Presidents and first ladies: The presents the ages of the last 10 U.S. presidents and their wives on the first day of their presidencies. Compute the least-squares regression line for predicting the president’s age from the first lady’s age. Compute the coefficient of determination- Construct a scatterplot of the presidents' ages (y) versus the first ladies' ages ( x ). Which point is an outlier? Remove die outlier and compute the least-squares regression line for predicting the president’s age from the first lady: s age. Is the outlier influential? Explain. Compute the coefficient of determination for the data set with the outlier removed. Is due proportion of variation explained by due least-squares regression he greater, less. or about the same without due outlier? Explain.

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Answer 1

Textbook Question

Chapter 4.3, Problem 24E

Presidents and first ladies: The presents the ages of the last 10 U.S. presidents and their wives on the first day of their presidencies.

Chapter 4.3, Problem 24E, Presidents and first ladies: The presents the ages of the last 10 U.S. presidents and their wives on

Compute the least-squares regression line for predicting the president’s age from the first lady’s age.
Compute the coefficient of determination-
Construct a scatterplot of the presidents' ages (y) versus the first ladies' ages (x).
Which point is an outlier?
Remove die outlier and compute the least-squares regression line for predicting the president’s age from the first lady: s age.
Is the outlier influential? Explain.
Compute the coefficient of determination for the data set with the outlier removed. Is due proportion of variation explained by due least-squares regression he greater, less. or about the same without due outlier? Explain.

a.

Expert Solution

To determine

To Calculate: The least-squares regression line for predicting the president’s age from lady’s age.

Answer to Problem 24E

The least-squares regression line is,

$y^=14+0.8426x$

Explanation of Solution

Given: The following table presents the age of the president’s and their wives on the first day of their presidencies.

Name	Her Age	His Age
Donald and Melania Trump	46	70
Barak and Mechelle Obama	45	47
George W. and Laura Bush	54	54
Bill and Hillary Clinton	45	46
George and Babara Bush	55	64
Ronald and Nancy Reagan	59	69
Jimmy and Rosalyn Carter	49	52
Gerald and Betty Ford	56	61
Richard and Pat Nixon	56	56
Lyndon and Lady Bird Johnson	50	55

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find these constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations

$x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

$n=10$ number of observations

$x¯=51.5$ average of lady’s age $(x)$

$y¯=57.4$ average of president’s age $(y)$

$sx=5.148$ standard deviation of $x$

$sy=8.409$ standard deviation of $y$

$r=0.5159$ correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=0.5159×8.4095.148 =0.8426$

And,

$b0=y¯−b1x¯ =57.4−(0.8426×51.5) b0=14.0061b0≈14$

By substituting above formula,

$y=b0+b1xy=14+0.8426x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=14+0.8426x$

b.

Expert Solution

To determine

To find: The correlation coefficient of the two variables.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.5159$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

Then,

$x¯=51.5sx=5.148y¯=57.4sy=8.409$

Then, a table should be constructed to calculate $r$ as follows.

$Her Age x His Age y x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 46 70 −1.0684 1.4984 −1.6009 45 47 −1.2626 −1.2368 1.5616 54 54 0.4856 −0.4043 −0.1963 45 46 −1.2626 −1.3557 1.7117 55 64 0.6799 0.7849 0.5337 59 69 1.4569 1.3795 2.0098 49 52 −0.4856 −0.6422 0.3119 56 61 0.8741 0.4281 0.3742 56 56 0.8741 −0.1665 −0.1455 50 55 −0.2914 −0.2854 0.0832 ∑( x− x ¯ s x )( y− y ¯ s y )=4.6434$

The correlation coefficient can be calculated as,

$r=110−1×4.6434=4.64349r=0.5159$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found tobe,

$r=0.5159$

c.

Expert Solution

To determine

To graph: The scatter plot of the given two quantitative variables.

Explanation of Solution

Graph:

Let $x$ be the lady’s age and $y$ be the president’s age. The scatter plot can be constructed by the ordered pairs using the Excel.

Elementary Statistics, Chapter 4.3, Problem 24E

Interpretation:

Out of all these $10$ ordered pairs, there is a weak positive relationship between president’s age and lady’s age.

d.

Expert Solution

To determine

To Identify: The outliers within the given data.

Answer to Problem 24E

There is one outlier which is age $70$

Explanation of Solution

Explain:

Here, Excel is used to calculate below statistics.

$Q1$ is the middle value in the first half of the rank-ordered data set.

$Q2$ is the median value in the set.

$Q3$ is the middle value in the second half of the rank-ordered data set.

The interquartile range is equal to $Q3$ minus $Q1$ .

To calculate upper bound,

$Q3+(IQR×1.5)=Q3+(( Q 3 − Q 1 )×1.5)=56.75+(56.75−48.5)×1.5=56.75+8.25×1.5=56.75+12.375Q3+(IQR×1.5)=69.125$

To calculate lower bound,

$Q1−(IQR×1.5)=Q1−(( Q 3 − Q 1 )×1.5)=48.5−(56.75−48.5)×1.5=48.5−8.25×1.5=48.5−12.375Q1−(IQR×1.5)=36.125$

An outlier is a data point that lies outside the upper bound and lower bound.

$Q1Q3IQR Lower Bound Upper Bound48.556.758.2536.12569.125$

Out of all $10$ ordered pairs, only one value can be identified as an outlier, where the age is not between the lower bound and upper bound.

$70>69.125$

e.

Expert Solution

To determine

To find: The least-squares regression line for predicting the president’s age from lady’s age with removing outlier.

Answer to Problem 24E

The least-squares regression is,

$y=−14.7+1.3567x$

Explanation of Solution

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations $x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

When the outlier is removed, the number of ordered pairs is $9$ .

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

$n=9$ ; number of observations.

$x¯=52.11$ ; average of lady’s age $(x)$

$y¯=56$ ; average of president’s age $(y)$

$sx=5.061$ ; standard deviation of $x$

$sy=7.583$ ; standard deviation of $y$

$r=0.9055$ ; correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=(0.9055)7.5835.061 =1.3567$

And,

$b0=y¯−b1x¯ =56−(1.3567)×52.11 =−14.6976 =−14.7$

By substituting into the above formula,

$y=b0+b1xy=−14.7+1.3567x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=−14.7+1.3567x$

f.

