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

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The joint CIs are referred to as Bon ferroni intervals. The method is easily

generalized to yield joint intervals for k different m

Yx

’s. Using the interval (12.7)

separately first for x ¼ x



then for x ¼ x



; ..., and finally for x ¼ x



yields a set of k

CIs for which the joint or simultaneous confidence level is guaranteed to be at least

100(1  ka)%.

Tests of hypotheses about b

+ b

x* are based on the test statistic T obtained

by replacing b

+ b

x* in the numerator of (12.6) by the null value m

. For example,

the assertion H

: b

+ b

(1200) ¼ 75 in Example 12.15 says that when the average

SAT is 1200, expected (i.e., true average) graduation rate is 75%. The test statistic

value is then t ¼½

1200ðÞ75=s

ð1200Þ

, and the test is upper-, lower-, or

two-tailed according to the inequality in H

A Prediction Interval for a Future Value of Y

Analogous to the CI (12.7) for m

Yx



, one frequently wishes to obtain an interval of

plausible values for the value of Y associated with some future observation when

the independent variable has value x*. In the scenario of Example 12.5, the CI

(12.7) can be used to provide an interval estimate of true average tree mass for all

trees exposed to CO

concentration x ¼ 600. Alternatively, we might wish an

interval of plausible values for the mass of a single such tree.

A CI refers to a parameter, or population characteristic, whose value is fixed

but unknown to us. In contrast, a future value of Y is not a parameter but instead a

random variabl e; for this reason we refer to an interval of plausible values for a

future Y as a prediction interval rather than a confidence interval. For the

confidence interval we use the error of estimation, b

þ b



ð



Þ,a

difference between a fixed (but unknown) quantity and a random variable. The

error of prediction is Y ð



Þ¼b

þ b



þ e ð



Þ, a difference

between two random variables. With the additional random e term, there is more

uncertainty in prediction than in estimation, so a PI will be wider than a CI. Because

the future value Y is independent of the observed Y

’s,

V½Y ð



Þ ¼ variance of prediction error

¼ VðYÞþVð



¼ s

þ s

ðx



 xÞ

¼ s

1 þ

ðx



 xÞ

Furthermore, because EðYÞ¼b

þ b



and Eð



Þ¼b

þ b



, the

expected value of the prediction error is E½Y ð



Þ ¼ 0. It can then be

shown that the standardized variable

T ¼

Y ð



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 þ

ðx



 xÞ

658 CHAPTER 12 Regression and Correlation

has a t distribution with n  2 df. Substituting this T into the probability statement

P(t

a/2,n2

< T < t

a/2,n2

) ¼ 1  a and manipulating to isolate Y between the two

inequalities yields the following interval.

A 100(1  a )% PI for a future Y observation to be made when x ¼ x* is



 t

a=2;n2

 s

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 þ

ðx



 xÞ



 t

a=2;n2



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

þ s



y  t

a=2;n2



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

þ s

ð12:8Þ

The interpretation of the predict ion level 100(1  a)% is identical to that of

previous confidence levels—if (12.8) is used repeatedly, in the long run the

resulting intervals will actually contain the observed y values 100(1  a)% of the

time. Notice that the 1 underneath the initial square root symbol makes the PI (12.8)

wider than the CI (12.7), although the intervals are both centered at



. Also,

as n !1the width of the CI approaches 0, whereas the width of the PI approaches

a/2

s (because even with perfect knowledge of b

and b

, there will still be

uncertainty in prediction).

Example 12.16 Let’s return to the university data of Example 12.15 and calculate a 95% prediction

interval for a graduation rate that would result from selecting a single university

whose average SAT is 1200. Re levant quantities from that example are

y ¼ 70:07 s

¼ 2:731 s ¼ 10:29

For a prediction level of 95% based on n  2 ¼ 18 df, the t critical value is 2.101,

exactly what we previously used for a 95% confidence level. The prediction

interval is then

70:07 ð2:101Þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

10:29

þ 2:731

¼ 70:07  2:101ðÞ10:646ðÞ

¼ 70:07  22:37 ¼ 47:70; 92:44ðÞ

Plausible values for a single observation on graduation rate when SAT is 1200 are

(at the 95% prediction level) between 47.70% and 92.44%. The 95% confidence

interval for graduation rate when SAT is 120 was (64.33, 75.81). The prediction

interval is much wider than this because of the extra 10.29

under the square root.

