_F>{\mu }_M$

(b)

To determine

To find: The two-sample t statistic.

Answer to Problem 21.25E

The two-sample t statistic for study 1 is –0.248.

The two-sample t statistic for study 2 is 1.507.

Explanation of Solution

Calculation:

From the given information, $n_M=56$, $n_F=56$, ${\overline{x}}_M=16,569$, ${\overline{x}}_F=16,177$, $s_M=9,108$, and $s_F=7,520$.

Test statistic for study 1:

The value of t statistic is obtained below:

$\begin{array}{c}t=\frac{\left({\overline{x}}_F-{\overline{x}}_M\right)}{\sqrt{\frac{s_F^2}{n_F}+\frac{s_M^2}{n_M}}}\\ =\frac{\left(16,177-16,569\right)}{\sqrt{\frac{{\left(7,520\right)}^2}{56}+\frac{{\left(9,108\right)}^2}{56}}}\\ =\frac{-392}{\text{1,578}\text{.35}}\\ =-0.248\end{array}$

Thus, the two-sample t statistic for study 1 is –0.248.

From the given information, $n_M=20$, $n_F=27$, ${\overline{x}}_M=12,867$, ${\overline{x}}_F=16,496$, $s_M=8,343$, and $s_F=7,914$.

Test statistic for study 2:

The value of t statistic is obtained below:

$\begin{array}{c}t=\frac{\left({\overline{x}}_F-{\overline{x}}_M\right)}{\sqrt{\frac{s_F^2}{n_F}+\frac{s_M^2}{n_M}}}\\ =\frac{\left(16,496-12,867\right)}{\sqrt{\frac{{\left(7,914\right)}^2}{27}+\frac{{\left(8,343\right)}^2}{20}}}\\ =\frac{\text{3,629}}{\text{2,408}\text{.31}}\\ =1.507\end{array}$

Thus, the two-sample t statistic for study 2 is 1.507.

(c)

To determine

To find: The degrees of freedom by using Option 2.

Answer to Problem 21.25E

The degree of freedom for study 1 is 55.

The degree of freedom for study 2 is 19.

Explanation of Solution

Calculation:

For study 1,

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

$\begin{array}{c}\text{df}=\text{Smaller of }{n}_1-1\text{ and }{n}_2-1\\ =56-1\\ =55\end{array}$

Thus, the degrees of freedom for study 1 is 55.

For study 2,

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

$\begin{array}{c}\text{df}=\text{Smaller of }{n}_1-1\text{ and }{n}_2-1\\ =20-1\\ =19\end{array}$

Thus, the degrees of freedom for study 2 is 19.

(d)

To determine

To find: The P-value.

Answer to Problem 21.25E

For study 1, the P-value for one-sided test is $P\text{-value}>0.25$.

For study 2, the P-value for one-sided test is $0.05<P\text{-value}<0.10$.

Explanation of Solution

Calculation:

For study 1,

The degree of freedom for study 1 is 55, which is not in the degrees of freedom column.

Thus, choose the next row with degrees of freedom smaller than 55. That is, the next row with smaller degrees of freedom would be 50.

From the table of “t Distribution critical values”, the t-value $\left|-0.248\right|$ is not located in the table. Hence, the P-value for one-sided test is $P\text{-value}>0.25$.

For study 2,

The degree of freedom for study 2 is 19.

From the table of “t Distribution critical values”, the t-value 1.507 is located between 1.328 and 1.729 with corresponding level 0.10 and 0.05. Hence, the P-value for one-sided test is $0.05<P\text{-value}<0.10$.

(e)

To determine

To identify: The conclusion.

Answer to Problem 21.25E

For study 1,

The conclusion is that, there is no evidence that the women talk more than men.

For study 2,

For study 2,

The conclusion is that, there is evidence that the women talk more than men.

Explanation of Solution

Conclusion:

Study 1:

Here, the P-value is very high.

Therefore, by the rejection rule, it can be concluded that there is no evidence to reject $H_0$. Thus, there is no evidence that the women talk more than men. That is, the study 1 does not support the idea.

Study 2:

Here, the P-value is $0.05<P\text{-value}<0.10$, which is less support about the null hypothesis

Therefore, by the rejection rule, it can be concluded that there is evidence to reject $H_0$. Thus, there is evidence that the women talk more than men. That is, the study 2 supports the idea.