Expert Solution

To determine

To Check: The influence of outlier.

Answer to Problem 24E

Yes. It is influenced.

Explanation of Solution

Explain:

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10
Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

To calculate above statistics, Excel is used. Here we can see statistics with outlier and without outlier. Second table shows statistics without outlier. Total number of observations, mean and standard deviation of dependent and independent variables are changed. So, it influences to least squares regression line for predicting the president’s age from lady’s age.

g.

Expert Solution

To determine

To find: The correlation coefficient of the two variables without outlier.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.9055$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

For the remaining $9$ pairs without the outlier,

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

Then,

$x¯=52.11sx=5.061y¯=56sy=7.583$

Then, a table should be constructed to calculate $r$ as follows.

$xy x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 45 47 −1.4051 −1.1869 1.6677 54 54 0.3732 −0.2637 −0.0984 45 46 −1.4051 −1.3187 1.8529 55 64 0.5708 1.0550 0.6022 59 69 1.3612 1.7144 2.3336 49 52 −0.6147 −0.5275 0.3243 56 61 0.7684 0.6594 0.5067 56 56 0.768400 50 55 −0.4171 −0.1319 0.0550 ∑( x− x ¯ s x )( y− y ¯ s y )=7.2438$

The correlation coefficient can be calculated as,

$r=19−1×7.2438=7.24388r=0.9055$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found to be,

$r=0.9055$

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Answer 2

Textbook Question

Chapter 4.3, Problem 24E

Presidents and first ladies: The presents the ages of the last 10 U.S. presidents and their wives on the first day of their presidencies.

Chapter 4.3, Problem 24E, Presidents and first ladies: The presents the ages of the last 10 U.S. presidents and their wives on

Compute the least-squares regression line for predicting the president’s age from the first lady’s age.
Compute the coefficient of determination-
Construct a scatterplot of the presidents' ages (y) versus the first ladies' ages (x).
Which point is an outlier?
Remove die outlier and compute the least-squares regression line for predicting the president’s age from the first lady: s age.
Is the outlier influential? Explain.
Compute the coefficient of determination for the data set with the outlier removed. Is due proportion of variation explained by due least-squares regression he greater, less. or about the same without due outlier? Explain.

a.

Expert Solution

To determine

To Calculate: The least-squares regression line for predicting the president’s age from lady’s age.

Answer to Problem 24E

The least-squares regression line is,

$y^=14+0.8426x$

Explanation of Solution

Given: The following table presents the age of the president’s and their wives on the first day of their presidencies.

Name	Her Age	His Age
Donald and Melania Trump	46	70
Barak and Mechelle Obama	45	47
George W. and Laura Bush	54	54
Bill and Hillary Clinton	45	46
George and Babara Bush	55	64
Ronald and Nancy Reagan	59	69
Jimmy and Rosalyn Carter	49	52
Gerald and Betty Ford	56	61
Richard and Pat Nixon	56	56
Lyndon and Lady Bird Johnson	50	55

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find these constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations

$x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

$n=10$ number of observations

$x¯=51.5$ average of lady’s age $(x)$

$y¯=57.4$ average of president’s age $(y)$

$sx=5.148$ standard deviation of $x$

$sy=8.409$ standard deviation of $y$

$r=0.5159$ correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=0.5159×8.4095.148 =0.8426$

And,

$b0=y¯−b1x¯ =57.4−(0.8426×51.5) b0=14.0061b0≈14$

By substituting above formula,

$y=b0+b1xy=14+0.8426x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=14+0.8426x$

b.

Expert Solution

To determine

To find: The correlation coefficient of the two variables.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.5159$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

Then,

$x¯=51.5sx=5.148y¯=57.4sy=8.409$

Then, a table should be constructed to calculate $r$ as follows.

$Her Age x His Age y x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 46 70 −1.0684 1.4984 −1.6009 45 47 −1.2626 −1.2368 1.5616 54 54 0.4856 −0.4043 −0.1963 45 46 −1.2626 −1.3557 1.7117 55 64 0.6799 0.7849 0.5337 59 69 1.4569 1.3795 2.0098 49 52 −0.4856 −0.6422 0.3119 56 61 0.8741 0.4281 0.3742 56 56 0.8741 −0.1665 −0.1455 50 55 −0.2914 −0.2854 0.0832 ∑( x− x ¯ s x )( y− y ¯ s y )=4.6434$

The correlation coefficient can be calculated as,

$r=110−1×4.6434=4.64349r=0.5159$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found tobe,

$r=0.5159$

c.

Expert Solution

To determine

To graph: The scatter plot of the given two quantitative variables.

Explanation of Solution

Graph:

Let $x$ be the lady’s age and $y$ be the president’s age. The scatter plot can be constructed by the ordered pairs using the Excel.

Elementary Statistics, Chapter 4.3, Problem 24E

Interpretation:

Out of all these $10$ ordered pairs, there is a weak positive relationship between president’s age and lady’s age.

d.

Expert Solution

To determine

To Identify: The outliers within the given data.

Answer to Problem 24E

There is one outlier which is age $70$

Explanation of Solution

Explain:

Here, Excel is used to calculate below statistics.

$Q1$ is the middle value in the first half of the rank-ordered data set.

$Q2$ is the median value in the set.

$Q3$ is the middle value in the second half of the rank-ordered data set.

The interquartile range is equal to $Q3$ minus $Q1$ .

To calculate upper bound,

$Q3+(IQR×1.5)=Q3+(( Q 3 − Q 1 )×1.5)=56.75+(56.75−48.5)×1.5=56.75+8.25×1.5=56.75+12.375Q3+(IQR×1.5)=69.125$

To calculate lower bound,

$Q1−(IQR×1.5)=Q1−(( Q 3 − Q 1 )×1.5)=48.5−(56.75−48.5)×1.5=48.5−8.25×1.5=48.5−12.375Q1−(IQR×1.5)=36.125$

An outlier is a data point that lies outside the upper bound and lower bound.

$Q1Q3IQR Lower Bound Upper Bound48.556.758.2536.12569.125$

Out of all $10$ ordered pairs, only one value can be identified as an outlier, where the age is not between the lower bound and upper bound.

