Find the correlation coefficient . Check whether a negative value of correlation coefficient is surprising or not. Interpret the results.

Question

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Answer 1

Question

Chapter 13, Problem 43CE

a.

To determine

Find the correlation coefficient.

Check whether a negative value of correlation coefficient is surprising or not.

Interpret the results.

a.

Expert Solution

Answer to Problem 43CE

The correlation coefficient is –0.384.

Explanation of Solution

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$33.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 1

The correlation coefficient is –0.384. Since the correlation coefficient is negative, there is a moderate negative correlation between x and y. The correlation coefficient indicates that as the value of x increases, the value of y will decrease. That is, if the PTS score increases then consequently PTS allowed will decrease. Thus, the negative correlation between the variables is not surprising.

b.

To determine

Find the value of coefficient of determination and explain the relationship.

b.

Expert Solution

Answer to Problem 43CE

The coefficient of determination is 0.147.

Explanation of Solution

The coefficient of determination is the square of the correlation coefficient. From Part (a), the correlation coefficient is –0.384.

The coefficient of determination is as follows:

Coefficient of determination=(correlation coefficient)2=(−0.384)2=0.147

The value of coefficient of determination is 0.147. Therefore, 14.7% of variation in the dependent variable is explained by the independent variable.

c.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ or not.

c.

Expert Solution

Answer to Problem 43CE

There is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

Explanation of Solution

Denote the population correlation as ρ.

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative.

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 32 and the correlation coefficient is –0.384.

The test statistic is as follows:

t=−0.38432−21−(−0.384)2=−0.384×300.852544=−2.2778≈−2.28

The degrees of freedom is as follows:

df=n−2=32−2=30

Thus, the level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 30)”.

Output obtained using the EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 2

From the EXCEL output, the critical value is –1.697(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –2.28 and the critical value is –1.697.

Here, t-calculated(=−2.28)< −tα(=−1.697).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

d.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ for each conference or not.

d.

Expert Solution

Answer to Problem 43CE

There is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

There is no evidence that a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Explanation of Solution

Denote the population correlation as ρ.

For conference AFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference AFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative in conference AFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 3

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.676.

The test statistic is as follows:

t=−0.67616−21−(−0.676)2=−0.676×140.543024=−3.432

The degrees of freedom is as follows:

df=n−2=16−2=14

The level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 14)”.

Output obtained using EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 4

From the EXCEL output, the critical value is –1.761(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –3.432 and the critical value is –1.761.

Here, t-calculated(=−3.432)< −tα(=−1.761).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

For conference NFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference NFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is a negative in conference NFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 5

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.197.

The test statistic is as follows:

t=−0.19716−21−(−0.197)2=−0.197×140.961191=−0.752

Conclusion:

The value of test statistic is –0.752 and the critical value is –1.761.

Here, t-calculated(=−0.752)> −tα(=−1.761).

By the rejection rule, fail to reject the null hypothesis.

Thus, there is no enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

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Answer 2

Question

Chapter 13, Problem 43CE

a.

To determine

Find the correlation coefficient.

Check whether a negative value of correlation coefficient is surprising or not.

Interpret the results.

a.

Expert Solution

Answer to Problem 43CE

The correlation coefficient is –0.384.

Explanation of Solution

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$33.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 1

The correlation coefficient is –0.384. Since the correlation coefficient is negative, there is a moderate negative correlation between x and y. The correlation coefficient indicates that as the value of x increases, the value of y will decrease. That is, if the PTS score increases then consequently PTS allowed will decrease. Thus, the negative correlation between the variables is not surprising.

b.

To determine

Find the value of coefficient of determination and explain the relationship.

b.

Expert Solution

Answer to Problem 43CE

The coefficient of determination is 0.147.

Explanation of Solution

The coefficient of determination is the square of the correlation coefficient. From Part (a), the correlation coefficient is –0.384.

The coefficient of determination is as follows:

Coefficient of determination=(correlation coefficient)2=(−0.384)2=0.147

The value of coefficient of determination is 0.147. Therefore, 14.7% of variation in the dependent variable is explained by the independent variable.

c.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ or not.

c.

Expert Solution

Answer to Problem 43CE

There is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

Explanation of Solution

Denote the population correlation as ρ.

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative.

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 32 and the correlation coefficient is –0.384.

The test statistic is as follows:

t=−0.38432−21−(−0.384)2=−0.384×300.852544=−2.2778≈−2.28

The degrees of freedom is as follows:

df=n−2=32−2=30

Thus, the level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 30)”.

