Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications

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Assuming that the pairs are drawn from a bivariate normal distribution allows

us to test hypotheses about r and to construct a CI. There is no completely

satisfactory way to check the plausibility of the bivariate normality assumption.

A partial check involves constructing two separate normal probability plots, one for

the sample x

’s and another for the sample y

’s, since bivariate normality implies

that the marginal distributions of both X and Y are normal. If either plot deviates

substantially from a straight-line pattern, the following inferential procedures

should not be used when the sample size n is small. Also, as discussed in Example

12.19, the scatter plot should show a roughly elliptical shape.

TESTING

FOR THE

ABSENCE

OF CORRE-

LATION

When H

: r ¼ 0 is true, the test statistic

T ¼

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

n  2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1  R

has a t distribution with n  2 df (see Exercise 65).

Alternative Hypothesis Rejection Region for Level a Test

: r > 0 t  t

a,n2

: r < 0 t t

a,n2

: r 6¼ 0 either t  t

a/2,n2

or t t

a/2,n2

A P-value based on n  2 df can be calculated as described previously.

Example 12.20 Neu rotoxic effects of manganese are well known and are usually caused by high

occupational exposure over long periods of time . In the fields of occupational

hygiene and environmental hygiene, the relationship between lipid peroxida tion,

which is responsible for deterioration of foods and damage to live tissue, and

occupational exposure had not been previously reported. The article “Lipid Perox-

idation in Workers Exposed to Manganese” (Scand. J. Work Environ. Health, 1996:

381–386) gave data on x ¼ manganese concentration in blood (ppb) and y ¼ con-

centration (mmol/L) of malondialdehyde, which is a stable product of lipid peroxi-

dation, both for a sample of 22 workers exposed to manganese and for a control

sample of 45 individuals. The value of r for the control sample was .29, from which

t ¼

ð:29Þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

45  2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1  :29

 2:0

The corresponding P-value for a two-tailed t test based on 43 df is roughly .052 (the

cited article reported only that the P-value > .05). We would not want to reject the

assertion that r ¼ 0 at either significance level .01 or .05. For the sample of

exposed workers, r ¼ .83 and t ¼ 6.7, clear evidence that there is a positive

relationship in the entire population of exposed workers from which the sam ple

was selected. Although in general correlation does not necessarily imply causation,

it is plausible here that higher levels of manganese cause higher levels of per-

oxidation.

■

668 CHAPTER 12 Regression and Correlation

Because r measures the extent to which there is a linear relationship between

the two variables in the population, the null hypothesis H

: r ¼ 0 states that there is

no such population relationship. In Section 12.3, we used the t ratio

to test for

a linear relationship between the two variables in the context of regression analysis.

It turns out that the two test procedures are completely equivalent because

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

n  2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1  r

(Exercise 65). When interest lies only in assessi ng the

strength of any linear relationship rather than in fitting a model and using it to

estimate or predict, the test statistic formula just presented requires fewer computa-

tions than does the t ratio.

Other Inferences Concerning r

The procedure for testing H

: r ¼ r

when r

6¼ 0 is not equivalent to any

procedure from regression analysis. The test statistic is based on a transformation

of R called the Fisher transformation.

PROPOSITION

When (X

, Y

), ...,(X

, Y

) is a sample from a bivariate normal distribution,

the rv

V ¼

1 þ R

1  R



ð12:10Þ

has approximately a normal distribution with mean and variance

1 þ r

1  r



n  3

The rationale for the transformation is to obtain a function of R that has a variance

independent of r; this would not be the case with R itself. Also, the approximation

will not be valid if n is quite small.

The test statistic for testing H

: r ¼ r

Z ¼

V 

ln½ð1 þ r

Þ=ð1  r

Þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

n  3

Alternative Hypothesis Rejection Region for Level a Test

: r > r

z  z

: r < r

z z

: r 6¼ r

either z  z

a/2

or z z

a/2

A P-value can be calculated in the same manner as for previous z tests.

12.5 Correlation 669

Example 12.21 As far back as Leonardo da Vinci, it was known that height and wingspan

(measured fingertip to fingertip between outstretched hands) are closely related.

For these measurements (in inches) from 16 students in a statistics class notice how

close the two values are.