$70>69.125$

e.

Expert Solution

To determine

To find: The least-squares regression line for predicting the president’s age from lady’s age with removing outlier.

Answer to Problem 24E

The least-squares regression is,

$y=−14.7+1.3567x$

Explanation of Solution

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations $x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

When the outlier is removed, the number of ordered pairs is $9$ .

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

$n=9$ ; number of observations.

$x¯=52.11$ ; average of lady’s age $(x)$

$y¯=56$ ; average of president’s age $(y)$

$sx=5.061$ ; standard deviation of $x$

$sy=7.583$ ; standard deviation of $y$

$r=0.9055$ ; correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=(0.9055)7.5835.061 =1.3567$

And,

$b0=y¯−b1x¯ =56−(1.3567)×52.11 =−14.6976 =−14.7$

By substituting into the above formula,

$y=b0+b1xy=−14.7+1.3567x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=−14.7+1.3567x$

f.

Expert Solution

To determine

To Check: The influence of outlier.

Answer to Problem 24E

Yes. It is influenced.

Explanation of Solution

Explain:

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10
Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

To calculate above statistics, Excel is used. Here we can see statistics with outlier and without outlier. Second table shows statistics without outlier. Total number of observations, mean and standard deviation of dependent and independent variables are changed. So, it influences to least squares regression line for predicting the president’s age from lady’s age.

g.

Expert Solution

To determine

To find: The correlation coefficient of the two variables without outlier.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.9055$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

For the remaining $9$ pairs without the outlier,

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

Then,

$x¯=52.11sx=5.061y¯=56sy=7.583$

Then, a table should be constructed to calculate $r$ as follows.

$xy x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 45 47 −1.4051 −1.1869 1.6677 54 54 0.3732 −0.2637 −0.0984 45 46 −1.4051 −1.3187 1.8529 55 64 0.5708 1.0550 0.6022 59 69 1.3612 1.7144 2.3336 49 52 −0.6147 −0.5275 0.3243 56 61 0.7684 0.6594 0.5067 56 56 0.768400 50 55 −0.4171 −0.1319 0.0550 ∑( x− x ¯ s x )( y− y ¯ s y )=7.2438$

The correlation coefficient can be calculated as,

$r=19−1×7.2438=7.24388r=0.9055$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found to be,

$r=0.9055$

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Answer 3

Textbook Question

Chapter 4.3, Problem 24E

Presidents and first ladies: The presents the ages of the last 10 U.S. presidents and their wives on the first day of their presidencies.

Chapter 4.3, Problem 24E, Presidents and first ladies: The presents the ages of the last 10 U.S. presidents and their wives on

Compute the least-squares regression line for predicting the president’s age from the first lady’s age.
Compute the coefficient of determination-
Construct a scatterplot of the presidents' ages (y) versus the first ladies' ages (x).
Which point is an outlier?
Remove die outlier and compute the least-squares regression line for predicting the president’s age from the first lady: s age.
Is the outlier influential? Explain.
Compute the coefficient of determination for the data set with the outlier removed. Is due proportion of variation explained by due least-squares regression he greater, less. or about the same without due outlier? Explain.

a.

Expert Solution

To determine

To Calculate: The least-squares regression line for predicting the president’s age from lady’s age.

Answer to Problem 24E

The least-squares regression line is,

$y^=14+0.8426x$

Explanation of Solution

Given: The following table presents the age of the president’s and their wives on the first day of their presidencies.

Name	Her Age	His Age
Donald and Melania Trump	46	70
Barak and Mechelle Obama	45	47
George W. and Laura Bush	54	54
Bill and Hillary Clinton	45	46
George and Babara Bush	55	64
Ronald and Nancy Reagan	59	69
Jimmy and Rosalyn Carter	49	52
Gerald and Betty Ford	56	61
Richard and Pat Nixon	56	56
Lyndon and Lady Bird Johnson	50	55

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find these constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations

$x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

$n=10$ number of observations

$x¯=51.5$ average of lady’s age $(x)$

$y¯=57.4$ average of president’s age $(y)$

$sx=5.148$ standard deviation of $x$

$sy=8.409$ standard deviation of $y$

$r=0.5159$ correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=0.5159×8.4095.148 =0.8426$

And,

$b0=y¯−b1x¯ =57.4−(0.8426×51.5) b0=14.0061b0≈14$

By substituting above formula,

$y=b0+b1xy=14+0.8426x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=14+0.8426x$

b.

Expert Solution

To determine

To find: The correlation coefficient of the two variables.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.5159$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

Then,

$x¯=51.5sx=5.148y¯=57.4sy=8.409$

Then, a table should be constructed to calculate $r$ as follows.

$Her Age x His Age y x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 46 70 −1.0684 1.4984 −1.6009 45 47 −1.2626 −1.2368 1.5616 54 54 0.4856 −0.4043 −0.1963 45 46 −1.2626 −1.3557 1.7117 55 64 0.6799 0.7849 0.5337 59 69 1.4569 1.3795 2.0098 49 52 −0.4856 −0.6422 0.3119 56 61 0.8741 0.4281 0.3742 56 56 0.8741 −0.1665 −0.1455 50 55 −0.2914 −0.2854 0.0832 ∑( x− x ¯ s x )( y− y ¯ s y )=4.6434$

The correlation coefficient can be calculated as,

$r=110−1×4.6434=4.64349r=0.5159$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found tobe,

$r=0.5159$

c.

Expert Solution

To determine

To graph: The scatter plot of the given two quantitative variables.

Explanation of Solution

Graph:

Let $x$ be the lady’s age and $y$ be the president’s age. The scatter plot can be constructed by the ordered pairs using the Excel.

Elementary Statistics, Chapter 4.3, Problem 24E

Interpretation:

Out of all these $10$ ordered pairs, there is a weak positive relationship between president’s age and lady’s age.

d.

Expert Solution

To determine

To Identify: The outliers within the given data.

Answer to Problem 24E

There is one outlier which is age $70$

Explanation of Solution

Explain:

Here, Excel is used to calculate below statistics.

$Q1$ is the middle value in the first half of the rank-ordered data set.

$Q2$ is the median value in the set.