Output obtained using the EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 2

From the EXCEL output, the critical value is –1.697(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –2.28 and the critical value is –1.697.

Here, t-calculated(=−2.28)< −tα(=−1.697).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

d.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ for each conference or not.

d.

Expert Solution

Answer to Problem 43CE

There is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

There is no evidence that a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Explanation of Solution

Denote the population correlation as ρ.

For conference AFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference AFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative in conference AFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 3

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.676.

The test statistic is as follows:

t=−0.67616−21−(−0.676)2=−0.676×140.543024=−3.432

The degrees of freedom is as follows:

df=n−2=16−2=14

The level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 14)”.

Output obtained using EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 4

From the EXCEL output, the critical value is –1.761(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –3.432 and the critical value is –1.761.

Here, t-calculated(=−3.432)< −tα(=−1.761).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

For conference NFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference NFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is a negative in conference NFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 5

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.197.

The test statistic is as follows:

t=−0.19716−21−(−0.197)2=−0.197×140.961191=−0.752

Conclusion:

The value of test statistic is –0.752 and the critical value is –1.761.

Here, t-calculated(=−0.752)> −tα(=−1.761).

By the rejection rule, fail to reject the null hypothesis.

Thus, there is no enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

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

Question

Chapter 13, Problem 43CE

a.

To determine

Find the correlation coefficient.

Check whether a negative value of correlation coefficient is surprising or not.

Interpret the results.

a.

Expert Solution

Answer to Problem 43CE

The correlation coefficient is –0.384.

Explanation of Solution

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$33.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 1

The correlation coefficient is –0.384. Since the correlation coefficient is negative, there is a moderate negative correlation between x and y. The correlation coefficient indicates that as the value of x increases, the value of y will decrease. That is, if the PTS score increases then consequently PTS allowed will decrease. Thus, the negative correlation between the variables is not surprising.

b.

To determine

Find the value of coefficient of determination and explain the relationship.

b.

Expert Solution

Answer to Problem 43CE

The coefficient of determination is 0.147.

Explanation of Solution

The coefficient of determination is the square of the correlation coefficient. From Part (a), the correlation coefficient is –0.384.

The coefficient of determination is as follows:

Coefficient of determination=(correlation coefficient)2=(−0.384)2=0.147

The value of coefficient of determination is 0.147. Therefore, 14.7% of variation in the dependent variable is explained by the independent variable.

c.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ or not.

c.

Expert Solution

Answer to Problem 43CE

There is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

Explanation of Solution

Denote the population correlation as ρ.

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative.

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 32 and the correlation coefficient is –0.384.

The test statistic is as follows:

t=−0.38432−21−(−0.384)2=−0.384×300.852544=−2.2778≈−2.28

The degrees of freedom is as follows:

df=n−2=32−2=30

Thus, the level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 30)”.

Output obtained using the EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 2

From the EXCEL output, the critical value is –1.697(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –2.28 and the critical value is –1.697.

Here, t-calculated(=−2.28)< −tα(=−1.697).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

d.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ for each conference or not.

d.

Expert Solution

Answer to Problem 43CE

There is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

There is no evidence that a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Explanation of Solution

Denote the population correlation as ρ.

For conference AFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference AFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative in conference AFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 3

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.676.

The test statistic is as follows:

t=−0.67616−21−(−0.676)2=−0.676×140.543024=−3.432

The degrees of freedom is as follows:

df=n−2=16−2=14

The level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 14)”.

Output obtained using EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 4

From the EXCEL output, the critical value is –1.761(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –3.432 and the critical value is –1.761.

Here, t-calculated(=−3.432)< −tα(=−1.761).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

For conference NFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference NFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is a negative in conference NFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 5

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.197.

The test statistic is as follows:

t=−0.19716−21−(−0.197)2=−0.197×140.961191=−0.752

Conclusion:

The value of test statistic is –0.752 and the critical value is –1.761.

Here, t-calculated(=−0.752)> −tα(=−1.761).

By the rejection rule, fail to reject the null hypothesis.

Thus, there is no enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

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

Question

Chapter 13, Problem 43CE

a.

To determine

Find the correlation coefficient.

Check whether a negative value of correlation coefficient is surprising or not.

Interpret the results.

a.

Expert Solution

Answer to Problem 43CE

The correlation coefficient is –0.384.

Explanation of Solution

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$33.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 1

The correlation coefficient is –0.384. Since the correlation coefficient is negative, there is a moderate negative correlation between x and y. The correlation coefficient indicates that as the value of x increases, the value of y will decrease. That is, if the PTS score increases then consequently PTS allowed will decrease. Thus, the negative correlation between the variables is not surprising.

b.