Student: 12345678

Height: 63.0 63.0 65.0 64.0 68.0 69.0 71.0 68.0

Wingspan: 62.0 62.0 64.0 64.5 67.0 69.0 70.0 72.0

Student: 910111213141516

Height: 68.0 72.0 73.0 73.5 70.0 70.0 72.0 74.0

Wingspan: 70.0 72.0 73.0 75.0 71.0 70.0 76.0 76.5

The scatter plot in Figure 12.26 shows an approximat ely linear shape, and the point

cloud is roughly elliptical. Also, the normal plots for the individual variables are

roughly linear, so the bivariate normal distribution can reasonably be assumed.

The correlation is compute d to be .9422. Can it be conclude that wingspan

and height are highly correlated, in the sense that r > .8? To carry out a test of

: r ¼ .8 versus H

: r > .8, we Fisher transform .9422 and .8:

1 þ :9422

1  :9422



¼ 1:757

1 þ :8

1  :8



¼ 1:099

The calculation is easily done on a calculator with hyperbolic functions,

because the inverse hyperbolic tangent is equivalent to the Fisher transformation.

That is, tanh

1

(.9422) ¼ 1.757 and tanh

1

(.8) ¼ 1.099. Compute z ¼

1:757  1:099ðÞ=ð1=

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

16  3

Þ¼2:37. Since 2.37  1.645, at level .05 we can

reject H

: r ¼ .8 in favor of H

: r  .8. Indeed, because 2.37  2.33, it is also

true that we can reject H

in this one-tailed test at the .01 level, and conclude that

wingspan is highly correlated with height.

■

62 64 66 68 70 72 74

Wingspan

Height

Figure 12.26 Wingspan plotted against height

670

CHAPTER 12 Regression and Correlation

To obtain a CI for r, we first derive an interval for

ln½ð1 þ rÞ=ð1  rÞ. Standardizing V, writing a probability statement, and

manipulating the resulting inequalities yields

v 

a=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

n  3

; v þ

a=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

n  3



ð12:11Þ

as a 100(1  a)% interval for m

, where v ¼

ln½ð1 þ rÞ=ð1  rÞ. This interval can

then be manipulated to yield a CI for r.

A 100(1  a)% confidence interval for r is

 1

þ 1

;

 1

þ 1



where c

and c

are the left and right endpoints, respectively, of the interval

(12.11).

Example 12.22

(Example 12.21

continued)

The sample correlation coefficient between wingspan and height was r ¼ .9422,

giving n ¼ 1.757. With n ¼ 16, a 95% confidence interval for m

1:757  1:96=

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

16  3

¼ 1:213; 2:301ðÞ¼c

; c

ðÞ.

The 95% interval for r is

21:213ðÞ

 1

21:213ðÞ

þ 1

;

22:301ðÞ

 1

22:301ðÞ

þ 1



¼ð:838;:980Þ

As before, this calculation can be done more easily using the hyperbolic

tangent, which is the inverse of the Fisher transformation. This gives (tanh(1.213),

tanh(2.301)) ¼ (.838, .980). Notice that this interval excludes .8, and that our

hypothesis test in Example 12.21 would have rejected H

: r ¼ .8 in favor of the

alternative H

: r > .8 at the .025 level. ■

Absent the assumption of bivariate normality, a bootstrap procedure can be

used to obtain a CI for r or test hypotheses.

In Chapter 5, we cautioned that a large value of the correlation coefficient

(near 1 or 1) implies only association and not causation. This applies to both

r and r. It is easy to find strong but weird correlations in which neither variable is

casually related to the other. For example, since prohibition ended in the 1930s,

beer consumption and church attendance have correlated very highly. Of course,

the reason is that both variables have incr eased in accord with population growth.

12.5 Correlation 671

Exercises Section 12.5 (56–67)

56. The article “Behavioural Effects of Mobile Tele-

phone Use During Simulated Driving” (Ergo-

nomics, 1995: 2536–2562) reported that for a

sample of 20 experimental subjects, the sample

correlation coefficient for x ¼ age and y ¼ time

since the subject had acquired a driving license

(yr) was .97. Why do you think the value of r is

so close to 1? (The article’s authors gave an

explanation.)

57. The Turbine Oil Oxidation Test (TOST) and the

Rotating Bomb Oxidation Test (RBOT) are two

different procedures for evaluating the oxidation

stability of steam turbine oils. The article

“Dependence of Oxidation Stability of Steam

Turbine Oil on Base Oil Composition” (J. Soc.

Tribologists Lubricat. Engrs., Oct. 1997: 19–24)

reported the accompanying observations on x ¼

TOST time (hr) and y ¼ RBOT time (min) for 12

oil specimens.