Figure 12.22, the MINITAB outpu t for Example 12.15, shows this interval as well

as the confidence interval.

■

The Bonferroni technique can be employed as in the case of confidence

intervals. If a PI with prediction level 100(1  a)% is calculated for each of k

different values of x, the simultaneous or joint prediction level for all k intervals is

at least 100(1  ka)%.

12.4 Inferences Concerning m

Y x



and the Prediction of Future Y Values 659

Exercises Section 12.4 (45–55)

45. Recall Example 12.5 and Example 12.6 of Sec-

tion 12.2, where the simple linear regression

model was applied to 8 observations on x ¼ CO

concentration and y ¼ mass in kilograms of pine

trees at age 11 months. Further calculations give

s ¼ .534 and

y ¼ 2:723, s

¼ :190 when

x ¼ 600, and

y ¼ 3:992, s

¼ :256 when

x ¼ 750.

a. Explain why s

is larger when x ¼ 750 than

when x ¼ 600.

b. Calculate a confidence interval with a confi-

dence level of 95% for the true average mass

of all trees grown with a CO

concentration of

600 parts per million.

c. Calculate a prediction interval with a predic-

tion level of 95% for the mass of a tree grown

with a CO

concentration of 600 parts per

million.

d. If a 95% CI is calculated for the true average

mass when CO

concentration is 750, what

will be the simultaneous confidence level for

both this interval and the interval calculated

in part (b)?

46. Reconsider the filtration rate–moisture content

data introduced in Example 12.7 (see also Exam-

ple 12.8).

a. Compute a 90% CI for b

+ 125b

, true aver-

age moisture content when the filtration rate

is 125.

b. Predict the value of moisture content for a

single experimental run in which the filtration

rate is 125 using a 90% prediction level. How

does this interval compare to the interval of

part (a)? Why is this the case?

c. How would the intervals of parts (a) and (b)

compare to a CI and PI when filtration rate is

115? Answer without actually calculating

these new intervals.

d. Interpret both H

: b

+ 125b

¼ 80 and

: b

+ 125b

< 80, and then carry out a

test at significance level .01.

47. Astringency is the quality in a wine that makes

the wine drinker’s mouth feel slightly rough, dry,

and puckery. The paper “Analysis of Tannins in

Red Wine Using Multiple Methods: Correlation

with Perceived Astringency” (Amer. J. Enol.

Vitic., 2006: 481–485) reported on an investiga-

tion to assess the relationship between perceived

astringency and tannin concentration using vari-

ous analytic methods. Here is data provided by

the authors on x ¼ tannin concentration by pro-

tein precipitation and y ¼ perceived astringency

as determined by a panel of tasters.

x 0.718 0.808 0.924 1.000 0.667 0.529 0.514 0.559

y 0.428 0.480 0.493 0.978 0.318 0.298 0.224 0.198

x 0.766 0.470 0.726 0.762 0.666 0.562 0.378 0.779

y 0.326 0.336 0.765 0.190 0.066 0.221 0.898 0.836

x 0.674 0.858 0.406 0.927 0.311 0.319 0.518 0.687

y 0.126 0.305 0.577 0.779 0.707 0.610 0.648 0.145

x 0.907 0.638 0.234 0.781 0.326 0.433 0.319 0.238

y 1.007 0.090 1.132 0.538 1.098 0.581 0.862 0.551

Relevant summary quantities are as follows:

¼ 19:404;

¼:549;

¼ 13:248032;

¼ 11:835795;

¼ 3:497811

¼ 13:248032 ð19:404Þ

=32 ¼ 1:48193150;

¼ 11:82637622

¼ 3:497811 ð19:404Þð:549Þ=32 ¼ 3:83071088

a. Fit the simple linear regression model to this

data. Then determine the proportion of

observed variation in astringency that can be

attributed to the model relationship between

astringency and tannin concentration.

b. Calculate and interpret a confidence interval

for the slope of the true regression line.

c. Estimate true average astringency when tan-

nin concentration is .6, and do so in a way that

conveys information about reliability and pre-

cision.

d. Predict astringency for a single wine sample

whose tannin concentration is .6, and do so in

a way that conveys information about reliabil-

ity and precision.

e. Is there compelling evidence for concluding

that true average astringency is positive when

tannin concentration is .7? State and test the

appropriate hypotheses.