To determine

Find the value of coefficient of determination and explain the relationship.

b.

Expert Solution

Answer to Problem 43CE

The coefficient of determination is 0.147.

Explanation of Solution

The coefficient of determination is the square of the correlation coefficient. From Part (a), the correlation coefficient is –0.384.

The coefficient of determination is as follows:

Coefficient of determination=(correlation coefficient)2=(−0.384)2=0.147

The value of coefficient of determination is 0.147. Therefore, 14.7% of variation in the dependent variable is explained by the independent variable.

c.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ or not.

c.

Expert Solution

Answer to Problem 43CE

There is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

Explanation of Solution

Denote the population correlation as ρ.

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative.

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 32 and the correlation coefficient is –0.384.

The test statistic is as follows:

t=−0.38432−21−(−0.384)2=−0.384×300.852544=−2.2778≈−2.28

The degrees of freedom is as follows:

df=n−2=32−2=30

Thus, the level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 30)”.

Output obtained using the EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 2

From the EXCEL output, the critical value is –1.697(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –2.28 and the critical value is –1.697.

Here, t-calculated(=−2.28)< −tα(=−1.697).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

d.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ for each conference or not.

d.

Expert Solution

Answer to Problem 43CE

There is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

There is no evidence that a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Explanation of Solution

Denote the population correlation as ρ.

For conference AFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference AFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative in conference AFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 3

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.676.

The test statistic is as follows:

t=−0.67616−21−(−0.676)2=−0.676×140.543024=−3.432

The degrees of freedom is as follows:

df=n−2=16−2=14

The level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 14)”.

Output obtained using EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 4

From the EXCEL output, the critical value is –1.761(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –3.432 and the critical value is –1.761.

Here, t-calculated(=−3.432)< −tα(=−1.761).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

For conference NFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference NFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is a negative in conference NFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 5

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.197.

The test statistic is as follows:

t=−0.19716−21−(−0.197)2=−0.197×140.961191=−0.752

Conclusion:

The value of test statistic is –0.752 and the critical value is –1.761.

Here, t-calculated(=−0.752)> −tα(=−1.761).

By the rejection rule, fail to reject the null hypothesis.

Thus, there is no enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

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Answer 5

Chapter 13, Problem 43CE

a.

To determine

Find the correlation coefficient.

Check whether a negative value of correlation coefficient is surprising or not.

Interpret the results.

a.

Expert Solution

Answer to Problem 43CE

The correlation coefficient is –0.384.

Explanation of Solution

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$33.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 1

The correlation coefficient is –0.384. Since the correlation coefficient is negative, there is a moderate negative correlation between x and y. The correlation coefficient indicates that as the value of x increases, the value of y will decrease. That is, if the PTS score increases then consequently PTS allowed will decrease. Thus, the negative correlation between the variables is not surprising.

b.

To determine

Find the value of coefficient of determination and explain the relationship.

b.

Expert Solution

Answer to Problem 43CE

The coefficient of determination is 0.147.

Explanation of Solution

The coefficient of determination is the square of the correlation coefficient. From Part (a), the correlation coefficient is –0.384.

The coefficient of determination is as follows:

Coefficient of determination=(correlation coefficient)2=(−0.384)2=0.147

The value of coefficient of determination is 0.147. Therefore, 14.7% of variation in the dependent variable is explained by the independent variable.

c.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ or not.

c.

Expert Solution

Answer to Problem 43CE

There is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

Explanation of Solution

Denote the population correlation as ρ.

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative.

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 32 and the correlation coefficient is –0.384.

The test statistic is as follows:

t=−0.38432−21−(−0.384)2=−0.384×300.852544=−2.2778≈−2.28

The degrees of freedom is as follows:

df=n−2=32−2=30

Thus, the level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 30)”.

Output obtained using the EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 2

From the EXCEL output, the critical value is –1.697(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –2.28 and the critical value is –1.697.

Here, t-calculated(=−2.28)< −tα(=−1.697).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

d.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ for each conference or not.

d.

Expert Solution

Answer to Problem 43CE

There is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

There is no evidence that a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Explanation of Solution

Denote the population correlation as ρ.

For conference AFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference AFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative in conference AFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 3

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.676.

The test statistic is as follows:

t=−0.67616−21−(−0.676)2=−0.676×140.543024=−3.432

The degrees of freedom is as follows:

df=n−2=16−2=14

The level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 14)”.

Output obtained using EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 4

From the EXCEL output, the critical value is –1.761(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –3.432 and the critical value is –1.761.