TOST 4200 3600 3750 3675 4050 2770

RBOT 370 340 375 310 350 200

TOST 4870 4500 3450 2700 3750 3300

RBOT 400 375 285 225 345 285

a. Calculate and interpret the value of the sample

correlation coefficient (as did the article’s

authors).

b. How would the value of r be affected if we had

let x ¼ RBOT time and y ¼ TOST time?

c. How would the value of r be affected if RBOT

time were expressed in hours?

d. Construct a scatter plot and normal probability

plots and comment.

e. Carry out a test of hypotheses to decide

whether RBOT time and TOST time are line-

arly related.

58. Torsion during hip external rotation and extension

may explain why acetabular labral tears occur in

professional athletes. The article “Hip Rotational

Velocities During the Full Golf Swing” (J. Sport

Sci. Med., 2009: 296 – 299) reported on an inves-

tigation in which lead hip internal peak rotational

velocity (x) and trailing hip peak external rota-

tional velocity (y) were determined for a sample of

15 golfers. Data provided by the article’s authors

was used to calculate the following summary

quantities:

¼ 64;732:83; S

¼ 130;566:96;

¼ 44;185:87

Separate normal probability plots showed very

substantial linear patterns.

a. Calculate a point estimate for the population

correlation coefficient.

b. If the simple linear regression model were fit

to the data, what proportion of variation in

external velocity could be attributed to the

model relationship? What would happen to

this proportion if the roles of x and y were

reversed? Explain.

c. Carry out a test at significance level .01 to

decide whether there is a linear relationship

between the two velocities in the sampled pop-

ulation; your conclusion should be based on a

P-value.

d. Would the conclusion of (c) have changed if

you had tested appropriate hypotheses to

decide whether there is a positive linear asso-

ciation in the population? What if a signifi-

cance level of .05 rather than .01 had been

used?

59. The authors of the paper “Objective Effects of a

Six Months’ Endurance and Strength Training

Program in Outpatients with Congestive Heart

Failure” (Med. Sci. Sports Exercise, 1999:

1102–1107) presented a correlation analysis to

investigate the relationship between maximal lac-

tate level x and muscular endurance y. The accom-

panying data was read from a plot in the paper.

x 400 750 770 800 850 1025 1200

y 3.80 4.00 4.90 5.20 4.00 3.50 6.30

x 1250 1300 1400 1475 1480 1505 2200

y 6.88 7.55 4.95 7.80 4.45 6.60 8.90

¼ 36.9839, S

¼ 2,628,930.357,

¼ 7377.704

A scatter plot shows a linear pattern.

a. Test to see whether there is a positive correla-

tion between maximal lactate level and muscu-

lar endurance in the population from which this

data was selected.

b. If a regression analysis were to be carried out

to predict endurance from lactate level, what

proportion of observed variation in endurance

could be attributed to the approximate linear

relationship? Answer the analogous question if

regression is used to predict lactate level from

672

CHAPTER 12 Regression and Correlation

endurance—and answer both questions with-

out doing any regression calculations.

60. Hydrogen content is conjectured to be an impor-

tant factor in porosity of aluminum alloy castings.

The article “The Reduced Pressure Test as a Mea-

suring Tool in the Evaluation of Porosity/Hydro-

gen Content in A1–7 Wt Pct Si-10 Vol Pct SiC(p)

Metal Matrix Composite” (Metallurg. Trans.,

1993: 1857–1868) gives the accompanying data

on x ¼ content and y ¼ gas porosity for one par-

ticular measurement technique.

x .18 .20 .21 .21 .21 .22 .23

y .46 .70 .41 .45 .55 .44 .24

x .23 .24 .24 .25 .28 .30 .37

y .47 .22 .80 .88 .70 .72 .75

MINITAB gives the following output in response

to a CORRELATION command:

Correlation of Hydrcon and

Porosity ¼ 0.449

a. Test at level .05 to see whether the population

correlation coefficient differs from 0.

b. If a simple linear regression analysis had been

carried out, what percentage of observed vari-

ation in porosity could be attributed to the

model relationship?

61. Physical properties of six flame-retardant fabric

samples were investigated in the article “Sensory

and Physical Properties of Inherently Flame-

Retardant Fabrics” (Textile Res., 1984: 61–68).