$Q3$ is the middle value in the second half of the rank-ordered data set.

The interquartile range is equal to $Q3$ minus $Q1$ .

To calculate upper bound,

$Q3+(IQR×1.5)=Q3+(( Q 3 − Q 1 )×1.5)=56.75+(56.75−48.5)×1.5=56.75+8.25×1.5=56.75+12.375Q3+(IQR×1.5)=69.125$

To calculate lower bound,

$Q1−(IQR×1.5)=Q1−(( Q 3 − Q 1 )×1.5)=48.5−(56.75−48.5)×1.5=48.5−8.25×1.5=48.5−12.375Q1−(IQR×1.5)=36.125$

An outlier is a data point that lies outside the upper bound and lower bound.

$Q1Q3IQR Lower Bound Upper Bound48.556.758.2536.12569.125$

Out of all $10$ ordered pairs, only one value can be identified as an outlier, where the age is not between the lower bound and upper bound.

$70>69.125$

e.

Expert Solution

To determine

To find: The least-squares regression line for predicting the president’s age from lady’s age with removing outlier.

Answer to Problem 24E

The least-squares regression is,

$y=−14.7+1.3567x$

Explanation of Solution

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations $x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

When the outlier is removed, the number of ordered pairs is $9$ .

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

$n=9$ ; number of observations.

$x¯=52.11$ ; average of lady’s age $(x)$

$y¯=56$ ; average of president’s age $(y)$

$sx=5.061$ ; standard deviation of $x$

$sy=7.583$ ; standard deviation of $y$

$r=0.9055$ ; correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=(0.9055)7.5835.061 =1.3567$

And,

$b0=y¯−b1x¯ =56−(1.3567)×52.11 =−14.6976 =−14.7$

By substituting into the above formula,

$y=b0+b1xy=−14.7+1.3567x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=−14.7+1.3567x$

f.

Expert Solution

To determine

To Check: The influence of outlier.

Answer to Problem 24E

Yes. It is influenced.

Explanation of Solution

Explain:

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10
Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

To calculate above statistics, Excel is used. Here we can see statistics with outlier and without outlier. Second table shows statistics without outlier. Total number of observations, mean and standard deviation of dependent and independent variables are changed. So, it influences to least squares regression line for predicting the president’s age from lady’s age.

g.

Expert Solution

To determine

To find: The correlation coefficient of the two variables without outlier.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.9055$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

For the remaining $9$ pairs without the outlier,

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

Then,

$x¯=52.11sx=5.061y¯=56sy=7.583$

Then, a table should be constructed to calculate $r$ as follows.

$xy x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 45 47 −1.4051 −1.1869 1.6677 54 54 0.3732 −0.2637 −0.0984 45 46 −1.4051 −1.3187 1.8529 55 64 0.5708 1.0550 0.6022 59 69 1.3612 1.7144 2.3336 49 52 −0.6147 −0.5275 0.3243 56 61 0.7684 0.6594 0.5067 56 56 0.768400 50 55 −0.4171 −0.1319 0.0550 ∑( x− x ¯ s x )( y− y ¯ s y )=7.2438$

The correlation coefficient can be calculated as,

$r=19−1×7.2438=7.24388r=0.9055$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found to be,

$r=0.9055$

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Answer 4

Textbook Question

Chapter 4.3, Problem 24E

Presidents and first ladies: The presents the ages of the last 10 U.S. presidents and their wives on the first day of their presidencies.

Chapter 4.3, Problem 24E, Presidents and first ladies: The presents the ages of the last 10 U.S. presidents and their wives on

Compute the least-squares regression line for predicting the president’s age from the first lady’s age.
Compute the coefficient of determination-
Construct a scatterplot of the presidents' ages (y) versus the first ladies' ages (x).
Which point is an outlier?
Remove die outlier and compute the least-squares regression line for predicting the president’s age from the first lady: s age.
Is the outlier influential? Explain.
Compute the coefficient of determination for the data set with the outlier removed. Is due proportion of variation explained by due least-squares regression he greater, less. or about the same without due outlier? Explain.

a.

Expert Solution

To determine

To Calculate: The least-squares regression line for predicting the president’s age from lady’s age.

Answer to Problem 24E

The least-squares regression line is,

$y^=14+0.8426x$

Explanation of Solution

Given: The following table presents the age of the president’s and their wives on the first day of their presidencies.

Name	Her Age	His Age
Donald and Melania Trump	46	70
Barak and Mechelle Obama	45	47
George W. and Laura Bush	54	54
Bill and Hillary Clinton	45	46
George and Babara Bush	55	64
Ronald and Nancy Reagan	59	69
Jimmy and Rosalyn Carter	49	52
Gerald and Betty Ford	56	61
Richard and Pat Nixon	56	56
Lyndon and Lady Bird Johnson	50	55

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find these constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations

$x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

$n=10$ number of observations

$x¯=51.5$ average of lady’s age $(x)$

$y¯=57.4$ average of president’s age $(y)$

$sx=5.148$ standard deviation of $x$

$sy=8.409$ standard deviation of $y$

$r=0.5159$ correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=0.5159×8.4095.148 =0.8426$

And,

$b0=y¯−b1x¯ =57.4−(0.8426×51.5) b0=14.0061b0≈14$

By substituting above formula,

$y=b0+b1xy=14+0.8426x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=14+0.8426x$

b.

Expert Solution

To determine

To find: The correlation coefficient of the two variables.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.5159$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

Then,

$x¯=51.5sx=5.148y¯=57.4sy=8.409$

Then, a table should be constructed to calculate $r$ as follows.

$Her Age x His Age y x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 46 70 −1.0684 1.4984 −1.6009 45 47 −1.2626 −1.2368 1.5616 54 54 0.4856 −0.4043 −0.1963 45 46 −1.2626 −1.3557 1.7117 55 64 0.6799 0.7849 0.5337 59 69 1.4569 1.3795 2.0098 49 52 −0.4856 −0.6422 0.3119 56 61 0.8741 0.4281 0.3742 56 56 0.8741 −0.1665 −0.1455 50 55 −0.2914 −0.2854 0.0832 ∑( x− x ¯ s x )( y− y ¯ s y )=4.6434$

The correlation coefficient can be calculated as,

$r=110−1×4.6434=4.64349r=0.5159$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found tobe,

$r=0.5159$

c.