Here, t-calculated(=−3.432)< −tα(=−1.761).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

For conference NFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference NFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is a negative in conference NFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 5

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.197.

The test statistic is as follows:

t=−0.19716−21−(−0.197)2=−0.197×140.961191=−0.752

Conclusion:

The value of test statistic is –0.752 and the critical value is –1.761.

Here, t-calculated(=−0.752)> −tα(=−1.761).

By the rejection rule, fail to reject the null hypothesis.

Thus, there is no enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Answer 6

Chapter 13, Problem 43CE

a.

To determine

Find the correlation coefficient.

Check whether a negative value of correlation coefficient is surprising or not.

Interpret the results.

a.

Expert Solution

Answer to Problem 43CE

The correlation coefficient is –0.384.

Explanation of Solution

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$33.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 1

The correlation coefficient is –0.384. Since the correlation coefficient is negative, there is a moderate negative correlation between x and y. The correlation coefficient indicates that as the value of x increases, the value of y will decrease. That is, if the PTS score increases then consequently PTS allowed will decrease. Thus, the negative correlation between the variables is not surprising.

Answer 7

Chapter 13, Problem 43CE

a.

To determine

Find the correlation coefficient.

Check whether a negative value of correlation coefficient is surprising or not.

Interpret the results.

Answer 8

a.

Expert Solution

Answer to Problem 43CE

The correlation coefficient is –0.384.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$33.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 1

The correlation coefficient is –0.384. Since the correlation coefficient is negative, there is a moderate negative correlation between x and y. The correlation coefficient indicates that as the value of x increases, the value of y will decrease. That is, if the PTS score increases then consequently PTS allowed will decrease. Thus, the negative correlation between the variables is not surprising.

Answer 13

b.

To determine

Find the value of coefficient of determination and explain the relationship.

b.

Expert Solution

Answer to Problem 43CE

The coefficient of determination is 0.147.

Explanation of Solution

The coefficient of determination is the square of the correlation coefficient. From Part (a), the correlation coefficient is –0.384.

The coefficient of determination is as follows:

Coefficient of determination=(correlation coefficient)2=(−0.384)2=0.147

The value of coefficient of determination is 0.147. Therefore, 14.7% of variation in the dependent variable is explained by the independent variable.

Answer 14

b.

To determine

Find the value of coefficient of determination and explain the relationship.

Answer 15

b.

Expert Solution

Answer to Problem 43CE

The coefficient of determination is 0.147.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The coefficient of determination is the square of the correlation coefficient. From Part (a), the correlation coefficient is –0.384.

The coefficient of determination is as follows:

Coefficient of determination=(correlation coefficient)2=(−0.384)2=0.147

The value of coefficient of determination is 0.147. Therefore, 14.7% of variation in the dependent variable is explained by the independent variable.

Answer 20

c.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ or not.

c.

Expert Solution

Answer to Problem 43CE

There is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

Explanation of Solution

Denote the population correlation as ρ.

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative.

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 32 and the correlation coefficient is –0.384.

The test statistic is as follows:

t=−0.38432−21−(−0.384)2=−0.384×300.852544=−2.2778≈−2.28

The degrees of freedom is as follows:

df=n−2=32−2=30

Thus, the level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 30)”.

Output obtained using the EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 2

From the EXCEL output, the critical value is –1.697(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –2.28 and the critical value is –1.697.

Here, t-calculated(=−2.28)< −tα(=−1.697).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

Answer 21

c.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ or not.

Answer 22

c.

Expert Solution

Answer to Problem 43CE

There is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Denote the population correlation as ρ.

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative.

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 32 and the correlation coefficient is –0.384.

The test statistic is as follows:

t=−0.38432−21−(−0.384)2=−0.384×300.852544=−2.2778≈−2.28

The degrees of freedom is as follows:

df=n−2=32−2=30

Thus, the level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 30)”.

Output obtained using the EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 2

From the EXCEL output, the critical value is –1.697(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –2.28 and the critical value is –1.697.

Here, t-calculated(=−2.28)< −tα(=−1.697).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’.

Answer 27

d.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ for each conference or not.

d.

Expert Solution

Answer to Problem 43CE

There is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

There is no evidence that a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Explanation of Solution

Denote the population correlation as ρ.

For conference AFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference AFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative in conference AFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 3

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.676.

The test statistic is as follows:

t=−0.67616−21−(−0.676)2=−0.676×140.543024=−3.432

The degrees of freedom is as follows:

df=n−2=16−2=14

The level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 14)”.

Output obtained using EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 4

From the EXCEL output, the critical value is –1.761(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –3.432 and the critical value is –1.761.

