Chapter 9, Problem 14CQ

a.

To determine

To identify: The claim and state $H_0$ and $H_1$

Answer to Problem 14CQ

The claim is that “there is a difference in the cholesterol levels between the two groups”.

Null hypothesis: $H_0:{\mu }_1={\mu }_2$

Alternative hypothesis: $H_1:{\mu }_1\ne {\mu }_2$ (claim)

Explanation of Solution

Given info:

Group 1 (25 and 40): ${\overline{X}}_1=\text{223}$, ${\sigma }_1=\text{6}$ and $n_1=30$

Group 1 (41 and 56): ${\overline{X}}_2=\text{229}$, ${\sigma }_2=\text{6}$ and $n_2=25$

Justification:

Here, the claim is that there is a difference in the cholesterol levels between the two groups. This can be written as ${\mu }_1\ne {\mu }_2$. The complement of the claim is ${\mu }_1={\mu }_2$. In the given experiment, the alternative hypothesis indicates the claim.

b.

To determine

To find: The critical value and rejection region for a two-tailed test with $\alpha =0.10$.

Answer to Problem 14CQ

The critical value at $\alpha =0.01$ is ±2.58.

The rejection region for a two-tailed test with $\alpha =0.01$ is, $z<-z_0\left(=-2.58\right)$, $z>+z_0\left(=+2.58\right)$.

Explanation of Solution

Calculation:

Here, the test is two tailed test.

From the Standard Normal Table, the critical value at $\alpha =0.01$ is ±2.58.

The rejection region for a two-tailed test with $\alpha =0.01$ is, $z<-z_0\left(=-2.58\right)$, $z>+z_0\left(=+2.58\right)$.

c.

To determine

To find: The standardized test statistic z.

To find: The 99% confidence interval for the difference of the two means.

Answer to Problem 14CQ

The standardized test statistic z is –3.7.

The 99% confidence interval for the difference of the two means is, $-10.2<{\mu }_1-{\mu }_2<-1.8$.

Explanation of Solution

Calculation:

Test statistic:

$\begin{array}{c}z=\frac{\left({\overline{X}}_1-{\overline{X}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_1}}}\\ =\frac{223-229}{\sqrt{\frac{{\left(6\right)}^2}{30}+\frac{{\left(6\right)}^2}{25}}}\\ =\frac{-\text{6}}{1.62}\\ =-3.7\end{array}$

Confidence interval:

$\begin{array}{c}\left({\overline{X}}_1-{\overline{X}}_2\right)\pm z_{\frac{\alpha }{2}}\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_1}}=\left(223-229\right)\pm 2.58\sqrt{\frac{{\left(6\right)}^2}{30}+\frac{{\left(6\right)}^2}{25}}\\ =-6\pm 2.58\left(1.62\right)\\ =\left(-6-\text{4}\text{.2,}-6+\text{4}\text{.2}\right)\\ =\left(-10.2,-1.8\right)\end{array}$

d.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of $\alpha =0.01$.

Answer to Problem 14CQ

The decision is “reject the null hypothesis”.

Explanation of Solution

Calculation:

Decision rule:

• If $z<z_0\left(=-2.58\right)$, then reject the null hypothesis.
• If $z>z_0\left(=+2.58\right)$, then reject the null hypothesis.

Here, the test statistic is greater than the critical value. That is, $z\left(=-3.7\right)<z_0\left(=-2.58\right)$.

Thus, the decision is “reject the null hypothesis”.

e.

To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 14CQ

The conclusion is that, there is enough evidence to support the claim that there is a difference in mean earnings between male and female college graduates.

Explanation of Solution

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that there is a difference in the cholesterol levels between the two groupsat 1% level of significance.