Chapter 9, Problem 9.2.5RE

For each exercise, perform these steps. Assume that all variables are normally or approximately normally distributed.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

5. Teachers’ Salaries A random sample of 15 teachers from Rhode Island has an average salary of $35,270, with a standard deviation of $3256. A random sample of 30 teachers from New York has an average salary of $29,512, with a standard deviation of $1432. Is there a significant difference in teachers’ salaries between the two states? Use α = 0.02. Find the 98% confidence interval for the difference of the two means.

To determine

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

Answer to Problem 9.2.5RE

The claim is that “there is a significant difference in teachers’ salaries between the two states”.

The hypotheses are given below:

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

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

Explanation of Solution

Given info:

${\overline{x}}_1=\$35,270$, ${\overline{x}}_2=10.3$, $s_1=\$3,256$, $s_2=\$1,432$, $n_1=15$, and $n_2=30$.

Justification:

Here, there is a significant difference in teachers’ salaries between the two states is tested. Hence, the claim is that there is a significant difference in teachers’ salaries between the two states. 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.

The hypotheses are given below:

Null hypothesis:

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

Alternative hypothesis:

$H_1:{\mu }_1\ne {\mu }_2$

To determine

To find: The critical value $\alpha =0.02$.

Answer to Problem 9.2.5RE

The critical value at $\alpha =0.02$ is $\pm 2.624$.

Explanation of Solution

Calculation:

Here, the test is two tailed test.

Critical value:

Here, variances are not equal. Hence, the degrees of freedom is,

$\begin{array}{c}\text{df}=\text{Smallar of }{n}_1-1\text{ and }{n}_2-1\\ =15-1\\ =14\end{array}$

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 ‘t’ distribution.
• In Degrees of freedom, enter 14.
• Click the Shaded Area tab.
• Choose Probability value and Two Tail for the region of the curve to shade.
• Enter the Probability value as 0.02.
• Click OK.

Output using the MINITAB software is given below:

From the output, the critical value is $\pm 2.624$.

To determine

To find: The test value.

To find: The 98% confidence interval for the difference of the two means.

Answer to Problem 9.2.5RE

The test value is –0.9.

The 98% confidence interval for the difference of the two means is $\underset{\_}{3,483<{\mu }_1-{\mu }_2<8,033}$.

Explanation of Solution

Calculation:

Test statistic:

Software Procedure:

Step by step procedure to obtain test statistic using the MINITAB software:

• Choose Stat > Basic Statistics > 2-Sample t.
• Choose Summarized data.
• In first, enter Sample size as15, Mean as 35,270, Standard deviation as 3,256.
• In second, enter Sample size as30, Mean as 29,512, Standard deviation as 1,432.
• Choose Options.
• In Confidence level, enter 98.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the test value is 6.54 and the 98% confidence interval for the difference of the two means is $\underset{\_}{3,483<{\mu }_1-{\mu }_2<8,033}$.

To determine

To make: The decision.

Answer to Problem 9.2.5RE

The decision is that, the null hypothesis isrejected.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to indicate the appropriate area and critical value using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 8.
• Click the Shaded Area tab.
• Choose Probability value and Two Tail for the region of the curve to shade.
• Enter the Probability value as 0.02.
• Enter 6.54 under show reference lines at X values
• Click OK.

Output using the MINITAB software is given below:

From the output, it can be observed that the test statistic value falls in the critical region. Therefore, the null hypothesis isrejected.

To determine

To summarize: The result.

Answer to Problem 9.2.5RE

The conclusion is that, there is enough evidence to support the claim that there is a significant difference in teachers’ salaries between the two states.

Explanation of Solution

Justification:

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that there is a significant difference in teachers’ salaries between the two states.