chapter 15 final (homework answers)

8/20/2019 Chapter 15 Final (Homework Answers)

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Chapter 15

15.1 The correlation is = 0.994, and the least-squares linear regression equation is

, where

r

ˆ 3.66 1.1969 y = − + x y = humerus length and x = femur length. The scatterplot with the

regression line below shows a strong, positive, linear relationship. Yes, femur length is a very

good predictor of humerus length.

Femur length (cm)

H u m e r u s l e n g h t ( c m )

7570656055504540

90

80

70

60

50

40

15.2 (a) The least-squares regression line is ˆ 11.547 0.84042 y x= + , where y = height (inches)

and x in arm span (inches). (b) Yes, the least-squares line is an appropriate model for the data

because the residual plot shows an unstructured horizontal band of points centered at zero.

Since 76 inches is within the range of arm spans examined in Mr. Shenk’s class, it is reasonable

to predict the height of a student with a 76 inch arm span.

15.3 (a) The observations are independent because they come from 13 unrelated colonies. (b)

The scatterplot of the residuals against the percent returning (below on the left) shows nosystematic deviations from the linear pattern. (c) The spread may be slightly wider in the middle,

but not markedly so. (d) The histogram (below on the right) shows no outliers or strong

skewness, so there are no clear deviations from Normality.

Percent return

r e s i d u a l

8070605040

5.0

2.5

0.0

-2.5

-5.0

-7.5

0

residual

C o u n t

5.02.50.0-2.5-5.0

5

4

3

2

1

0

15.4 (a) The observations are independent because they come from 16 different individuals. (b)

The scatterplot of the residuals against nonexercise activity (below on the left) shows no

systematic deviations from the linear pattern. One residual, about 1.6, is slightly larger than theothers, but this is nothing to get overly concerned about. (c) The spread is slightly higher for

larger values of nonexercise acitvity, but not markedly so. (d) The histogram (below on the right)

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shows no outliers and a slight skewness to the right, but this does not suggest a lack of

Normality.

Nonexercise act ivity (calories)

R e s i d u a l

7006005004003002001000-100

2.0

1.5

1.0

0.5

0.0

-0.5

-1.0

0

Residual

C o u n t

1.51.00.50.0-0.5-1.0

5

4

3

2

1

0

15.5 (a) The slope parameter β represents the change in the mean humerus length when femur

length increases by 1 cm. (b) The estimate of β is 1.1969b = , and the estimate of α is

. (c) The residuals are −0.8226, −0.3668, 3.0425, −0.9420, and −0.9110, and their sum

is −0.0001. The standard deviation is estimated by

3.66a = −

( )2 11.791.982

2 3

resid s

n= =

−

∑ .

15.6 (a) The scatterplot (below on the left) shows a strong, positive linear relationship between

x = speed (feet/second) and y = steps (per second). The correlation is 0.999r = and the least-

squares regression line is . (b) The residuals (rounded to 4 decimal

places) are 0.0106, −0.0013, −0.0010, −0.0110, −0.0093, 0.0031, and 0.0088, and their sum is

−0.0001 (essentially 0, except for rounding error). (c) The estimate of

ˆ 1.76608 0.080284 y = + x

α is , the

estimate of

1.76608a =

β is , and the estimate of0.080284b = σ is0.00041

0.00915

s = .

Speed (ft/s)

S t e p s ( p e r s e c o n d )

2221201918171615

3.6

3.5

3.4

3.3

3.2

3.1

3.0

15.7 (a) The scatterplot below shows a strong, positive linear relationship. (b) The slope β

gives this rate. The estimate of β is listed as the coefficient of “year” in the output, b =

9.31868 tenths of a millimeter. (c) We are not able to make an inference for the tilt rate from asimple linear regression model, because the observations are not independent.