Use the accompanying data and a .05 significance

level to determine whether there is a significant

correlation between stiffness x (mg-cm) and thick-

ness y (mm). Is the result of the test surprising in

light of the value of r?

x 7.98 24.52 12.47 6.92 24.11 35.71

y .28 .65 .32 .27 .81 .57

62. The article “Increases in Steroid Binding

Globulins Induced by Tamoxifen in Patients with

Carcinoma of the Breast” (J. Endocrinol., 1978:

219–226) reports data on the effects of the drug

tamoxifen on change in the level of cortisol-bind-

ing globulin (CBG) of patients during treatment.

With age ¼ x and DCBG ¼ y, summary values

are n ¼ 26,

¼ 1613,

ðx

xÞ

¼3756:96,

¼281:9,

ðy

yÞ

¼ 465:34, and

¼16;731

a. Compute a 90% CI for the true correlation

coefficient r.

b. Test H

: r ¼.5 versus H

: r < .5 at level

.05.

c. In a regression analysis of y on x, what propor-

tion of variation in change of cortisol-binding

globulin level could be explained by variation

in patient age within the sample?

d. If you decide to perform a regression analysis

with age as the dependent variable, what pro-

portion of variation in age is explainable by

variation in DCBG?

63. A sample of n ¼ 500 (x, y) pairs was collected

and a test of H

: r ¼ 0 versus H

: r 6¼ 0 was

carried out. The resulting P-value was computed

to be .00032.

a. What conclusion would be appropriate at level

of significance .001?

b. Does this small P-value indicate that there is a

very strong relationship between x and y (a

value of r that differs considerably from 0)?

Explain.

c. Now suppose a sample of n ¼ 10,000 (x, y)

pairs resulted in r ¼ .022. Test H

: r ¼0 versus

: r 6¼ 0 at level .05. Is the result statistically

significant? Comment on the practical signifi-

cance of your analysis.

64. Let x be number of hours per week of studying and

y be grade point average. Suppose we have one

sample of (x, y) pairs for females and another for

males. Then we might like to test the hypothesis

: r

r

¼ 0 against the alternative that the two

population correlation coefficients are different.

a. Use properties of the transformed variable V ¼

.5ln[(1 + R)/(1  R)] to propose an appropriate

test statistic and rejection region (let R

and

denote the two sample correlation coeffi-

cients).

b. The paper “Relational Bonds and Customer’s

Trust and Commitment: A Study on the Mod-

erating Effects of Web Site Usage” (Serv. Ind.

J., 2003: 103–124) reported that n

¼ 261,

¼ .59, n

¼ 557, r

¼ .50, where the first

sample consisted of corporate website users

and the second of non-users; here r is the

correlation between an assessment of the

strength of economic bonds and performance.

Carry out the test for this data (as did the

authors of the cited paper).

65. Verify that the t ratio for testing H

: b

¼ 0in

Section 12.3 is identical to t for testing H

: r ¼ 0.

66. Verify Property 2 of the correlation coefficient:

the value of r is independent of the units in which

x and y are measured; that is, if x

¼ ax

+ c and

¼ by

+ d, a > 0, b > 0, then r for the (x

, y

)

pairs is the same as r for the (x

, y

) pairs.

12.5 Correlation 673

67. Consider a time series—that is, a sequence of

observations X

, X

, ...on some response variable

(e.g., concentration of a pollutant) over time—

with observed values x

, x

, ..., x

over n time

periods. Then the lag 1 autocorrelation coefficient

is defined as

n1

i¼1

ðx

 xÞðx

iþ1

 xÞ

i¼1

ðx

 xÞ

Autocorrelation coefficients r

, r

, ... for lags 2,

3, ... are defined analogously.

a. Calculate the values of r

, r

, and r

for the

temperature data from Exercise 79 of Chapter 1.

b. Consider the n  1 pairs (x

, x

), (x

, x

), ...,

n  1

, x

). What is the difference between the

formula for the sample correlation coefficient r

applied to these pairs and the formula for r

What if n, the length of the series, is large?

What about r

compared to r for the n 2 pairs

, x

), (x

, x

), ...,(x

n  2

, x

c. Analogous to the population correlation coeffi-

cient r,letr

(i ¼ 1, 2, 3, ... ) denote the theo-

retical or long-run autocorrelation coefficients at

the various lags. If all these r’s are zero, there is

no (linear) relationship between observations in

the series at any lag. In this case, if n is large, each

has approximately a normal distribution with

mean 0 and standard deviation 1=

ﬃﬃﬃ

and differ-

ent R

’s are almost independent. Thus H

: r

¼ 0

can be rejected at a significance level of approxi-

mately .05 if either r

r2=

ﬃﬃﬃ

or r

2=

ﬃﬃﬃ

.If

n ¼ 100 and r

¼ .16, r

¼.09, r

¼.15, is

there evidence of theoretical autocorrelation at

any of the first three lags?

d. If you are testing the null hypothesis in (c) for

more than one lag, why might you want to

increase the cutoff constant 2 in the rejection

region? [Hint: What about the probability of

committing at least one type I error?]