48. The simple linear regression model provides a

very good fit to the data on rainfall and runoff

volume given in Exercise 17 of Section 12.2. The

equation of the least squares line is

y ¼1:128 þ :82697x, r

¼ :975, and s ¼ 5:24.

a. Use the fact that s

¼ 1:44 when rainfall

volume is 40 m

to predict runoff in a way

660

CHAPTER 12 Regression and Correlation

that conveys information about reliability and

precision. Does the resulting interval suggest

that precise information about the value of

runoff for this future observation is available?

Explain your reasoning.

b. Calculate a PI for runoff when rainfall is 50

using the same prediction level as in part (a).

What can be said about the simultaneous pre-

diction level for the two intervals you have

calculated?

49. You are told that a 95% CI for expected lead

content when traffic flow is 15, based on a sam-

ple of n ¼ 10 observations, is (462.1, 597.7).

Calculate a CI with confidence level 99% for

expected lead content when traffic flow is 15.

50. Refer to Exercise 21 in which x ¼ available

travel space in feet and y ¼ separation distance

in feet between a bicycle and a passing car.

a. MINITAB gives s

ð15Þ

¼ :186 and

ð20Þ

¼ :360. Explain why one is much

larger than the other.

b. Calculate a 95% CI for expected separation

distance when available travel space is 15 ft.

(Use s

ð15Þ

¼ :186.)

c. Calculate a 95% PI for a single instance of

separation distance when available travel

space is 20 ft. (Use s

ð20Þ

¼ :360.)

51. Plasma etching is essential to the fine-line pat-

tern transfer in current semiconductor processes.

The article “Ion Beam-Assisted Etching of Alu-

minum with Chlorine” (J. Electrochem. Soc.,

1985: 2010–2012) gives the accompanying data

(read from a graph) on chlorine flow (x,in

SCCM) through a nozzle used in the etching

mechanism and etch rate (y, in 100 A/min).

x 1.5 1.5 2.0 2.5 2.5 3.0 3.5 3.5 4.0

y 23.0 24.5 25.0 30.0 33.5 40.0 40.5 47.0 49.0

The summary statistics are

¼ 24:0,

¼ 312:5,

¼ 70:50,

¼902:25;

¼11;626:75,

¼6:448718,

10:602564.

a. Does the simple linear regression model spec-

ify a useful relationship between chlorine

flow and etch rate?

b. Estimate the true average change in etch rate

associated with a 1-SCCM increase in flow

rate using a 95% confidence interval, and

interpret the interval.

c. Calculate a 95% CI for m

Y·3.0

, the true average

etch rate when flow ¼ 3.0. Has this average

been precisely estimated?

d. Calculate a 95% PI for a single future obser-

vation on etch rate to be made when flow

¼ 3.0. Is the prediction likely to be accurate?

e. Would the 95% CI and PI when flow ¼ 2.5 be

wider or narrower than the corresponding

intervals of parts (c) and (d)? Answer without

actually computing the intervals.

f. Would you recommend calculating a 95% PI

for a flow of 6.0? Explain.

g. Calculate simultaneous CI’s for true average

etch rate when chlorine flow is 2.0, 2.5, and

3.0, respectively. Your simultaneous confi-

dence level should be at least 97%.

52. Consider the following four intervals based on

the data of Exercise 20 (Section 12.2):

a. A 95% CI for lichen nitrogen when NO



is .5

b. A 95% PI for lichen nitrogen when NO



is .5

c. A 95% CI for lichen nitrogen when NO



is .8

d. A 95% PI for lichen nitrogen when NO



is .8

e. Without computing any of these intervals,

what can be said about their widths relative

to each other?

53. The decline of water supplies in certain areas of

the United States has created the need for

increased understanding of relationships

between economic factors such as crop yield

and hydrologic and soil factors. The article

“Variability of Soil Water Properties and Crop

Yield in a Sloped Watershed” (Water Resources

Bull., 1988: 281–288) gives data on grain sor-

ghum yield (y, in g/m-row) and distance upslope

(x, in m) on a sloping watershed. Selected obser-

vations are given in the accompanying table.

x 0 102030455070

y 500 590 410 470 450 480 510

x 80 100 120 140 160 170 190

y 450 360 400 300 410 280 350

a. Construct a scatter plot. Does the simple lin-

ear regression model appear to be plausible?

b. Carry out a test of model utility.

c. Estimate true average yield when distance

upslope is 75 by giving an interval of plausi-

ble values.