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Year (coded as last two digits)

L e a n ( c o d e d f r o m 2 . 9 m e t e r s )

87858381797775

750

725

700

675

650

15.8 (a) The least-squares regression line is ˆ 0.12049 0.008569 y x= + , where y = the proportion

of perch killed and x = the number of perch. The fact that the slope is positive tells us that as the

number of perch increases, the proportion being killed by bass also increases. (b) The regression

standard error is s = 0.1886, which estimates the standard deviation σ . (c) Who? The

individuals are kelp perch. What? The response variable is the proportion of perch killed and

the explanatory variable is the number of perch available (or in the pen); both variables aquantitative. Why? The researcher was interested in examining the relationship between

predators and available prey. When, where, how, and by whom? Todd Anderson published the

data obtained from the ocean floor off the coast of southern California in 2001.Graphs: Thescatterplot provided clearly shows that the proportion of perch killed increases as the number of

perch increases. Numerical Summaries The mean proportions of perch killed are 0.175, 0.283,

0.425, and 0.646, in order from smallest to largest number of perch available. Model The least-squares regression model is provided in part (a). Interpretation The data clearly support the

predator-prey principle provided. (Students will soon learn how to formally test this hypothesis.)

(d) Using df = 16 − 2 = 14 and , a 95% confidence interval for* 2.145t = β is

= (0.0033, 0.0138). We are 95% confident that the proportion of

perch killed increases on average between 0.0033 and 0.0138 for each addition perch added to

the pen.

0.008569 2.145 0.002456± ×

15.9 The regression equation is , where =calories andˆ 560.65 3.0771 y = − x y x =time. The

scatterplot with regression line (below) shows that the longer a child remains at the table, the

fewer calories he or she will consume. The conditions for inference are satisfied. Using df = 18

and , a 95% confidence interval for* 2.101t = β is 3.0771 2.101 0.8498− ± × = (−4.8625,

−1.2917). With 95% confidence, we estimate that for every extra minute a child sits the table, heor she will consume an average of between 1.29 and 4.86 calories less during lunch.

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Time ( average number of minutes)

C a l o r i e s ( a v e r a g e n u m b e r )

454035302520

520

500

480

460

440

420

400

15.10 (a) Excel’s 95% confidence interval for β is (0.0033, 0.0138). This matches the

confidence interval calculated in Exercise 15.8. We are 95% confident that the proportion of

perch killed increases on average between 0.0033 and 0.0138 for each addition perch added to

the pen. (b) See Exercise 15.8 part (d) for a verification using the Minitab output. Using df =

16 − 2 = 14 and with the Excel output, a 95% confidence interval for* 2.145t = β is

= (0.0032, 0.0140). (c) Using df = 16 − 2 = 14 and , a 90%

confidence interval for

0.0086 2.145 0.0025± × * 1.761t =

β is 0.0086 1.761 0.0025± × = (0.0042, 0.0130).

15.11 (a) The least-squares regression line from the S-PLUS output is ˆ 3.6596 1.1969 y x= − + ,

where y = humerus length and x = femur length. (b) The test statistic is

1.196915.9374

0.0751b

bt

SE = = . (c) The test statistic t has df = 5 − 2 =3. The largest value in Table D

is 12.92. Since 15.9374 > 12.92, we know that P-value < 0.0005. (d) There is very strong

evidence that β > 0, that is, the line is useful for predicting the length of the humerus given the

length of the femur. (e) Using df = 3 and , a 99% confidence interval for*

5.841t = β is= (0.7582,1.6356). We are 99% confident that for every extra centimeter

in femur length, the length of the humerus will increase on average between 0.7582 cm and

1.6356 cm.

1.1969 5.841 0.0751± ×

15.12 (a) The value of or 99.8% is very close to one (or 100%), which indicates

perfect linear association. (b) The slope parameter

2 0.998r =

β gives this rate. Using df = 5 and

, a 99% confidence interval for* 4.032t = β is 0.080284 4.032 0.0016± × = (0.0738, 0.0867).

We are 99% confident that the rate at which steps per second increase as running speed increases

by 1 ft/s is on average between 0.0738 and 0.0867.

15.13 (a) The scatterplot (below) with regression line shows a strong, positive linear association between the number of jet skis in use (explanatory variable) and the number of accidents

(response variable). (b) We want to test 0 : H 0 β = (there is no association between number of jet

skis in use and number of accidents) versus :a H 0 β > (there is a positive association between

number of jet skis in use and number of accidents). (c) The conditions are independence, themean number of accidents should have a linear relationship with the number of jet skis in use,

the standard deviation should be the same for each number of jet skis in use, and the number of

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accidents should follow a Normal distribution. The conditions are satisfied except for having

independent observations, so we will proceed with caution. (d) LinRegTTest reports that t =21.079 with df = 8 and P-value is 0.000. With the earlier caveat, there is very strong evidence to

reject and conclude that there is a significant positive association between number of

accidents and number of jet skis in use. As the number of jet skis in use increases, the number of

accidents significantly increases. (e) Using df = 8 and , a 98% confidence interval for

0 H

*

2.896t = β is 0.0048 = (0.0042, 0.0054). With 98% confidence, we estimate that for

every extra thousand jet skis in use, the number of accidents increase by a mean of between 4.2

and 5.4 per year.