Expert Solution

To determine

To graph: The scatter plot of the given two quantitative variables.

Explanation of Solution

Graph:

Let $x$ be the lady’s age and $y$ be the president’s age. The scatter plot can be constructed by the ordered pairs using the Excel.

Elementary Statistics, Chapter 4.3, Problem 24E

Interpretation:

Out of all these $10$ ordered pairs, there is a weak positive relationship between president’s age and lady’s age.

d.

Expert Solution

To determine

To Identify: The outliers within the given data.

Answer to Problem 24E

There is one outlier which is age $70$

Explanation of Solution

Explain:

Here, Excel is used to calculate below statistics.

$Q1$ is the middle value in the first half of the rank-ordered data set.

$Q2$ is the median value in the set.

$Q3$ is the middle value in the second half of the rank-ordered data set.

The interquartile range is equal to $Q3$ minus $Q1$ .

To calculate upper bound,

$Q3+(IQR×1.5)=Q3+(( Q 3 − Q 1 )×1.5)=56.75+(56.75−48.5)×1.5=56.75+8.25×1.5=56.75+12.375Q3+(IQR×1.5)=69.125$

To calculate lower bound,

$Q1−(IQR×1.5)=Q1−(( Q 3 − Q 1 )×1.5)=48.5−(56.75−48.5)×1.5=48.5−8.25×1.5=48.5−12.375Q1−(IQR×1.5)=36.125$

An outlier is a data point that lies outside the upper bound and lower bound.

$Q1Q3IQR Lower Bound Upper Bound48.556.758.2536.12569.125$

Out of all $10$ ordered pairs, only one value can be identified as an outlier, where the age is not between the lower bound and upper bound.

$70>69.125$

e.

Expert Solution

To determine

To find: The least-squares regression line for predicting the president’s age from lady’s age with removing outlier.

Answer to Problem 24E

The least-squares regression is,

$y=−14.7+1.3567x$

Explanation of Solution

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations $x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

When the outlier is removed, the number of ordered pairs is $9$ .

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

$n=9$ ; number of observations.

$x¯=52.11$ ; average of lady’s age $(x)$

$y¯=56$ ; average of president’s age $(y)$

$sx=5.061$ ; standard deviation of $x$

$sy=7.583$ ; standard deviation of $y$

$r=0.9055$ ; correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=(0.9055)7.5835.061 =1.3567$

And,

$b0=y¯−b1x¯ =56−(1.3567)×52.11 =−14.6976 =−14.7$

By substituting into the above formula,

$y=b0+b1xy=−14.7+1.3567x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=−14.7+1.3567x$

f.

Expert Solution

To determine

To Check: The influence of outlier.

Answer to Problem 24E

Yes. It is influenced.

Explanation of Solution

Explain:

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10
Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

To calculate above statistics, Excel is used. Here we can see statistics with outlier and without outlier. Second table shows statistics without outlier. Total number of observations, mean and standard deviation of dependent and independent variables are changed. So, it influences to least squares regression line for predicting the president’s age from lady’s age.

g.

Expert Solution

To determine

To find: The correlation coefficient of the two variables without outlier.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.9055$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

For the remaining $9$ pairs without the outlier,

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

Then,

$x¯=52.11sx=5.061y¯=56sy=7.583$

Then, a table should be constructed to calculate $r$ as follows.

$xy x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 45 47 −1.4051 −1.1869 1.6677 54 54 0.3732 −0.2637 −0.0984 45 46 −1.4051 −1.3187 1.8529 55 64 0.5708 1.0550 0.6022 59 69 1.3612 1.7144 2.3336 49 52 −0.6147 −0.5275 0.3243 56 61 0.7684 0.6594 0.5067 56 56 0.768400 50 55 −0.4171 −0.1319 0.0550 ∑( x− x ¯ s x )( y− y ¯ s y )=7.2438$

The correlation coefficient can be calculated as,

$r=19−1×7.2438=7.24388r=0.9055$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found to be,

$r=0.9055$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

To Calculate: The least-squares regression line for predicting the president’s age from lady’s age.

Answer to Problem 24E

The least-squares regression line is,

$y^=14+0.8426x$

Explanation of Solution

Given: The following table presents the age of the president’s and their wives on the first day of their presidencies.

Name	Her Age	His Age
Donald and Melania Trump	46	70
Barak and Mechelle Obama	45	47
George W. and Laura Bush	54	54
Bill and Hillary Clinton	45	46
George and Babara Bush	55	64
Ronald and Nancy Reagan	59	69
Jimmy and Rosalyn Carter	49	52
Gerald and Betty Ford	56	61
Richard and Pat Nixon	56	56
Lyndon and Lady Bird Johnson	50	55

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find these constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations

$x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

$n=10$ number of observations

$x¯=51.5$ average of lady’s age $(x)$

$y¯=57.4$ average of president’s age $(y)$

$sx=5.148$ standard deviation of $x$

$sy=8.409$ standard deviation of $y$

$r=0.5159$ correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=0.5159×8.4095.148 =0.8426$

And,

$b0=y¯−b1x¯ =57.4−(0.8426×51.5) b0=14.0061b0≈14$

By substituting above formula,

$y=b0+b1xy=14+0.8426x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=14+0.8426x$

b.

Expert Solution

To determine

To find: The correlation coefficient of the two variables.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.5159$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

Then,

$x¯=51.5sx=5.148y¯=57.4sy=8.409$

Then, a table should be constructed to calculate $r$ as follows.

$Her Age x His Age y x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 46 70 −1.0684 1.4984 −1.6009 45 47 −1.2626 −1.2368 1.5616 54 54 0.4856 −0.4043 −0.1963 45 46 −1.2626 −1.3557 1.7117 55 64 0.6799 0.7849 0.5337 59 69 1.4569 1.3795 2.0098 49 52 −0.4856 −0.6422 0.3119 56 61 0.8741 0.4281 0.3742 56 56 0.8741 −0.1665 −0.1455 50 55 −0.2914 −0.2854 0.0832 ∑( x− x ¯ s x )( y− y ¯ s y )=4.6434$

The correlation coefficient can be calculated as,

$r=110−1×4.6434=4.64349r=0.5159$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found tobe,

$r=0.5159$

c.

