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.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

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.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

To identify: The claim and state H0 and Ha .

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: H0:μ1=μ2

Alternative hypothesis: H1:μ1≠μ2 (claim)

Explanation of Solution

Given info:

x¯1=$35,270 , x¯2=10.3 , s1=$3,256 , s2=$1,432 , n1=15 , and n2=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 μ1≠μ2 . The complement of the claim is μ1=μ2 . In the given experiment, the alternative hypothesis indicates the claim.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

b.

Expert Solution

To determine

To find: The critical value α=0.02 .

Answer to Problem 9.2.5RE

The critical value at α=0.02 is ±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,

df=Smallar of n1−1 and n2−1=15−1=14

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 1

From the output, the critical value is ±2.624 .

c.

Expert Solution

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 3,483<μ1−μ2<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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 2

From the MINITAB output, the test value is 6.54 and the 98% confidence interval for the difference of the two means is 3,483<μ1−μ2<8,033_ .

d.

Expert Solution

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 3

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

e.

Expert Solution

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

What does the margin of error include? When a margin of error is reported for a survey, it includes a. random sampling error and other practical difficulties like undercoverage and non-response b. random sampling error, but not other practical difficulties like undercoverage and nonresponse c. practical difficulties like undercoverage and nonresponse, but not random smapling error d. none of the above is corret

solve part a on paper

solve on paper

Answer 2

Textbook Question

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.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

To identify: The claim and state H0 and Ha .

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: H0:μ1=μ2

Alternative hypothesis: H1:μ1≠μ2 (claim)

Explanation of Solution

Given info:

x¯1=$35,270 , x¯2=10.3 , s1=$3,256 , s2=$1,432 , n1=15 , and n2=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 μ1≠μ2 . The complement of the claim is μ1=μ2 . In the given experiment, the alternative hypothesis indicates the claim.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

b.

Expert Solution

To determine

To find: The critical value α=0.02 .

Answer to Problem 9.2.5RE

The critical value at α=0.02 is ±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,

df=Smallar of n1−1 and n2−1=15−1=14

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 1

From the output, the critical value is ±2.624 .

c.

Expert Solution

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 3,483<μ1−μ2<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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 2

From the MINITAB output, the test value is 6.54 and the 98% confidence interval for the difference of the two means is 3,483<μ1−μ2<8,033_ .

d.

Expert Solution

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 3

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

e.

Expert Solution

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

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.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

To identify: The claim and state H0 and Ha .

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: H0:μ1=μ2

Alternative hypothesis: H1:μ1≠μ2 (claim)

Explanation of Solution

Given info:

x¯1=$35,270 , x¯2=10.3 , s1=$3,256 , s2=$1,432 , n1=15 , and n2=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 μ1≠μ2 . The complement of the claim is μ1=μ2 . In the given experiment, the alternative hypothesis indicates the claim.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

b.

Expert Solution

To determine

To find: The critical value α=0.02 .

Answer to Problem 9.2.5RE

The critical value at α=0.02 is ±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,

df=Smallar of n1−1 and n2−1=15−1=14

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 1

From the output, the critical value is ±2.624 .

c.

Expert Solution

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 3,483<μ1−μ2<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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 2

From the MINITAB output, the test value is 6.54 and the 98% confidence interval for the difference of the two means is 3,483<μ1−μ2<8,033_ .

d.

Expert Solution

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 3

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

e.

Expert Solution

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

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.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

To identify: The claim and state H0 and Ha .

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: H0:μ1=μ2

Alternative hypothesis: H1:μ1≠μ2 (claim)

Explanation of Solution

Given info:

x¯1=$35,270 , x¯2=10.3 , s1=$3,256 , s2=$1,432 , n1=15 , and n2=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 μ1≠μ2 . The complement of the claim is μ1=μ2 . In the given experiment, the alternative hypothesis indicates the claim.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

b.

Expert Solution

To determine

To find: The critical value α=0.02 .

Answer to Problem 9.2.5RE

The critical value at α=0.02 is ±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,

df=Smallar of n1−1 and n2−1=15−1=14

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 1

From the output, the critical value is ±2.624 .

c.