2.896 0.0002± ×

Number of jet skis in use

N u m b e r o f a c c i d e n t s

9000008000007000006000005000004000003000002000001000000

4000

3000

2000

1000

0

15.14 (a) We want to test 0 : H 0 β = (there is no association between yearly consumption of wine

and deaths from heart disease) versus :a H 0 β < (there is a negative association between yearly

consumption of wine and deaths from heart disease). The data are obtained from different

nations, so independence seems reasonable. The other conditions of constant variance, linear

relationship and Normality are also satisfied. The test statistic is22.969

6.463.557

t −

= − with df =

17 and P-value < 0.0005. Since the P-value is smaller than any reasonable significance level,say 1%, we reject 0 H . We have very strong evidence of a significant negative association

between the consumption of wine and deaths from heart disease. (b) Using df = 17 and

, a 95% confidence interval for* 2.110t = β is 22.969 2.110 3.557− ± × = (−30.4743, −15.4637).

With 95% confidence, we estimate that the number of deaths from heart disease (per 100,000

people) decreases on average between 15.46 and 30.47 for each additional liter of wineconsumed (per person).

Wine consumption (liters per person)

D e a t h s f r o m h e a r t d i s e a s e ( p e r 1 0 0 , 0 0 0 p e o p l e )

9876543210

300

250

200

150

100

50



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15.15 (a) The scatterplot below shows a moderately strong, positive linear association between y = number of beetle larvae clusters and x = number of beaver-caused stumps. (b) The least-

squares regression line is . =83.9%, so regression on stump counts explains

83.9% of the variation in the number of beetle larvae. (c) We want to test

ˆ 1.286 11.894 y = − + x

0

2r

0: H β = versus

:a H 0 β ≠ . The conditions for inference are met, and the test statistic is 10.47t = with df = 21.The output shows P-value = 0.000, so we have very strong evidence that beaver stump countshelp explain beetle larvae counts.

Number of beaver -caused stumps

N u m b e r o f b e e t l e l a r v a e c l u s t e r s

54321

60

50

40

30

20

10

0

15.16 (a) The mean of the standardized residuals is 0.00174 and the standard deviation is 1.014.Since the residuals are standardized, we expect the mean and standard deviation to be close to 0

and 1, respectively. (b) A stemplot is shown below on the left. The distribution is slightly

skewed to the left, but this is not unusual for a small data set. There are no striking departures

from Normality. For a standard Normal distribution, we would expect 95% of the observationsto fall between −2.0 and 2.0. Thus, −1.99 is quite reasonable. (c) The residual plot on the right

below shows no obvious patterns.

Stem- and- l eaf of Resi dual s N = 23Leaf Uni t = 0. 10

3 - 1 9655 - 1 306 - 0 710 - 0 4422

( 4) 0 02249 0 567894 1 2233

Number of beave r-caused st umps

S t a n d a r d i z e d r e s i d u a l s

54321

1.5

1.0

0.5

0.0

-0.5

-1.0

-1.5

-2.0

0

CASE CLOSED!(1) Descriptive statistics for x = number of three-point shots taken and y = percent made are

shown below. The average number of three-point shots taken per game is 15.684 and the

standard deviation is 2.865. The average percent of three-point shots made per game is 35.379

and the standard deviation is 1.425. The correlation is 0.958r = − and the scatterplot belowshows a negative association between these two variables. Notice that the cluster of points in the

bottom right corner shows some positive association, but the overall association between x and y is clearly negative.

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Vari abl e N Mean St Dev Mi ni mum Q1 Medi an Q3 Maxi mum Taken 19 15. 684 2. 865 9. 200 13. 800 17. 100 17. 700 18. 300Percent 19 35. 379 1. 425 34. 100 34. 400 34. 600 36. 200 38. 400

Number of 3-pointers taken

P e r c e n t o f 3 - p o i n t e r s m a d e

1816141210

39

38

37

36

35

34

(2) The least-squares regression line is ˆ 42.8477 0.4762 y x= −

2 0.917r =

0

with or 91.7%. The

linear model provides a reasonably good fit for these data. However, the residual plot shows aclear pattern with positive residuals for small and large numbers of 3-pointers taken and negativeresiduals in between the two extremes.