12.4 Inferences Concerning m

Y x



and the Prediction of Future Y Values 661

54. Infestation of crops by insects has long been of

great concern to farmers and agricultural scien-

tists. The article “Cotton Square Damage by the

Plant Bug, Lygus hesperus, and Abscission Rates”

(J. Econ. Entomol., 1988: 1328–1337) reports data

on x ¼ age of a cotton plant (days) and y ¼ %

damaged squares. Consider the accompanying

n ¼ 12 observations (read from a scatter plot in

the article).

x 9 1212151818

y 11 12 23 30 29 52

x 21 21 27 30 30 33

y 41 65 60 72 84 93

a. Why is the relationship between x and y not

deterministic?

b. Does a scatter plot suggest that the simple

linear regression model will describe the rela-

tionship between the two variables?

c. The summary statistics are

¼ 246,

¼ 5742,

¼ 572,

¼ 35;634

and

¼ 14;022. Determine the equation

of the least squares line.

d. Predict the percentage of damaged squares

when the age is 20 days by giving an interval

of plausible values.

55. Verify that Vð

xÞ is indeed given by

the expression in the text. [Hint:

Vð

Þ¼

 VðY

Þ.]

12.5

Correlation

In many situations the objective in studying the joint behavior of two variables is to

see whether they are related, rather than to use one to predict the value of the other.

In this section, we first develop the sample correlation coefficient r as a measure of

how strongly related two variables x and y are in a sample and then relate r to the

correlation coefficient r defined in Chapter 5.

The Sample Correlation Coefﬁcient r

Given n pairs of observations (x

, y

), (x

, y

), ...,(x

, y

), it is natural to speak of x

and y having a positive relationship if large x’s are paired with large y’s and small

x’s with small y’s. Similarly, if large x’s are paired with small y’s and small x’s with

large y’s, then a negative relationship between the variables is implied. Consider

the quantity

i¼1

ðx

 xÞðy

 yÞ¼

i¼1



i¼1



i¼1



Then if the relationship is strongly positive, an x

above the mean x will tend to be

paired with a y

above the mean y, so that ðx

 xÞðy

 yÞ > 0, and this product will

also be positive whenever both x

and y

are below their respective means. Thus a

positive relationship implies that S

will be positive. An analogous argument

shows that when the relationship is negative, S

will be negative, since most of

the products ðx

 xÞðy

 yÞ will be negative. This is illustrated in Figure 12.23 .

Although S

seems a plausible measure of the strength of a relationship, we

do not yet have any idea of how positive or negative it can be. Unfortunately, S

has a serious defect: By changi ng the unit of measurement for either x or y, S

can

be made either arbitrarily large in magnitu de or arbitrarily close to zero. For

example, if S

¼ 25 when x is measured in meters, then S

¼ 25,000 when x is

662 CHAPTER 12 Regression and Correlation

measured in millimeters and .025 when x is expressed in kilometers. A reasonable

condition to impose on any measure of how strongly x and y are related is that the

calculated measure should not depend on the particular unit used to measure them. This

condition is achieved by modifying S

to obtain the sample correlation coefficient.

DEFINITION

The sample correlation coefficient for the n pairs (x

, y

), ...,(x

, y

)is

r ¼

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðx

 xÞ

q ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðy

 yÞ

ﬃﬃﬃﬃﬃﬃ

ð12:9Þ

Example 12.17 An accurate assessment of soil productivity is critical to rational land-use planning.

Unfortunately, as the author of the article “Productivity Ratings Based on Soil Series”

(Prof. Geographer, 1980: 158–163) argues, an acceptable soil productivity index is not

so easy to come by. One difficulty is that productivity is determined partly by which

crop is planted, and the relationship between yield of two different crops planted in the

same soil may not be very strong. To illustrate, the article presents the accompanying

data on corn yield x and peanut yield y (mT/ha) for eight different types of soil.

x 2.4 3.4 4.6 3.7 2.2 3.3 4.0 2.1

y 1.33 2.12 1.80 1.65 2.00 1.76 2.11 1.63

With

¼ 25:7,

¼ 14:40,

¼ 88:31,

¼ 46:856,

26:4324,

¼ 88:31 

25:7

¼ 88:31  82:56 ¼ 5:75

¼ 26:4324 

14:40

¼ :5124

¼ 46:856 

ð25:7Þð14:40Þ

¼ :5960

x x

Figure 12.23 (a) Scatter plot with S

positive; (b) scatter plot with S

negative

[+ means ðx





xÞðy





yÞ > 0, and  means ðx





xÞðy





yÞ < 0]

12.5 Correlation 663

from which

r ¼

:5960

ﬃﬃﬃﬃﬃﬃﬃﬃﬃ

5:75

p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

:5124

¼ :347

■

Properties of r

The most important properties of r are as follows:

1. Th e value of r does not depend on which of the two variables is labeled x and

which is labeled y.

