Chapter 9, Problem 9.60SE

i.

To determine

To identify: The provided data is sufficient to indicate that there is difference in average number of minutes spent doing household chores for men and women.

Answer to Problem 9.60SE

  $z\left(34.14\right)<z_{\alpha /2}\left(2.58\right)$so $H_0$can be rejected and there is evidence for indicating that the difference between the average no of minutes spent doing household for men and women.

Explanation of Solution

Given:

1. No of sample size, $n_1=795$
2. No of sample size, $n_2=1136$
3. ${\overline{x}}_1=$ Sample mean $=72$
4. ${\overline{x}}_2=$ Sample mean $=54$
5. $s_1=$ Sample standard deviation $=10.4$
6. $s_2=$ Sample standard deviation $=12.7$
7. $\alpha =$ significance level $=1\%=0.01$

Formula Used:

Test statistic:

  $z=\frac{\left(\overline{x_1}-\overline{x_2}\right)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Calculation:

An advertising age white paper reported the no of minutes spent doing household chores for women at the time when compared to men and women.

Null hypothesis,

  $H_0:$ There is no significance difference between the average no of minutes spent doing household chores for men and women.

  $H_0:\left({\mu }_1-{\mu }_2\right)=D_0$

Alternative hypothesis,

  $H_a:$ There is a significance difference between the average no of minutes spent doing household chores for men and women.

  $H_a:\left({\mu }_1-{\mu }_2\right)\ne D_0$

Now, for finding out the test statistic:

The test statistic used to test the hypothesis is,

  $\begin{array}{l}z=\frac{\left(\overline{x_1}-\overline{x_2}\right)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{\left(72-54\right)-0}{\sqrt{\frac{{\left(10.4\right)}^2}{795}+\frac{{\left(12.7\right)}^2}{1136}}}\\ =34.14\end{array}$

Thus, test statistic is $34.14$ .

Critical value:

Since, level of significance is $\alpha =0.01$ and the test is two-tailed test, rejection region is determined with tail area equals to the $\alpha =0.01$ .

  $H_0$ can be rejected if $z\left(34.14\right)<z_{\alpha /2}\left(2.58\right)$

Conclusion:

Thus, $z\left(34.14\right)<z_{\alpha /2}\left(2.58\right)$ so $H_0$ can be rejected and there is evidence for indicating that the difference between the average no of minutes spent doing household for men and women.

ii.

To determine

To identify: The provided data is sufficient to indicate that there is difference in average number of minutes spent doing household chores for millennial men and classified as a boomers.

Answer to Problem 9.60SE

  $z\left(2.58\right)<z_{\alpha /2}\left(-2.58\right)$so $H_0$can be rejected and there is evidence for indicating that the difference between the average no of minutes spent doing household for millennial men and men is classified as boomers.

Explanation of Solution

Given:

1. No of sample size, $n_1=345$
2. No of sample size, $n_2=475$
3. ${\overline{x}}_1=$ Sample mean $=72$
4. ${\overline{x}}_2=$ Sample mean $=54$
5. $s_1=$ Sample standard deviation $=9.2$
6. $s_2=$ Sample standard deviation $=13.9$
7. $\alpha =$ significance level $=1\%=0.01$

Formula Used:

Test statistic:

  $z=\frac{\left(\overline{x_1}-\overline{x_2}\right)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Calculation:

An advertising age white paper reported the no of minutes spent doing household chores for women at the time when compared for millennial men and men is classified as boomers.

Null hypothesis,

  $H_0:$ There is no significance difference between the average no of minutes spent doing household chores for millennial men and men is classified as boomers.

  $H_0:\left({\mu }_1-{\mu }_2\right)=D_0$

Alternative hypothesis,

  $H_a:$ There is a significance difference between the average no of minutes spent doing household chores for millennial men and men is classified as boomers.

  $H_a:\left({\mu }_1-{\mu }_2\right)\ne D_0$

Now, for finding out the test statistic:

The test statistic used to test the hypothesis is,

  $\begin{array}{l}z=\frac{\left(\overline{x_1}-\overline{x_2}\right)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{\left(72-54\right)-0}{\sqrt{\frac{{\left(9.2\right)}^2}{345}+\frac{{\left(13.9\right)}^2}{475}}}\\ =22.29\end{array}$

Thus, test statistic is $22.29$ .