(3) The point is tagged as being influential because it may have a considerable impact on the

regression line. Influential points often pull the regression line in their direction so the residualstend to be small for influential points.

(4) We want to test 0 : H β = versus :a H 0 β ≠ . Independence is reasonable because the data are

from different seasons. The linear relationship condition is met, but the constant variance

condition and the Normality are both questionable so we will proceed with caution. A histogramof the percent made below shows that the distribution is skewed to the right. The test statistic is

with df = 17 and P-value = 0.000. We have very strong evidence of a significant

association between the number of three-pointers taken and the percent made.

13.7t = −

Percent of thee-pointers made

C o u n t

3837363534

9

8

7

6

5

4

3

2

1

0

(5) Using df = 17 and , a 95% confidence interval for* 2.110t = β is 0.4762 2.110 0.03475− ± ×

= (−0.5495, −0.4029). With 95% confidence, we estimate that for every additional three-pointer

taken, the percent made will decrease on average between 0.40 and 0.55.

15.17 Regression of fuel consumption on speed gives 0.01466b = − , , and

with df = 13 and P-value= 0.541. Thus, we have no evidence to suggest a straight-

0.02334bSE =

0.63t = −



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line relationship between speed and fuel use. The scatterplot below shows a strong relationship

between speed and fuel use, but the relationship is not linear. See Exercise 3.9 for more details.

Speed (km/h)

F u e l c o n s u m p t i o n

160140120100806040200

22.5

20.0

17.5

15.0

12.5

10.0

7.5

5.0

15.18 Repeated measurements of Sarah’s height are clearly not independent.

15.19 (a) The slope β tells us the mean change in the percent of forest lost for a 1 unit (1 cent

per pound) increase in the price of coffee. The estimate of β is 0.05525b = and the estimate of

α is . (b) This says that the straight-line relationship described by the least-squares

line is very strong. = 0.907 or 91% indicates that 91% of the total variation in the percent of

forest lost is accounted for by the straight-line relationship with prices paid to coffee growers.

(c) The P-value refers to the two-sided alternative:

1.0134a = −2r

0 : H 0 β = versus :a H 0 β ≠ . The small P-

value indicates that we have very strong evidence of a significant association between the

percent of forest lost and the price paid for coffee. (d) The residuals are −0.0988, 0.3934,

−0.2800, −0.2053, and 0.1907, and their sum is 0. The standard deviationσ is estimated by

0.32150.3274

3s = . (e) A scatterplot (on the left) and a residual plot (on the right) are shown

below. Even though the number of observations is small, there are no obvious problems with thelinear regression model. Coffee price appears to be a very good predictor of forest lost for this

range of values.

Price (cents per pound)

F o r e s t l o s t ( p e r c e n t )

7060504030

3.0

2.5

2.0

1.5

1.0

0.5

Price (cents per pound)

R e s i d u a l

7060504030

0.4

0.3

0.2

0.1

0.0

-0.1

-0.2

-0.3

0

15.20 (a) The scatterplot below, with the regression line ˆ 70.436874 274.7821 y x= + , shows a

moderate, positive, linear association. The linear relationship explains or 49.3% of

the variation in gate velocity. (b) We want to test

2 0.493r

0 : H 0 β = versus :a H 0 β ≠ . The test statistic

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is274.7821

3.116388.17712

t = with df = 10 and P-value = 0.011. (Table C indicates that 0.01 < P-

value < 0.02.) Since the P-value < 0.05, we reject 0 H and conclude that there is a significant

linear relationship between thickness and gate velocity. The regression formula might be used asa rule of thumb for new workers to follow, but the wide spread in the scatterplot below suggests

that there may be other factors that should be taken into account in choosing the gate velocity.

Cylinder wall thickness (inches)

G a t e v e l o c i t y ( f t / s e c )

0.90.80.70.60.50.40.30.2

350

300

250

200

150

100

15.21 (a) A scatterplot with the regression line is shown below. or 99.2%. (b) The

estimates of

2 0.992r =

α , β , and σ are a = −2.3948 cm, b = 0.1585 cm / min, and s = 0.8059 cm. (c) The

least-squares regression line is , where = length andˆ 2.3948 0.1585 y x= − + y x = time.