2. Th e value of r is independent of the units in which x and y are measured.

3. 1  r  1

4. r ¼ 1 if and only if (iff) all (x

, y

) pairs lie on a straight line with positive slope,

and r ¼1 iff all (x

, y

) pairs lie on a straight line with negative slope.

5. Th e square of the sample correlation coefficient gives the value of the

coefficient of determination that woul d result from fitting the simple linear

regression model—in symbols, (r)

¼ r

Property 1 should be evident. Exercise 66 asks you to verify Property 2. To

derive Property 5, recall the regression analysis of variance identity (12.4),

[SST ¼ SSE þ SSR ¼ SSE þ

 yÞ

]. It is easily shown [Exercise 24(b)]

that

 y ¼

ðx

 xÞ, and therefore

yÞ

ðx

xÞ

ðx

xÞðy

yÞ

ðx

xÞ

ðx

xÞ

ðx

xÞðy

yÞ½

ðx

xÞ

ðy

yÞ

ðy

yÞ

¼ðrÞ

SST

Here (r)

is the square of the correlation coefficient. Substituting this result into the

identity (12.4) gives SST ¼ SSE + (r)

SST, so (r)

¼ (SST  SSE)/SST, com-

pleting the derivation of Property 5.

Because (r)

¼ (SST  SSE)/SST, and the numerator cannot be bigger than

the denominator, Property 3 follows immediately. Furthermore, because the ratio

can be 1 if and only if SSE ¼ 0, we conclude that r

¼ 1 if and only if all the points

fall on a straight line. If the correlation is positive this will be a line with positive

slope, and if the correlation is negative it will be a line with negative slope, so we

have verified Property 4.

Property 1 stands in marked contrast to what happens in regress ion analysis,

where virtually all quantities of interest (the estimated slope, estimated y-intercept,

, etc.) depend on which of the two variables is treated as the depend ent variable.

However, Property 5 shows that the proportion of variation in the dependent

variable explained by fitting the simple linear regression model does not depend

on which variable plays this role.

Property 2 is equivalent to saying that r is unchanged if each x

is replaced by

and if each y

is replaced by dy

(where c and d are positive, giving a change in

the scale of measurement), as well as if each x

is replaced by x

 a and y

by y

 b

664 CHAPTER 12 Regression and Correlation

(which changes the location of zero on the measurement axis). This implies, for

example, that r is the same whether temperature is measured in



For



Property 3 tells us that the maximum value of r, corresponding to the largest

possible degree of positive relationship, is r ¼ 1, whereas the most negative

relationship is identified with r ¼1. According to Property 4, the largest positive

and largest negative correlations are achieved only when all points lie along a

straight line. Any other configuration of points, even if the configuration suggests a

deterministic relationship between variables, will yield an r value less than 1 in

absolute magnitude. Thus r measures the degree of linear relationship among

variables. A value of r near 0 is not evidence of the lack of a strong relationship,

but only the absence of a linear relation, so that such a value of r must be interpreted

with caution. Figure 12.24 illustrates several configurations of points associated

with different values of r.

A frequently asked question is, “When can it be said that there is a

strong correlation between the variables, and when is the correlation weak?”

A reasonable rule of thumb is to say that the correlation is weak if 0 

|r|  .5,

strong if .8 

|r|  1, and moderate otherwise. It may surprise you that r ¼ .5 is

considered weak, but r

¼ .25 implies that in a regression of y on x, only 25% of

observed y variation would be expl ained by the model. In Example 12.17, the

correlation between corn yield and peanut yield would be described as weak.