12.6

Assessing Model Adequacy

A plot of the observed pairs (x

, y

) is a necessary first step in deciding on the form

of a mathematical relationship between x and y. It is possible to fit many functions

other than a linear one (y ¼ b

+ b

x) to the data, using either the principle of least

squares or another fitting method. Once a function of the chosen form has been

fitted, it is important to check the fit of the model to see whether it is in fact

appropriate. One way to study the fit is to superimpose a graph of the best-fit

function on the scatter plot of the data. However, any tilt or curvature of the best-fit

function may obscure some aspects of the fit that should be investigated. Further-

more, the scale on the vertical axis may make it difficult to assess the extent to

which observed values deviate from the best-fit functions.

Residuals and Standardized Residuals

A more effective approach to assessment of model adequacy is to compute the

fitted or predicted values

and the residuals e

¼ y



and then plot various

functions of these computed quantities. We then examine the plots either to confirm

our choice of model or for indications that the model is not appropriate. Suppose the

simple linear regression model is correct, and let y ¼

x be the equation of the

estimated regression line. Then the ith residual is e

¼ y

ð

Þ. To derive

properties of the residuals, let e

¼ Y



represent the ith residual as a random

variable (rv) (before observations are actually made). Then

EðY



Þ¼EY

ðÞEð

Þ¼b

þ b

ðb

þ b

Þ¼0 ð12:12Þ

Because

ð¼

Þis a linear function of the Y

’s, so is Y



(the coefficients

depend on the x

’s). Thus the normality of the Y

’s implies that each residual is

normally distributed. It can also be shown (Exercise 76) that

674 CHAPTER 12 Regression and Correlation

VðY



Þ¼s

1 



ðx

 xÞ

ð12:13Þ

Replacing s

by s

and taking the square root of Equation (12.13) gives the

estimated standard deviation of a residual.

Let’s now standardize each residual by subtracting the mean value (zero) and

then dividing by the estimated standard deviation.

The standardized residuals are given by





ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 



ðx

 xÞ

i ¼ 1; ...; n ð12:14Þ

Notice that the variances of the residuals d iffer from one another. If n is reasonably

large, though, the bracketed term in (12.13) will be approximately 1, so some

sources use e

/s as the standardized residual. Computation of the e

*’s can be

tedious, but the most widely used statistical computer packages automatically

provide these values and (upon request) can construct various plots involving them.

Example 12.23 Example 12.12 presented data on x ¼ average SAT for entering freshmen and

y ¼ six-year percentage graduation rate. Here we reproduce the data along with the

fitted valu es and their estim ated standard deviations, residuals and their estimated

standard deviations, and standardized residuals. The estimated regression line is

y ¼36.18 + .08855x, and r

¼ .729. Notice that estimated standar d deviations

of the residuals (in the s

column) differ somewhat, so e* 6¼ e/s. The standard

deviations of the residuals are higher near

x, in contrast to the standard deviations of

the predicted values, which are lower near

722.21 38 27.7651 4.9554 10.2349 9.020 1.135

833.32 31 37.6034 3.8016 6.6034 9.564 0.690

877.77 44 41.5392 3.3838 2.4608 9.719 0.253

899.99 42 43.5067 3.1894 1.5067 9.785 0.154

944.43 45 47.4416 2.8394 2.4416 9.892 0.247

944.43 51 47.4416 2.8394 3.5584 9.892 0.360

1005.00 70 52.8048 2.4785 17.1952 9.989 1.721

1011.10 58 53.3449 2.4517 4.6551 9.995 0.466

1055.54 48 57.2799 2.3208 9.2799 10.026 0.926

1055.54 76 57.2799 2.3208 18.7201 10.026 1.867

1060.00 42 57.6748 2.3143 15.6748 10.028 1.563

1080.00 54 59.4457 2.3012 5.4457 10.031 0.543

1099.99 42 61.2157 2.3142 19.2157 10.028 1.916

1155.00 54 66.0866 2.4780 12.0866 9.989 1.210

1166.65 67 67.1181 2.5345 0.1181 9.974 0.012

1215.00 65 71.3993 2.8346 6.3993 9.893 0.647

1235.00 80 73.1702 2.9845 6.8298 9.849 0.693

1380.00 88 86.0092 4.3392 1.9908 9.332 0.213

1395.00 96 87.3374 4.4963 8.6626 9.257 0.936

1465.00 98 93.5356 5.2522 4.4644 8.850 0.504

■

12.6 Assessing Model Adequacy 675

Diagnostic Plots

The basic plots that many statisticians recommend for an assessment of model

validity and usefulness are the following:

1. y

on the vertical axis versus x

on the horizontal axis

2. y

on the vertical axis versus

on the horizontal axis

3. e

* (or e

) on the vertical axis versus x

on the horizontal axis

4. e

* (or e

) on the vertical axis versus

on the horizontal axis

5. A normal probability plot of the standardized residuals (or residuals)

Plots 3 and 4 are called residual plots against the independent variable and fitted

(predicted) values, respective ly.

If Plot 2 yields points close to the 45



line [slope +1 through (0, 0)], then the

estimated regression function gives accurate predictions of the values actually

observed. Thus Plot 2 provides a visual assessment of model effectiveness in

making predictions. Provided that the model is correct, neither residua l plot should

exhibit distinct patterns. The residuals should be randomly distributed about

0 according to a normal distribution, so all but a very few standardized residuals

should lie between 2 and +2 (i.e., all but a few residuals within 2 standard

deviations of their expected value 0). The plot of standardized residuals versus

is really a combination of the two other plots, showing implicitly both how

residuals vary with x and how fitted values compare with observed values. This

latter plot is the single one most often recommended for multiple regression

analysis. Plot 5 allows the analyst to assess the plausibility of the assumpt ion that

e has a normal distribution.

Example 12.24

(Example 12.23

continued)

Figure 12.27 presents the five plots just recommended along with a sixth plot. The

plot of y versus

y confirms the impression given by r

that x is fairly effective in

predicting y. The residual plots show no unusual pattern or discrepant values. The

normal probability plot of the standardized residuals is quite straight. In summary,

the first five plots leave us with no qualms about either the appropriateness of a

simple linear relationship or the fit to the given data.

Notice that plotting against x yields the same shape as a plot against the

predicted values. Is this surprising? The predicted value is a linear function of x,so

the plots will have the same appearance. Given that the plots look the same, why

include both? This is preparation for the next section, where more than one

predictor is allowed, and plotting against x is not the same as plotting against the

predicted values.

The sixth plot in Figure 12.27 is in accord with what was found graphically in

Example 12.12. In that example, Figure 12.18 showed that private universities

might tend to have better graduation rates than state universities. For another

graphical view of this, we show in the last plot of Figure 12.27 the standardized

residuals plotted against a variable that is 0 for sta te universit ies and 1 for private

universities. In this graph the private universities do seem to have an advantage, but

we will need to wait until the next section for a hypothesis test, which requires

including this new variable as a second predictor in the model.

676 CHAPTER 12 Regression and Correlation

Difﬁculties and Remedies

Although we hope that our analysis will yield plots like the first five of

Figure 12.27, quite frequently the plots will suggest one or more of the following

difficulties:

1. A nonlinear probabilistic relationship between x and y is appropriate.

2. Th e variance of e (and of Y) is not a constant s

but depends on x.

3. Th e selected model fits the data well except for a very few discrepant or outlying

data values, which may have greatly influenced the choice of the best-fit

function.

4. Th e error term e does not have a normal distribution (this is related to item 3).

5. When the subscript i indicates the time order of the observations, the e

’s exhibit

dependence over time.

6. One or mor e relevant independent variables have been omitted from the model.

1200800 1000 1400

SAT

20 40 60 80 100

Predicted Value

-2

-36.18

+.08855

Standardized

residuals vs.

predicted

60 8020 40 100

Predicted Value

800 1000 1200 1400

SAT

-2

vs.

predicted

Standardized

residuals vs.

-2 2-1

z Score

-2

0.00 0.25 0.50 0.75 1.00

State = 0 Private = 1

-2

Normal

probability

plot

Standardized

residuals vs.

another variable

Graduation Rate

Standardized Residual

Graduation Rate

Standardized Residual

Figure 12.27 Plots for the data from Example 12.24 ■

12.6 Assessing Model Adequacy 677