Time (min)

L e n g t h ( c m )

200150100500

30

25

20

15

10

5

0

15.22 (a) A scatterplot with the least-squares regression line ˆ 3.5051 0.0034 y x= − is shown

below. We want to test 0 : H 0 β = versus :a H 0 β < . The test statistic is 4.64t = − with df = 14

and P-value < 0.0005. We have very strong evidence that people with higher NEA gain less fat.

(b) To find this interval, we need , which is given in the Minitab output below as

0.0007414. Using df = 14 and , a 90% confidence interval for

bSE

* 1.761t = β is

= (−

0.0047,−

0.0021).0.00344 1.761 0.0007414− ± ×



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NEA change (cal)

F a t g a i n ( k g )

7006005004003002001000-100

4

3

2

1

0

The r egressi on equat i on i sFat gai n ( kg) = 3. 51 - 0. 00344 NEA change (cal )

Predi ct or Coef SE Coef T PConst ant 3. 5051 0. 3036 11. 54 0. 000NEA change ( cal ) - 0. 0034415 0. 0007414 - 4. 64 0. 000

S = 0. 739853 R- Sq = 60. 6% R- Sq( adj ) = 57. 8%

15.23 (a) A scatterplot is shown below. There is a moderate, positive, linear association betweeninvestment returns in the U.S. and investments overseas. (b) The test statistic is

0.61812.6091

0.2369b

bt

SE = = with df = 25 and 0.01 < P-value < 0.02. Thus, we have fairly strong

evidence that there is a significant linear relationship between the two returns. That is, the slope

is nonzero. (c) or 21.4%, so only 21.4% of the variation in the overseas returns is

explained by using linear regression with U.S. returns as the explanatory variable. Using this

linear regression model for prediction will not be very useful in practice.

2 0.214r =

U.S. return (%)

O v e r s e a s r e t u r n ( % )

403020100-10-20-30

70

60

50

40

30

20

10

0

-10

-20

15.24 (a) The residual plot (below on the left) shows that the variability about the regression

line increases as the U.S. return increases. (b) The histogram (below on the right) indicates that

the distribution of the residuals is skewed to the right. The outlier is from 1986, when theoverseas return was much higher than our regression model predicts.

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U.S. return (%)

R e s i d u a l

403020100-10-20-30

60

50

40

30

20

10

0

-10

-20

-30

0

Residual

C o u n t

40200-20

6

5

4

3

2

1

0

15.25 (a) The scatterplot below (on the left) shows a weak, negative association between corn

yield and weeds. The least-squares regression line is ˆ 166.483 1.0987 y x= − , where y = corn yield

(bushels per acre) and x = weeds (per meter). or 20.9%, so the linear relationship

explains about 20.9% of the variation in yield. (b) The t statistic for testing

2 0.209r =

0: H 0 β = versus

: 0a

H β < is with df = 14 and P-value = 0.0375. Since 0.0375 < 0.05, there is

sufficient evidence to conclude that more weeds reduce corn yields. (c) The small number of

observations for each value of the explanatory variable (weeds/meter), the large variability inthose observations, and the small value of r

1.92t = −

2 will make prediction with this model imprecise. A

residual plot below (on the right) also shows that the linear model is quite imprecise.

Weeds (per meter)

C o r n y i e l d ( b u s h e l s p e r a c r e )

9876543210

180

170

160

150

140

Weeds per meter

R e s i d u a l

9876543210

10

5

0

-5

-10

-15



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15.26 Using df = 21 and , a 90% confidence interval for* 1.721t = β is

= (−12.9454, −6.4444). With 90% confidence, we estimate that for

each one minute increase in time (a slower, more leisurely swim) the professor’s pulse will drop

on average between 6 and 13 beats per minute. There is a negative relationship between the professor’s swimming time and heart rate. A scatterplot is shown below.

9.6949 1.721 1.8887− ± ×

Time (in minutes)

P u l s e ( b e a t s p e r m i n u t e )

36.536.035.535.034.534.0

160

150

140

130

120

324 Chapter 15

chapter 15 final (homework answers)

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