The Population Correlation Coefﬁcient r

and Inferences About Correlation

The correlation coefficient r is a measure of how strongly related x and y are in the

observed sample. We can think of the pairs (x

, y

) as having been drawn from a

bivariate population of pairs, with (X

, Y

) having joint probability distribution

f(x, y). In Chapter 5, we defined the correlation coefficient r(X, Y)by

r near + 1 r near - 1

r near 0, no

arent relationshi

r near 0, nonlinear

relationshi

Figure 12.24 Data plots for different values of r

12.5 Correlation 665

r ¼ rðX; YÞ¼

CovðX; YÞ

where

CovðX; YÞ¼

ðx  m

Þðy  m

Þf ðx; yÞðX; YÞ discrete

1

ðx  m

Þðy  m

Þf ðx; yÞdx dy ðX; YÞ continuous

(

If we think of f(x, y) as describing the distribution of pairs of values within the

entire population, r becomes a measure of how strongly related x and y are in that

population. Properties of r analogous to those for r were given in Chapter 5 .

The population correlation coefficient r is a parameter or population charac-

teristic, just as m

, m

, s

, and s

are, and we can use the sample correlation

coefficient to make various inferences about r. In particular, r is a point est imate

for r, and the corresponding estimator is

r ¼ R ¼

ðX

 XÞðY

 YÞ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðX

 XÞ

q ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ðY

 YÞ

Example 12.18 In some locations, there is a strong association between concentrations of two

different pollutants. The article “The Carbon Compon ent of the Los Angeles

Aerosol: Source Apportionment and Contributions to the Visibility Budget”

(J. Air Pollution Contr. Fed., 1984: 643–650) reports the accompanying data on

ozone concentration x (ppm) and secondary carbon concentration y (mg/m

x .066 .088 .120 .050 .162 .186 .057 .100

y 4.6 11.6 9.5 6.3 13.8 15.4 2.5 11.8

x .112 .055 .154 .074 .111 .140 .071 .110

y 8.0 7.0 20.6 16.6 9.2 17.9 2.8 13.0

The summary quantities are n ¼ 16,

¼ 1:656,

¼ 170:6,

¼ :196912,

¼ 20:0397,

¼ 2253:56, from which

r ¼

20:0397 ð1:656Þð170 :6Þ=16

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

:196912 ð1 :656Þ

=16

q ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2253:56 ð170:6Þ

=16

2:3826

ð:1597Þð20:8456Þ

¼ :716

The point estimate of the population correlation coefficient r between ozone

concentration and secondary carbon concentration is

r ¼ r ¼ :716.

■

666 CHAPTER 12 Regression and Correlation

The small-sample intervals and test procedures presented in Chapters 8–10

were based on an assumption of popul ation normality. To test hypotheses about r,

we must make an analogous assumpt ion about the distribution of pairs of (x, y)

values in the population. We are now assuming that both X and Y are random, with

joint distribution given by the bivariate normal pdf introduced in Section 5.3.

If X ¼ x, recall that the (conditional) distribution of Y is normal with mean

Yx

¼ m

þðrs

Þðx  m

Þ and variance ð1  r

Þs

. This is exactly the model

used in simple linear regression with b

¼ m

 rm

; b

¼ rs

, and

¼ð1  r

Þs

independent of x. The implication is that if the observed pairs

, y

) are actually drawn from a bivariate normal distribution, then the simple

linear regression model is an appropriate way of studying the behavior of Y for

fixed x.Ifr ¼ 0, then m

Y·x

¼ m

independent of x; in fact, when r ¼ 0 the joint

probability density function f(x, y) can be factored into a part involving x only and a

part involving y only, which implies that X and Y are independent variables.

Example 12.19 As discussed in Section 5.3, contours of the bivariate normal distribution are

elliptical, and this suggests that a scatter plot of observed (x, y) pairs from such a

joint distribution should have a roughly elliptical shape. The accompa nying scatter

plot of y ¼ visceral fat (cm

) by the CT method versus x ¼ visceral fat (cm

)by

the US method for a sample of n ¼ 100 obese women appeared in the paper

“Methods of Estimation of Visceral Fat: Advantages of Ultrasonography” (Obes.

Res., 2003: 1488–1494). Computerized tomography is considered the most accu-

rate technique for body fat measurement, but is costly, time consuming, and

involves exposure to ionizing radiation; the US method is noninvasive and less

expensive.

The pattern in the scatter plot seems consistent with an assumption of

bivariate normality . Here r ¼ .71, which is not all that impressive ( r

¼ .50), but

the investigators reported that a test of H

: r ¼ 0 (to be introduced shortly) gives

P-value < .001. Of course we would want values from the two methods to be very

highly correlated before regarding one as an adequate substitute for the other.

■

100

250

300

200

150

350

81012

Fat by CT

Fat by US

Figure 12.25 Scatter Plot for Example 12.19

12.5 Correlation 667