Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

Question

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Answer 1

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 11, Problem 48DE

a.

To determine

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

a.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean salary of teams in the American League and μ2 denote the mean salary of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=14+16−2=28

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 28.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 1

From the MINITAB output, the critical values are ±1.983.

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.94.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

b.

To determine

Check whether there is evidence that there is a difference in the mean home attendance of team in the American League versus teams in the National League.

b.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean home attendance of teams in the American League and μ2 denote the mean home attendance of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

From the MINITAB output in Part (a), the critical values are ±2.048.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 3

From the given MINITAB output, the value of the test statistic is 0.88.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

c.

To determine

Check whether there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

c.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of difference in the mean number of wins for teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denotes the mean number of wins for teams in the American League and μ2 denotes the mean number of wins for teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=10+10−2=18

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 18.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 4

From the MINITAB output, the critical values are ±2.101.

The decision rule is as follows:

If t <−2.101, then reject the null hypothesis H0.

If t > 2.101, then reject the null hypothesis H0.

If −2.101<t<2.101, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 5

From the given MINITAB output, the value of the test statistic is –13.

Decision:

The value of test statistic does not lie between the critical values.

That is, Test statistic(−13)< Critical value(±2.101)

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

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Answer 2

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 11, Problem 48DE

a.

To determine

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

a.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean salary of teams in the American League and μ2 denote the mean salary of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=14+16−2=28

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 28.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 1

From the MINITAB output, the critical values are ±1.983.

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.94.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

b.

To determine

Check whether there is evidence that there is a difference in the mean home attendance of team in the American League versus teams in the National League.

b.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean home attendance of teams in the American League and μ2 denote the mean home attendance of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

From the MINITAB output in Part (a), the critical values are ±2.048.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 3

From the given MINITAB output, the value of the test statistic is 0.88.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

c.

To determine

Check whether there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

c.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of difference in the mean number of wins for teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denotes the mean number of wins for teams in the American League and μ2 denotes the mean number of wins for teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=10+10−2=18

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 18.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 4

From the MINITAB output, the critical values are ±2.101.

The decision rule is as follows:

If t <−2.101, then reject the null hypothesis H0.

If t > 2.101, then reject the null hypothesis H0.

If −2.101<t<2.101, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 5

From the given MINITAB output, the value of the test statistic is –13.

Decision:

The value of test statistic does not lie between the critical values.

That is, Test statistic(−13)< Critical value(±2.101)

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

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

Answer 3

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 11, Problem 48DE

a.

To determine

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

a.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean salary of teams in the American League and μ2 denote the mean salary of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=14+16−2=28

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 28.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 1

From the MINITAB output, the critical values are ±1.983.

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.94.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

b.

To determine

Check whether there is evidence that there is a difference in the mean home attendance of team in the American League versus teams in the National League.

b.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean home attendance of teams in the American League and μ2 denote the mean home attendance of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

From the MINITAB output in Part (a), the critical values are ±2.048.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 3

From the given MINITAB output, the value of the test statistic is 0.88.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

c.

To determine

Check whether there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

c.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of difference in the mean number of wins for teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denotes the mean number of wins for teams in the American League and μ2 denotes the mean number of wins for teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=10+10−2=18

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 18.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 4

From the MINITAB output, the critical values are ±2.101.

The decision rule is as follows:

If t <−2.101, then reject the null hypothesis H0.

If t > 2.101, then reject the null hypothesis H0.

If −2.101<t<2.101, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 5

From the given MINITAB output, the value of the test statistic is –13.

Decision:

The value of test statistic does not lie between the critical values.

That is, Test statistic(−13)< Critical value(±2.101)

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

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

Answer 4

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 11, Problem 48DE

a.

To determine

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

a.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean salary of teams in the American League and μ2 denote the mean salary of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=14+16−2=28

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 28.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 1

From the MINITAB output, the critical values are ±1.983.

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.94.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

b.

To determine

Check whether there is evidence that there is a difference in the mean home attendance of team in the American League versus teams in the National League.

b.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean home attendance of teams in the American League and μ2 denote the mean home attendance of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

From the MINITAB output in Part (a), the critical values are ±2.048.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 3

From the given MINITAB output, the value of the test statistic is 0.88.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

c.

To determine

Check whether there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

c.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of difference in the mean number of wins for teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denotes the mean number of wins for teams in the American League and μ2 denotes the mean number of wins for teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=10+10−2=18

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 18.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 4

From the MINITAB output, the critical values are ±2.101.

The decision rule is as follows:

If t <−2.101, then reject the null hypothesis H0.

If t > 2.101, then reject the null hypothesis H0.

If −2.101<t<2.101, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 5

From the given MINITAB output, the value of the test statistic is –13.

Decision:

The value of test statistic does not lie between the critical values.

That is, Test statistic(−13)< Critical value(±2.101)

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

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Answer 5

Chapter 11, Problem 48DE

a.

To determine

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

a.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean salary of teams in the American League and μ2 denote the mean salary of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=14+16−2=28

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 28.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 1

From the MINITAB output, the critical values are ±1.983.

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.94.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

b.

