Chapter 9.1, Problem 6E

a.

To determine

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

Answer to Problem 6E

The claim is that “there is a difference in means of the salaries”.

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

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

Explanation of Solution

The given information is the annual average salary of the California teachers is $64,421 and for New York teachers is $62,332. The random sample of 45 teachers in the two places and their average salaries and the standard deviations is,

For California, ${\overline{X}}_1=\text{\$64,510}$, ${\sigma }_1=\text{8,200}$

For New York, ${\overline{X}}_2=\text{\$62,900}$, ${\sigma }_2=\text{7,800}$.

Here, the claim is that “there is a difference in means of the salaries”. 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 values at $\alpha =0.10$.

Answer to Problem 6E

The critical value at $\alpha =0.10$ is $\underset{\_}{\pm 1.645}$.

Explanation of Solution

Calculation:

Here, the test is two tailed test.

Critical value:

Software Procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘normal’ distribution.
• Click the Shaded Area tab.
• Choose Probability value and Two Tail for the region of the curve to shade.
• Enter the Probability value as 0.10.
• Click OK.

Output using the MINITAB software is given below:

From the output, the critical value is $\underset{\_}{\pm 1.645}$.

c.

To determine

To find: The test statistic.

Answer to Problem 6E

The test statistic z is –0.266.

Explanation of Solution

Calculation:

The test statistic is,

$\begin{array}{c}z=\frac{\left({\overline{X}}_1-{\overline{X}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_1}}}\\ =\frac{\left(64,510-62,900\right)-\left(64,421-62,332\right)}{\sqrt{\frac{{\left(8,200\right)}^2}{45}+\frac{{\left(7,800\right)}^2}{45}}}\\ =\frac{\left(1,610\right)-\left(2,089\right)}{\sqrt{1,494,222.222+1,352,000}}\\ =\frac{-449}{1,687.08}\\ =-0.266\end{array}$

Thus, the test statistic is –0.266.

d.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of $\alpha =0.10$.

Answer to Problem 6E

The decision is “reject the null hypothesis”.

Explanation of Solution

Calculation:

Decision:

If $\left|z\right|>z_{\frac{\alpha }{2}}\Rightarrow z>z_{\frac{\alpha }{2}}\text{and}z<-z_{\frac{\alpha }{2}}$, reject the null hypothesis H₀

Conclusion:

Here test statistic value is greater than the critical value

That is $-0.266\left(=z\right)>-1.645\left(=-z_{\frac{\alpha }{2}}\right)$

By rejection rule the null hypothesis is not rejected.

e.

To determine

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

Answer to Problem 6E

The conclusion is that there is enough evidence to support the claim that there is a difference in means of the salaries.

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 means of the salariesat 10% level of significance.