Expert Solution

To determine

To graph: The scatter plot of the given two quantitative variables.

Explanation of Solution

Graph:

Let $x$ be the lady’s age and $y$ be the president’s age. The scatter plot can be constructed by the ordered pairs using the Excel.

Elementary Statistics, Chapter 4.3, Problem 24E

Interpretation:

Out of all these $10$ ordered pairs, there is a weak positive relationship between president’s age and lady’s age.

d.

Expert Solution

To determine

To Identify: The outliers within the given data.

Answer to Problem 24E

There is one outlier which is age $70$

Explanation of Solution

Explain:

Here, Excel is used to calculate below statistics.

$Q1$ is the middle value in the first half of the rank-ordered data set.

$Q2$ is the median value in the set.

$Q3$ is the middle value in the second half of the rank-ordered data set.

The interquartile range is equal to $Q3$ minus $Q1$ .

To calculate upper bound,

$Q3+(IQR×1.5)=Q3+(( Q 3 − Q 1 )×1.5)=56.75+(56.75−48.5)×1.5=56.75+8.25×1.5=56.75+12.375Q3+(IQR×1.5)=69.125$

To calculate lower bound,

$Q1−(IQR×1.5)=Q1−(( Q 3 − Q 1 )×1.5)=48.5−(56.75−48.5)×1.5=48.5−8.25×1.5=48.5−12.375Q1−(IQR×1.5)=36.125$

An outlier is a data point that lies outside the upper bound and lower bound.

$Q1Q3IQR Lower Bound Upper Bound48.556.758.2536.12569.125$

Out of all $10$ ordered pairs, only one value can be identified as an outlier, where the age is not between the lower bound and upper bound.

$70>69.125$

e.

Expert Solution

To determine

To find: The least-squares regression line for predicting the president’s age from lady’s age with removing outlier.

Answer to Problem 24E

The least-squares regression is,

$y=−14.7+1.3567x$

Explanation of Solution

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations $x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

When the outlier is removed, the number of ordered pairs is $9$ .

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

$n=9$ ; number of observations.

$x¯=52.11$ ; average of lady’s age $(x)$

$y¯=56$ ; average of president’s age $(y)$

$sx=5.061$ ; standard deviation of $x$

$sy=7.583$ ; standard deviation of $y$

$r=0.9055$ ; correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=(0.9055)7.5835.061 =1.3567$

And,

$b0=y¯−b1x¯ =56−(1.3567)×52.11 =−14.6976 =−14.7$

By substituting into the above formula,

$y=b0+b1xy=−14.7+1.3567x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=−14.7+1.3567x$

f.

Expert Solution

To determine

To Check: The influence of outlier.

Answer to Problem 24E

Yes. It is influenced.

Explanation of Solution

Explain:

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10
Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

To calculate above statistics, Excel is used. Here we can see statistics with outlier and without outlier. Second table shows statistics without outlier. Total number of observations, mean and standard deviation of dependent and independent variables are changed. So, it influences to least squares regression line for predicting the president’s age from lady’s age.

g.

Expert Solution

To determine

To find: The correlation coefficient of the two variables without outlier.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.9055$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

For the remaining $9$ pairs without the outlier,

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

Then,

$x¯=52.11sx=5.061y¯=56sy=7.583$

Then, a table should be constructed to calculate $r$ as follows.

$xy x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 45 47 −1.4051 −1.1869 1.6677 54 54 0.3732 −0.2637 −0.0984 45 46 −1.4051 −1.3187 1.8529 55 64 0.5708 1.0550 0.6022 59 69 1.3612 1.7144 2.3336 49 52 −0.6147 −0.5275 0.3243 56 61 0.7684 0.6594 0.5067 56 56 0.768400 50 55 −0.4171 −0.1319 0.0550 ∑( x− x ¯ s x )( y− y ¯ s y )=7.2438$

The correlation coefficient can be calculated as,

$r=19−1×7.2438=7.24388r=0.9055$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found to be,

$r=0.9055$

Answer 8

a.

Expert Solution

To determine

To Calculate: The least-squares regression line for predicting the president’s age from lady’s age.

Answer to Problem 24E

The least-squares regression line is,

$y^=14+0.8426x$

Explanation of Solution

Given: The following table presents the age of the president’s and their wives on the first day of their presidencies.

Name	Her Age	His Age
Donald and Melania Trump	46	70
Barak and Mechelle Obama	45	47
George W. and Laura Bush	54	54
Bill and Hillary Clinton	45	46
George and Babara Bush	55	64
Ronald and Nancy Reagan	59	69
Jimmy and Rosalyn Carter	49	52
Gerald and Betty Ford	56	61
Richard and Pat Nixon	56	56
Lyndon and Lady Bird Johnson	50	55

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find these constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations

$x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

$n=10$ number of observations

$x¯=51.5$ average of lady’s age $(x)$

$y¯=57.4$ average of president’s age $(y)$

$sx=5.148$ standard deviation of $x$

$sy=8.409$ standard deviation of $y$

$r=0.5159$ correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=0.5159×8.4095.148 =0.8426$

And,

$b0=y¯−b1x¯ =57.4−(0.8426×51.5) b0=14.0061b0≈14$

By substituting above formula,

$y=b0+b1xy=14+0.8426x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=14+0.8426x$

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To Calculate: The least-squares regression line for predicting the president’s age from lady’s age.

Answer 13

Answer to Problem 24E

The least-squares regression line is,

$y^=14+0.8426x$

Answer 14

Explanation of Solution

Given: The following table presents the age of the president’s and their wives on the first day of their presidencies.