To determine

Check whether there is evidence that there is a difference in the mean home attendance of team in the American League versus teams in the National League.

b.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean home attendance of teams in the American League and μ2 denote the mean home attendance of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

From the MINITAB output in Part (a), the critical values are ±2.048.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 3

From the given MINITAB output, the value of the test statistic is 0.88.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

c.

To determine

Check whether there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

c.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of difference in the mean number of wins for teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denotes the mean number of wins for teams in the American League and μ2 denotes the mean number of wins for teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=10+10−2=18

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 18.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 4

From the MINITAB output, the critical values are ±2.101.

The decision rule is as follows:

If t <−2.101, then reject the null hypothesis H0.

If t > 2.101, then reject the null hypothesis H0.

If −2.101<t<2.101, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 5

From the given MINITAB output, the value of the test statistic is –13.

Decision:

The value of test statistic does not lie between the critical values.

That is, Test statistic(−13)< Critical value(±2.101)

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

Answer 6

Chapter 11, Problem 48DE

a.

To determine

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

a.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean salary of teams in the American League and μ2 denote the mean salary of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=14+16−2=28

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 28.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 1

From the MINITAB output, the critical values are ±1.983.

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.94.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Answer 7

Chapter 11, Problem 48DE

a.

To determine

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

Answer 8

a.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

In this context, let μ1 denote the mean salary of teams in the American League and μ2 denote the mean salary of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=14+16−2=28

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 28.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 1

From the MINITAB output, the critical values are ±1.983.

The decision rule is as follows:

If t <−1.983, then reject the null hypothesis H0.

If t > 1.983, then reject the null hypothesis H0.

If −1.983<t<1.983, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 2

From the given MINITAB output, the value of the test statistic is –0.94.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Answer 13

b.

To determine

Check whether there is evidence that there is a difference in the mean home attendance of team in the American League versus teams in the National League.

b.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denote the mean home attendance of teams in the American League and μ2 denote the mean home attendance of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

From the MINITAB output in Part (a), the critical values are ±2.048.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 3

From the given MINITAB output, the value of the test statistic is 0.88.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Answer 14

b.

To determine

Check whether there is evidence that there is a difference in the mean home attendance of team in the American League versus teams in the National League.

Answer 15

b.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

In this context, let μ1 denote the mean home attendance of teams in the American League and μ2 denote the mean home attendance of teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

From the MINITAB output in Part (a), the critical values are ±2.048.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 3

From the given MINITAB output, the value of the test statistic is 0.88.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Answer 20

c.

To determine

Check whether there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

c.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of difference in the mean number of wins for teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let μ1 denotes the mean number of wins for teams in the American League and μ2 denotes the mean number of wins for teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=10+10−2=18

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 18.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 4

From the MINITAB output, the critical values are ±2.101.

The decision rule is as follows:

If t <−2.101, then reject the null hypothesis H0.

If t > 2.101, then reject the null hypothesis H0.

If −2.101<t<2.101, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 5

From the given MINITAB output, the value of the test statistic is –13.

Decision:

The value of test statistic does not lie between the critical values.

That is, Test statistic(−13)< Critical value(±2.101)

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

Answer 21

c.

To determine

Check whether there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

Answer 22

c.

Expert Solution

Answer to Problem 48DE

The conclusion is that there is no evidence of difference in the mean number of wins for teams in the American League versus teams in the National League.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

In this context, let μ1 denotes the mean number of wins for teams in the American League and μ2 denotes the mean number of wins for teams in the National League.

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1≠μ2

Significance level, α:

It is given that the significance level, α=0.05.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n2−2=10+10−2=18

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

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 18.
Click the Shaded Area tab.
Choose Probability and Both Tails for the region of the curve to shade.
Enter the Probability as 0.05.
Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 4

From the MINITAB output, the critical values are ±2.101.

The decision rule is as follows:

If t <−2.101, then reject the null hypothesis H0.

If t > 2.101, then reject the null hypothesis H0.

If −2.101<t<2.101, then fail to reject the null hypothesis H0.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

Choose Stat > Basic Statistics > 2 sample t.
Choose Each sample in its own column.
In Sample 1, enter the column of American.
In Sample 2, enter the column of National.
Choose Assume equal variance.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 11, Problem 48DE , additional homework tip 5

From the given MINITAB output, the value of the test statistic is –13.

Decision:

The value of test statistic does not lie between the critical values.

That is, Test statistic(−13)< Critical value(±2.101)

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

Answer 27

Ch. 11 - Tom Sevits is the owner of the Appliance Patch....Ch. 11 - Prob. 1E Ch. 11 - Prob. 2E Ch. 11 - Prob. 3E Ch. 11 - As part of a study of corporate employees, the...Ch. 11 - Prob. 5E Ch. 11 - Mary Jo Fitzpatrick is the vice president for...Ch. 11 - Prob. 2SR Ch. 11 - Prob. 7E Ch. 11 - For Exercises 7 and 8: (a) state the decision...

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

Videos

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

Answer to Problem 48DE

Explanation of Solution

Check whether there is evidence that there is a difference in the mean home attendance of team in the American League versus teams in the National League.

Answer to Problem 48DE

Explanation of Solution

Check whether there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

Answer to Problem 48DE

Explanation of Solution

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Chapter 11 Solutions