EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
6th Edition
ISBN: 9781319287573
Author: Starnes
Publisher: MPS PUB
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Chapter 3.2, Problem 57E

(a)

To determine

To calculate and interpret the residual for the 2010 gold winner Yu-Na Kim, who scored 78.50 in the short program and 150.06 in the fee skate.

(a)

Expert Solution
Check Mark

Answer to Problem 57E

The residual is 3.765 .

Explanation of Solution

The relationship between the short program score and the free skate score is given in the question. And the regression line is as:

  y^=16.2+2.07x

And also given that Yu-Na Kim scored 78.50 in the short program and 150.06 in the fee skate. Thus, we have,

  x=16.2y=2.07

Thus the predicted free skate score is as:

  y^=16.2+2.07x=16.2+2.07(78.50)=146.295

Thus, the residual will be calculated as:

  Residual=yy^=15.06146.295=3.765

This then implies that we underestimated the free skate score by 3.765 for Yu-Na Kim when making a prediction using the regression line.

(b)

To determine

To interpret the slope of the least square regression line.

(b)

Expert Solution
Check Mark

Explanation of Solution

The relationship between the short program score and the free skate score is given in the question. And the regression line is as:

  y^=16.2+2.07x

And also given that Yu-Na Kim scored 78.50 in the short program and 150.06 in the fee skate. Thus, we have,

  x=16.2y=2.07

As we know that the slope is coefficient of x in the least square regression equation and represents the average increase or decrease of y per unit of x . Thus,

  b1=2.07

This then implies that on average, the free skate score increase by 2.07 points whn the short program score increased by one point.

(c)

To determine

To interpret the value of s .

(c)

Expert Solution
Check Mark

Explanation of Solution

The relationship between the short program score and the free skate score is given in the question. And the regression line is as:

  y^=16.2+2.07x

And also given that Yu-Na Kim scored 78.50 in the short program and 150.06 in the fee skate. Thus, we have,

  x=16.2y=2.07

And the value of s is, s=10.2 . As we know that the standard error of the estimate s represents the average error of predictions thus the average deviation between actual and the predicted values. Thus the predicted free skate score using the equation of least square regression line deviated on average by 10.2 from the actual free skate score.

(d)

To determine

To interpret the value of r2 .

(d)

Expert Solution
Check Mark

Explanation of Solution

The relationship between the short program score and the free skate score is given in the question. And the regression line is as:

  y^=16.2+2.07x

And also given that Yu-Na Kim scored 78.50 in the short program and 150.06 in the fee skate. Thus, we have,

  x=16.2y=2.07

And the value of r2 is,

  r2=0.736=73.6%

As we know that the coefficient of determination measures the proportion of variation in the responses y variable that is explained by the least square regression model using the explanatory x variable. Thus, we can say that 73.6% of the variation between in the free skate score is explained by the least square regression line using the short program score as explanatory variable.

Chapter 3 Solutions

EBK PRACTICE OF STAT.F/AP EXAM,UPDATED

Ch. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.1 - Prob. 18ECh. 3.1 - Prob. 19ECh. 3.1 - Prob. 20ECh. 3.1 - Prob. 21ECh. 3.1 - Prob. 22ECh. 3.1 - Prob. 23ECh. 3.1 - Prob. 24ECh. 3.1 - Prob. 25ECh. 3.1 - Prob. 26ECh. 3.1 - Prob. 27ECh. 3.1 - Prob. 28ECh. 3.1 - Prob. 29ECh. 3.1 - Prob. 30ECh. 3.1 - Prob. 31ECh. 3.1 - Prob. 32ECh. 3.1 - Prob. 33ECh. 3.1 - Prob. 34ECh. 3.1 - Prob. 35ECh. 3.1 - Prob. 36ECh. 3.2 - Prob. 37ECh. 3.2 - Prob. 38ECh. 3.2 - Prob. 39ECh. 3.2 - Prob. 40ECh. 3.2 - Prob. 41ECh. 3.2 - Prob. 42ECh. 3.2 - Prob. 43ECh. 3.2 - Prob. 44ECh. 3.2 - Prob. 45ECh. 3.2 - Prob. 46ECh. 3.2 - Prob. 47ECh. 3.2 - Prob. 48ECh. 3.2 - Prob. 49ECh. 3.2 - Prob. 50ECh. 3.2 - Prob. 51ECh. 3.2 - Prob. 52ECh. 3.2 - Prob. 53ECh. 3.2 - Prob. 54ECh. 3.2 - Prob. 55ECh. 3.2 - Prob. 56ECh. 3.2 - Prob. 57ECh. 3.2 - Prob. 58ECh. 3.2 - Prob. 59ECh. 3.2 - Prob. 60ECh. 3.2 - Prob. 61ECh. 3.2 - Prob. 62ECh. 3.2 - Prob. 63ECh. 3.2 - Prob. 64ECh. 3.2 - Prob. 65ECh. 3.2 - Prob. 66ECh. 3.2 - Prob. 67ECh. 3.2 - Prob. 68ECh. 3.2 - Prob. 69ECh. 3.2 - Prob. 70ECh. 3.2 - Prob. 71ECh. 3.2 - Prob. 72ECh. 3.2 - Prob. 73ECh. 3.2 - Prob. 74ECh. 3.2 - Prob. 75ECh. 3.2 - Prob. 76ECh. 3.2 - Prob. 77ECh. 3.2 - Prob. 78ECh. 3.2 - Prob. 79ECh. 3.2 - Prob. 80ECh. 3 - Prob. R3.1RECh. 3 - Prob. R3.2RECh. 3 - Prob. R3.3RECh. 3 - Prob. R3.4RECh. 3 - Prob. R3.5RECh. 3 - Prob. R3.6RECh. 3 - Prob. T3.1SPTCh. 3 - Prob. T3.2SPTCh. 3 - Prob. T3.3SPTCh. 3 - Prob. T3.4SPTCh. 3 - Prob. T3.5SPTCh. 3 - Prob. T3.6SPTCh. 3 - Prob. T3.7SPTCh. 3 - Prob. T3.8SPTCh. 3 - Prob. T3.9SPTCh. 3 - Prob. T3.10SPTCh. 3 - Prob. T3.11SPTCh. 3 - Prob. T3.12SPTCh. 3 - Prob. T3.13SPT
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