Critical value:

Since, level of significance is $\alpha =0.01$ and the test is two-tailed test, rejection region is determined with tail area equals to the $\alpha =0.01$ by critical value approach.

  $H_0$ can be rejected if $z\left(2.58\right)<z_{\alpha /2}\left(-2.58\right)$

Conclusion:

Thus, $z\left(2.58\right)<z_{\alpha /2}\left(-2.58\right)$ so $H_0$ can be rejected and there is evidence for indicating that the difference between the average no of minutes spent doing household for millennial men and men is classified as boomers.

iii.

To determine

To identify: The provided data is sufficient to indicate that there is difference in average number of minutes spent doing household chores for men classified as a Xers and Boomers.

Answer to Problem 9.60SE

Thus, $z\left(0\right)<z_{\alpha /2}\left(2.58\right)$, so one cannot reject the null hypothesis.

So, there is no evidence to indicate that the difference between the average no of minutes spent doing household chores for men classified as Xers and boomers.

Explanation of Solution

Given:

1. No of sample size, $n_1=475$
2. No of sample size, $n_2=316$
3. ${\overline{x}}_1=$ Sample mean $=54$
4. ${\overline{x}}_2=$ Sample mean $=54$
5. $s_1=$ Sample standard deviation $=13.9$
6. $s_2=$ Sample standard deviation $=10.5$
7. $\alpha =$ significance level $=1\%=0.01$

Formula Used:

Test statistic:

  $z=\frac{\left(\overline{x_1}-\overline{x_2}\right)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Calculation:

An advertising age white paper reported the no of minutes spent doing household chores for women at the time when compared for men classified as a Xers and Boomers.

Null hypothesis,

  $H_0:$ There is no significance difference between the average no of minutes spent doing household chores for men classified as a Xers and Boomers.

  $H_0:\left({\mu }_1-{\mu }_2\right)=D_0$

Alternative hypothesis,

  $H_a:$ There is a significance difference between the average no of minutes spent doing household chores for men classified as a Xers and Boomers.

  $H_a:\left({\mu }_1-{\mu }_2\right)\ne D_0$

Now, for finding out the test statistic:

The test statistic used to test the hypothesis is,

  $\begin{array}{l}z=\frac{\left(\overline{x_1}-\overline{x_2}\right)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{\left(54-54\right)-0}{\sqrt{\frac{{\left(13.9\right)}^2}{475}+\frac{{\left(10.5\right)}^2}{316}}}\\ =0\end{array}$

Thus, test statistic is $0$ .

Critical value:

Since, level of significance is $\alpha =0.01$ and the test is two-tailed test, rejection region is determined with tail area equals to the $\alpha =0.01$ by critical value approach.

  $H_0$ can be rejected if $z>2.58$ or $z<-2.58$

So, $z\left(0\right)<z_{\alpha /2}\left(2.58\right)$ , so one cannot reject the null hypothesis.

So, there is no evidence to indicate that the difference between the average no of minutes spent doing household chores for men classified as Xers and boomers.

Conclusion:

Thus, $z\left(0\right)<z_{\alpha /2}\left(2.58\right)$ , so one cannot reject the null hypothesis.

So, there is no evidence to indicate that the difference between the average no of minutes spent doing household chores for men classified as Xers and boomers.

iv.

To determine

To identify: Practical conclusion that can be drawn from part (i), (ii), and (iii).

Answer to Problem 9.60SE

In part (i): Null hypothesis is rejected.

In part (ii): Null hypothesis is rejected.

In part (iii): Null hypothesis is not rejected in this case.

Explanation of Solution

From Part (i):

The calculated $z\left(34.14\right)>z_{\alpha /2}\left(2.58\right)$ , so null hypothesis will be rejected.

Hence, there is evidence to indicate that the difference between the average no of minutes spent doing household chores for men and women.

From Part (ii):

The calculated $z\left(22.29\right)>z_{\alpha /2}\left(2.58\right)$ , so null hypothesis will be rejected.

Hence, there is evidence to indicate that the difference between the average no of minutes spent doing household chores for millennial men and men classified as boomers.

From Part (iii):

The calculated $z\left(0\right)<z_{\alpha /2}\left(2.58\right)$ , so null hypothesis will not be rejected in this case.

Hence, there is no evidence to indicate that the difference between the average no of minutes spent doing household chores for men classified as Xers and men classified as Boomers.