Expert Solution

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 3,483<μ1−μ2<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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 2

From the MINITAB output, the test value is 6.54 and the 98% confidence interval for the difference of the two means is 3,483<μ1−μ2<8,033_ .

d.

Expert Solution

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 3

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

e.

Expert Solution

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

To identify: The claim and state H0 and Ha .

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: H0:μ1=μ2

Alternative hypothesis: H1:μ1≠μ2 (claim)

Explanation of Solution

Given info:

x¯1=$35,270 , x¯2=10.3 , s1=$3,256 , s2=$1,432 , n1=15 , and n2=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 μ1≠μ2 . The complement of the claim is μ1=μ2 . In the given experiment, the alternative hypothesis indicates the claim.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

b.

Expert Solution

To determine

To find: The critical value α=0.02 .

Answer to Problem 9.2.5RE

The critical value at α=0.02 is ±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,

df=Smallar of n1−1 and n2−1=15−1=14

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 1

From the output, the critical value is ±2.624 .

c.

Expert Solution

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 3,483<μ1−μ2<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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 2

From the MINITAB output, the test value is 6.54 and the 98% confidence interval for the difference of the two means is 3,483<μ1−μ2<8,033_ .

d.

Expert Solution

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 3

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

e.

Expert Solution

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.

Answer 8

a.

Expert Solution

To determine

To identify: The claim and state H0 and Ha .

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: H0:μ1=μ2

Alternative hypothesis: H1:μ1≠μ2 (claim)

Explanation of Solution

Given info:

x¯1=$35,270 , x¯2=10.3 , s1=$3,256 , s2=$1,432 , n1=15 , and n2=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 μ1≠μ2 . The complement of the claim is μ1=μ2 . In the given experiment, the alternative hypothesis indicates the claim.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To identify: The claim and state H0 and Ha .

Answer 13

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: H0:μ1=μ2

Alternative hypothesis: H1:μ1≠μ2 (claim)

Answer 14

Explanation of Solution

Given info:

x¯1=$35,270 , x¯2=10.3 , s1=$3,256 , s2=$1,432 , n1=15 , and n2=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 μ1≠μ2 . The complement of the claim is μ1=μ2 . In the given experiment, the alternative hypothesis indicates the claim.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Answer 15

b.

Expert Solution

To determine

To find: The critical value α=0.02 .

Answer to Problem 9.2.5RE

The critical value at α=0.02 is ±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,

df=Smallar of n1−1 and n2−1=15−1=14

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 1

From the output, the critical value is ±2.624 .

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To find: The critical value α=0.02 .

Answer 20

Answer to Problem 9.2.5RE

The critical value at α=0.02 is ±2.624 .

Answer 21

Explanation of Solution

Calculation:

Here, the test is two tailed test.

Critical value:

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

df=Smallar of n1−1 and n2−1=15−1=14

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 1

From the output, the critical value is ±2.624 .

Answer 22

c.

Expert Solution

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 3,483<μ1−μ2<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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 2

From the MINITAB output, the test value is 6.54 and the 98% confidence interval for the difference of the two means is 3,483<μ1−μ2<8,033_ .

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

To find: The test value.

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

Answer 27

Answer to Problem 9.2.5RE

The test value is –0.9.

The 98% confidence interval for the difference of the two means is 3,483<μ1−μ2<8,033_ .

Answer 28

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 2

From the MINITAB output, the test value is 6.54 and the 98% confidence interval for the difference of the two means is 3,483<μ1−μ2<8,033_ .

Answer 29

d.

Expert Solution

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 3

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

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

To make: The decision.

Answer 34

Answer to Problem 9.2.5RE

The decision is that, the null hypothesis isrejected.

Answer 35

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:

ALEKS 360 ELEM STATISTICS, Chapter 9, Problem 9.2.5RE , additional homework tip 3

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

Answer 36

e.

Expert Solution

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.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

To summarize: The result.

Answer 41

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.

Answer 42

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.