Here, t-calculated(=−3.432)< −tα(=−1.761).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

For conference NFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference NFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is a negative in conference NFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 5

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.197.

The test statistic is as follows:

t=−0.19716−21−(−0.197)2=−0.197×140.961191=−0.752

Conclusion:

The value of test statistic is –0.752 and the critical value is –1.761.

Here, t-calculated(=−0.752)> −tα(=−1.761).

By the rejection rule, fail to reject the null hypothesis.

Thus, there is no enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Answer 28

d.

To determine

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ for each conference or not.

Answer 29

d.

Expert Solution

Answer to Problem 43CE

There is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

There is no evidence that a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Denote the population correlation as ρ.

For conference AFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference AFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is negative in conference AFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 3

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.676.

The test statistic is as follows:

t=−0.67616−21−(−0.676)2=−0.676×140.543024=−3.432

The degrees of freedom is as follows:

df=n−2=16−2=14

The level of significance is 0.05.

Critical value:

Software procedure:

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

Open an EXCEL file.
In cell A1, enter the formula “=T.INV (0.05, 14)”.

Output obtained using EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 4

From the EXCEL output, the critical value is –1.761(=−tα) .

Decision rule:

Reject the null hypothesis H₀, if t-calculated< −tα. Otherwise, fail to reject H₀.

Conclusion:

The value of test statistic is –3.432 and the critical value is –1.761.

Here, t-calculated(=−3.432)< −tα(=−1.761).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference AFC.

For conference NFC:

The hypotheses are given below:

Null hypothesis:

H0:ρ≥0

That is, the correlation between ‘point scored’ and ‘point allowed’ is greater than or equal to zero in conference NFC.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between ‘point scored’ and ‘point allowed’ is a negative in conference NFC.

Step-by-step procedure to obtain the correlation coefficient using MegaStat software:

In an EXCEL sheet enter the data values of x and y.
Go to Add-Ins > MegaStat > Correlation/Regression > Correlation matrix.
Enter Input Range as $A$1:$B$17.
Click on OK.

Output obtained using MegaStat is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 43CE , additional homework tip 5

Test statistic:

The test statistic is as follows:

t=rn−21−r2

Here, the sample size is 16 and the correlation coefficient is –0.197.

The test statistic is as follows:

t=−0.19716−21−(−0.197)2=−0.197×140.961191=−0.752

Conclusion:

The value of test statistic is –0.752 and the critical value is –1.761.

Here, t-calculated(=−0.752)> −tα(=−1.761).

By the rejection rule, fail to reject the null hypothesis.

Thus, there is no enough evidence to infer that there is a negative association between the ‘point scored’ and ‘point allowed’ in conference NFC.

Answer 34

Ch. 13 - Prob. 1SR Ch. 13 - Prob. 1E Ch. 13 - Prob. 2E Ch. 13 - Bi-lo Appliance Super-Store has outlets in several...Ch. 13 - Prob. 4E Ch. 13 - Prob. 5E Ch. 13 - The owner of Maumee Ford-Volvo wants to study the...Ch. 13 - Prob. 2SR Ch. 13 - Prob. 7E Ch. 13 - Prob. 8E

Find the correlation coefficient . Check whether a negative value of correlation coefficient is surprising or not. Interpret the results.

Concept explainers

Videos

Find the correlation coefficient.

Check whether a negative value of correlation coefficient is surprising or not.

Interpret the results.

Answer to Problem 43CE

Explanation of Solution

Find the value of coefficient of determination and explain the relationship.

Answer to Problem 43CE

Explanation of Solution

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ or not.

Answer to Problem 43CE

Explanation of Solution

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ for each conference or not.

Answer to Problem 43CE

Explanation of Solution

Want to see more full solutions like this?

Chapter 13 Solutions

Find the correlation coefficient . Check whether a negative value of correlation coefficient is surprising or not. Interpret the results.

Concept explainers

Videos

Find the correlation coefficient. Check whether a negative value of correlation coefficient is surprising or not. Interpret the results.

Answer to Problem 43CE

Explanation of Solution

Find the value of coefficient of determination and explain the relationship.

Answer to Problem 43CE

Explanation of Solution

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ or not.

Answer to Problem 43CE

Explanation of Solution

Test whether there is a negative association between the ‘point scored’ and ‘point allowed’ for each conference or not.

Answer to Problem 43CE

Explanation of Solution

Want to see more full solutions like this?

Chapter 13 Solutions

Find the correlation coefficient.

Check whether a negative value of correlation coefficient is surprising or not.

Interpret the results.