Name	Her Age	His Age
Donald and Melania Trump	46	70
Barak and Mechelle Obama	45	47
George W. and Laura Bush	54	54
Bill and Hillary Clinton	45	46
George and Babara Bush	55	64
Ronald and Nancy Reagan	59	69
Jimmy and Rosalyn Carter	49	52
Gerald and Betty Ford	56	61
Richard and Pat Nixon	56	56
Lyndon and Lady Bird Johnson	50	55

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find these constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations

$x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

$n=10$ number of observations

$x¯=51.5$ average of lady’s age $(x)$

$y¯=57.4$ average of president’s age $(y)$

$sx=5.148$ standard deviation of $x$

$sy=8.409$ standard deviation of $y$

$r=0.5159$ correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=0.5159×8.4095.148 =0.8426$

And,

$b0=y¯−b1x¯ =57.4−(0.8426×51.5) b0=14.0061b0≈14$

By substituting above formula,

$y=b0+b1xy=14+0.8426x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=14+0.8426x$

Answer 15

b.

Expert Solution

To determine

To find: The correlation coefficient of the two variables.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.5159$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

Then,

$x¯=51.5sx=5.148y¯=57.4sy=8.409$

Then, a table should be constructed to calculate $r$ as follows.

$Her Age x His Age y x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 46 70 −1.0684 1.4984 −1.6009 45 47 −1.2626 −1.2368 1.5616 54 54 0.4856 −0.4043 −0.1963 45 46 −1.2626 −1.3557 1.7117 55 64 0.6799 0.7849 0.5337 59 69 1.4569 1.3795 2.0098 49 52 −0.4856 −0.6422 0.3119 56 61 0.8741 0.4281 0.3742 56 56 0.8741 −0.1665 −0.1455 50 55 −0.2914 −0.2854 0.0832 ∑( x− x ¯ s x )( y− y ¯ s y )=4.6434$

The correlation coefficient can be calculated as,

$r=110−1×4.6434=4.64349r=0.5159$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found tobe,

$r=0.5159$

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To find: The correlation coefficient of the two variables.

Answer 20

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.5159$

Answer 21

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10

Then,

$x¯=51.5sx=5.148y¯=57.4sy=8.409$

Then, a table should be constructed to calculate $r$ as follows.

$Her Age x His Age y x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 46 70 −1.0684 1.4984 −1.6009 45 47 −1.2626 −1.2368 1.5616 54 54 0.4856 −0.4043 −0.1963 45 46 −1.2626 −1.3557 1.7117 55 64 0.6799 0.7849 0.5337 59 69 1.4569 1.3795 2.0098 49 52 −0.4856 −0.6422 0.3119 56 61 0.8741 0.4281 0.3742 56 56 0.8741 −0.1665 −0.1455 50 55 −0.2914 −0.2854 0.0832 ∑( x− x ¯ s x )( y− y ¯ s y )=4.6434$

The correlation coefficient can be calculated as,

$r=110−1×4.6434=4.64349r=0.5159$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found tobe,

$r=0.5159$

Answer 22

c.

Expert Solution

To determine

To graph: The scatter plot of the given two quantitative variables.

Explanation of Solution

Graph:

Let $x$ be the lady’s age and $y$ be the president’s age. The scatter plot can be constructed by the ordered pairs using the Excel.

Elementary Statistics, Chapter 4.3, Problem 24E

Interpretation:

Out of all these $10$ ordered pairs, there is a weak positive relationship between president’s age and lady’s age.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

To graph: The scatter plot of the given two quantitative variables.

Answer 27

Explanation of Solution

Graph:

Let $x$ be the lady’s age and $y$ be the president’s age. The scatter plot can be constructed by the ordered pairs using the Excel.

Elementary Statistics, Chapter 4.3, Problem 24E

Interpretation:

Out of all these $10$ ordered pairs, there is a weak positive relationship between president’s age and lady’s age.

Answer 28

d.

Expert Solution

To determine

To Identify: The outliers within the given data.

Answer to Problem 24E

There is one outlier which is age $70$

Explanation of Solution

Explain:

Here, Excel is used to calculate below statistics.

$Q1$ is the middle value in the first half of the rank-ordered data set.

$Q2$ is the median value in the set.

$Q3$ is the middle value in the second half of the rank-ordered data set.

The interquartile range is equal to $Q3$ minus $Q1$ .

To calculate upper bound,

$Q3+(IQR×1.5)=Q3+(( Q 3 − Q 1 )×1.5)=56.75+(56.75−48.5)×1.5=56.75+8.25×1.5=56.75+12.375Q3+(IQR×1.5)=69.125$

To calculate lower bound,

$Q1−(IQR×1.5)=Q1−(( Q 3 − Q 1 )×1.5)=48.5−(56.75−48.5)×1.5=48.5−8.25×1.5=48.5−12.375Q1−(IQR×1.5)=36.125$

An outlier is a data point that lies outside the upper bound and lower bound.

$Q1Q3IQR Lower Bound Upper Bound48.556.758.2536.12569.125$

Out of all $10$ ordered pairs, only one value can be identified as an outlier, where the age is not between the lower bound and upper bound.

$70>69.125$

Answer 29

d.

Expert Solution

Answer 30

d.

Expert Solution

Answer 31

Expert Solution

Answer 32

To determine

To Identify: The outliers within the given data.

Answer 33

Answer to Problem 24E

There is one outlier which is age $70$

Answer 34

Explanation of Solution

Explain:

Here, Excel is used to calculate below statistics.

$Q1$ is the middle value in the first half of the rank-ordered data set.

$Q2$ is the median value in the set.

$Q3$ is the middle value in the second half of the rank-ordered data set.

The interquartile range is equal to $Q3$ minus $Q1$ .

To calculate upper bound,

$Q3+(IQR×1.5)=Q3+(( Q 3 − Q 1 )×1.5)=56.75+(56.75−48.5)×1.5=56.75+8.25×1.5=56.75+12.375Q3+(IQR×1.5)=69.125$

To calculate lower bound,

$Q1−(IQR×1.5)=Q1−(( Q 3 − Q 1 )×1.5)=48.5−(56.75−48.5)×1.5=48.5−8.25×1.5=48.5−12.375Q1−(IQR×1.5)=36.125$

An outlier is a data point that lies outside the upper bound and lower bound.

$Q1Q3IQR Lower Bound Upper Bound48.556.758.2536.12569.125$

Out of all $10$ ordered pairs, only one value can be identified as an outlier, where the age is not between the lower bound and upper bound.

$70>69.125$

Answer 35

e.

Expert Solution

To determine

To find: The least-squares regression line for predicting the president’s age from lady’s age with removing outlier.

