Elementary Statistics ( 3rd International Edition ) Isbn:9781260092561
Elementary Statistics ( 3rd International Edition ) Isbn:9781260092561
3rd Edition
ISBN: 9781259969454
Author: William Navidi Prof.; Barry Monk Professor
Publisher: McGraw-Hill Education
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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

  1. Compute the least-squares regression line for predicting the president’s age from the first lady’s age.
  2. Compute the coefficient of determination-
  3. Construct a scatterplot of the presidents' ages (y) versus the first ladies' ages (x).
  4. Which point is an outlier?
  5. Remove die outlier and compute the least-squares regression line for predicting the president’s age from the first lady: s age.
  6. Is the outlier influential? Explain.
  7. 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
Check Mark
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.

    NameHer AgeHis Age
    Donald and Melania Trump4670
    Barak and Mechelle Obama4547
    George W. and Laura Bush5454
    Bill and Hillary Clinton4546
    George and Babara Bush5564
    Ronald and Nancy Reagan5969
    Jimmy and Rosalyn Carter4952
    Gerald and Betty Ford5661
    Richard and Pat Nixon5656
    Lyndon and Lady Bird Johnson5055

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=1n1( 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
    NMeanStd. Deviation
    His age1057.408.409
    Her age1051.505.148
    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=14.0061b014

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
Check Mark
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=1n1( 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
    NMeanStd. Deviation
    His age1057.408.409
    Her age1051.505.148
    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.

   HerAge x HisAge 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=1101×4.6434=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
Check Mark
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 ( 3rd International Edition ) Isbn:9781260092561, 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
Check Mark
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+(56.7548.5)×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(56.7548.5)×1.5=48.58.25×1.5=48.512.375Q1(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
Check Mark
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=1n1( x x ¯ s x )( y y ¯ s y )

Where,

  n is the number of observationsx¯ 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
    NMeanStd. Deviation
    His age956.007.583
    Her age952.115.061
    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
Check Mark
To determine

To Check: The influence of outlier.

Answer to Problem 24E

Yes. It is influenced.

Explanation of Solution

Explain:

    Descriptive Statistics
    NMeanStd. Deviation
    His age1057.408.409
    Her age1051.505.148
    Valid N (listwise)10
    Descriptive Statistics
    NMeanStd. Deviation
    His age956.007.583
    Her age952.115.061
    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
Check Mark
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=1n1( 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
    NMeanStd. Deviation
    His age956.007.583
    Her age952.115.061
    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=191×7.2438=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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Chapter 4 Solutions

Elementary Statistics ( 3rd International Edition ) Isbn:9781260092561

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