Answer to Problem 24E

The least-squares regression is,

$y=−14.7+1.3567x$

Explanation of Solution

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations $x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

When the outlier is removed, the number of ordered pairs is $9$ .

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

$n=9$ ; number of observations.

$x¯=52.11$ ; average of lady’s age $(x)$

$y¯=56$ ; average of president’s age $(y)$

$sx=5.061$ ; standard deviation of $x$

$sy=7.583$ ; standard deviation of $y$

$r=0.9055$ ; correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=(0.9055)7.5835.061 =1.3567$

And,

$b0=y¯−b1x¯ =56−(1.3567)×52.11 =−14.6976 =−14.7$

By substituting into the above formula,

$y=b0+b1xy=−14.7+1.3567x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=−14.7+1.3567x$

Answer 36

e.

Expert Solution

Answer 37

e.

Expert Solution

Answer 38

Expert Solution

Answer 39

To determine

To find: The least-squares regression line for predicting the president’s age from lady’s age with removing outlier.

Answer 40

Answer to Problem 24E

The least-squares regression is,

$y=−14.7+1.3567x$

Answer 41

Explanation of Solution

Calculation:

Here,

$y$ = president’s age

$x$ = lady’s age

From below formula we can find least regression line.

$y^=b0+b1x$

where $b0$ and $b1$ are constants.

We can find constants from below formulas.

$b1=rsysxb0=y¯−b1x¯r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where,

$n$ is the number of observations $x¯$ is the average of $x$

$y¯$ is the average of $y$

$sx$ is the standard deviation of $x$

$sy$ is the standard deviation of $y$

When the outlier is removed, the number of ordered pairs is $9$ .

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

$n=9$ ; number of observations.

$x¯=52.11$ ; average of lady’s age $(x)$

$y¯=56$ ; average of president’s age $(y)$

$sx=5.061$ ; standard deviation of $x$

$sy=7.583$ ; standard deviation of $y$

$r=0.9055$ ; correlation between $x$ and $y$

To find constants,

$b1=rsysxb1=(0.9055)7.5835.061 =1.3567$

And,

$b0=y¯−b1x¯ =56−(1.3567)×52.11 =−14.6976 =−14.7$

By substituting into the above formula,

$y=b0+b1xy=−14.7+1.3567x$

Conclusion:

The least-squares regression line for predicting the president’s age from lady’s age is,

$y=−14.7+1.3567x$

Answer 42

f.

Expert Solution

To determine

To Check: The influence of outlier.

Answer to Problem 24E

Yes. It is influenced.

Explanation of Solution

Explain:

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10
Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

To calculate above statistics, Excel is used. Here we can see statistics with outlier and without outlier. Second table shows statistics without outlier. Total number of observations, mean and standard deviation of dependent and independent variables are changed. So, it influences to least squares regression line for predicting the president’s age from lady’s age.

Answer 43

f.

Expert Solution

Answer 44

f.

Expert Solution

Answer 45

Expert Solution

Answer 46

To determine

To Check: The influence of outlier.

Answer 47

Answer to Problem 24E

Yes. It is influenced.

Answer 48

Explanation of Solution

Explain:

Descriptive Statistics
	N	Mean	Std. Deviation
His age	10	57.40	8.409
Her age	10	51.50	5.148
Valid N (listwise)	10
Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

To calculate above statistics, Excel is used. Here we can see statistics with outlier and without outlier. Second table shows statistics without outlier. Total number of observations, mean and standard deviation of dependent and independent variables are changed. So, it influences to least squares regression line for predicting the president’s age from lady’s age.

Answer 49

g.

Expert Solution

To determine

To find: The correlation coefficient of the two variables without outlier.

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.9055$

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

For the remaining $9$ pairs without the outlier,

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

Then,

$x¯=52.11sx=5.061y¯=56sy=7.583$

Then, a table should be constructed to calculate $r$ as follows.

$xy x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 45 47 −1.4051 −1.1869 1.6677 54 54 0.3732 −0.2637 −0.0984 45 46 −1.4051 −1.3187 1.8529 55 64 0.5708 1.0550 0.6022 59 69 1.3612 1.7144 2.3336 49 52 −0.6147 −0.5275 0.3243 56 61 0.7684 0.6594 0.5067 56 56 0.768400 50 55 −0.4171 −0.1319 0.0550 ∑( x− x ¯ s x )( y− y ¯ s y )=7.2438$

The correlation coefficient can be calculated as,

$r=19−1×7.2438=7.24388r=0.9055$

Conclusion:

The correlation coefficient between the interest rates for two mortgage plans is found to be,

$r=0.9055$

Answer 50

g.

Expert Solution

Answer 51

g.

Expert Solution

Answer 52

Expert Solution

Answer 53

To determine

To find: The correlation coefficient of the two variables without outlier.

Answer 54

Answer to Problem 24E

The correlation coefficient is found to be,

$r=0.9055$

Answer 55

Explanation of Solution

Calculation:

The correlation coefficient $(r)$ is given by the formula,

$r=1n−1∑( x− x ¯ s x )( y− y ¯ s y )$

Where $sx$ and $sy$ is the standard deviation of $x$ and $y$ .

The means and the standard deviations of the both variables can be obtained by using the Excel.

For the remaining $9$ pairs without the outlier,

Descriptive Statistics
	N	Mean	Std. Deviation
His age	9	56.00	7.583
Her age	9	52.11	5.061
Valid N (listwise)	9

Then,

$x¯=52.11sx=5.061y¯=56sy=7.583$

Then, a table should be constructed to calculate $r$ as follows.

$xy x− x ¯ s x y− y ¯ s y ( x− x ¯ s x )( y− y ¯ s y ) 45 47 −1.4051 −1.1869 1.6677 54 54 0.3732 −0.2637 −0.0984 45 46 −1.4051 −1.3187 1.8529 55 64 0.5708 1.0550 0.6022 59 69 1.3612 1.7144 2.3336 49 52 −0.6147 −0.5275 0.3243 56 61 0.7684 0.6594 0.5067 56 56 0.768400 50 55 −0.4171 −0.1319 0.0550 ∑( x− x ¯ s x )( y− y ¯ s y )=7.2438$