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Unformatted text preview: 11.2 SIMPLE LINEAR REGRESSION 399 Estimating (T: There is actually another unknown parameter in our regression model, 03 (the variance of the
error term 6). The residuals e, : _v, i )7, are used to obtain an estimate of 0'2. The sum of
squares of the residuals, often calted the error sum of squares. is H SSE : E 97 : 2i}? ’ fill
i=1 1:! (ll12) We can show that the expected value of the error sum of squares is 151555) I (r: — 2)rr2.
Therefore an unbiased estimator of 0'2 is lisliiualur i ul‘ Variance ‘ SSE l (1113)  Computing SSAE using Equation 1112 would be fairly tedious. A more convenient computing
formula can be obtained by substituting if" I [30 + 81x, into Equation 1112 and simplifying.
The resulting computing formula is SSE : SST — an}... (1144) where SST = 2?:101 — i702 : 221 1y? — rifz is the total sum of squares of the response
trariableltr. Formulas such as this are presented in Section E t4. The error sum of squares and
the estimate oftr2 for the oxygen purity data, 62 : 1.18, are highlighted in the Minitab output in Table 1 12. EXERCISES. FOR HECT!O.\' l 12 111. An article in Concrete Research (“Near Surface Characteristics of Concrete: Intrinsic Permeability." Vol. 41, 1989). presented data on compressive strength x and intrinsic per meability}: of various concrete mixes and cures. Summary quan tities are rt = 14. 21} : 572, 2y? : 23.530, 2x, : 43. 21,; : 157.42, and Exy; : 1697.80. Assume that the two variables are related according to the simple linear regression model. (a) Calculate the least squares estimates of the slope and intercept.
Estimate oz. Graph the regression line. (b) Use the equation of the ﬁtted line to predict What perme—
ability would be observed when the compressive strength
is x : 4.3. it) Give a point estimate of the mean permeability when
compressive strength is x = 3.7. (d) Suppose that the observed value of permeability at .r :
3.7 is y = 46.1. Calculate the value of the corresponding
residual. l !—3. Regression methods were used to analyze the data
from a study investigating the relationship between roadway
surface temperature (x) and pavement deflection (y). Summary
quantities were it : 20, 2y, : 12.75. 2y? : 8.86. 2x, I
1478, Ex? : 143,215.s,andzx,y, = 1083.67, (a) Calculate the least squares estimates ofthe slope and in
tercept. Graph the regression line. Estimate oz. (b) Use the equation of the ﬁtted line to predict what pave
ment deﬂection would be observed when the surface
temperature is 85311 (c) What is the mean pavement deflection when the surface
temperature is 90°F? (d) What change in mean pavement deflection would be ex
pected for a 13F change in surface temperature? I 1—5. The following table presents data on the ratings of
quarterbacks for the 2004 National Football League season
(source: The Sports .‘\"etwor7r). [t is suspected that the rating (r) 404 ll , J CHAPTER 11 SIMPLE LINEAR REGRESSION AND CORRELATION PROPERTIES 0]" THE LEAST SQUARES IiSTlMA'i'ORS lastimnlcd 3
Mn miard 5
lirrm's ! The statistical properties of the least squares estimators E30 and {31 may be easily described
Recall that we have assumed that the error term 6 in the model Y = BO + le + e is a random
variable with mean zero and variance 02. Since the values of): are ﬁxed, Yis a random vari
able with mean uYL : [30 + BI): and variance 0'2. Therefore, the values of EU and [31 depend
on the observed y’s; thus, the least squares estimators of the regression coefﬁcients maybe
viewed as random variables. We will investigate the bias and variance properties of the least n squares estimators B.) and B]. Consider ﬁrst [31. Because 81 is a linear combination of the observations Y” we can use
properties of expectation to show that the expected value of B] is 13(81): B] (1115) Thus, BI ‘15 an unbiased estimator ofthe true slope [31.
Now consider the variance of BI. Since we have assumed that V(eI) = oz, it follows that V(Y,) : (:2. Because 8] is a linear combination of the observations Yi, the results in Section
55 can be applied to show that V<Bl) : g3 (1116) For the intercept, we can show in a similar manner that
A A 2 1 12
Elﬁn) : Bu and V(Bn) : a t * (1117) Thus, S0 is an unbiased estimator of the intercept B”. The covariance of the random vari
ables 80 and B1 is not zero. It can be shown {see Exercise 1192) that cov(f‘}0a 81) =
—(T2)_C/SH. The estimate of (I2 could be used in Equations 1116 and 1 117 to provide estimates of
the variance of the slope and the intercept. We call the square roots of the resulting variance
estimators the estimated standard errors of the slope and intercept, respectively. In simple linear regression the estimated standard error of the slope and the
estimated standard error of the intercept are A 62 A l“ 1 3c? seam): and se(BO)= 02[R+E} 2 respectively, where c‘r is computed from Equation 1113. ___ ___... . . iiiiiw ..__.____._.. . ., The Minitab computer output in Table 1 12 reports the estimated standard errors of the slope
and intercept under the column heading “SE coeff.” 410 plasma and ccrebrospinal ﬂuid lcptin measurements. The data
follow: '1'ZBMl(kgim2): 19.92 20.59 29.02 20.78 25.97 20.39 23.29 [7.27 35.24 45.5 34.6 40.6 32.9 28.2
52.] 33.3 47.0 .r 2 Age (yr): 30.] (a) Test for signiﬁcance of regression using L‘t : 0.05. Find
the Pvaluc for this test. Can you conclude that the model
speciﬁes a useful linear relationship between these two
variables? (b) Estimate o2 and the standard deviation of B1. (c) What is the standard error of the intercept in this model?
iii Suppose that each value ofx, is multiplied by a pos itive constant a. and each value of}; is multiplied by another positive constant [7. Show that the rstatistic for testing H0: BI = 0 versus H,: B[ i 0 is unchanged in value. 5;,5,‘t7‘vi‘li§lf7‘~i{.'li {N‘l'liRVALS CHAPTER I 1 SIMPLE LINEAR REURESSION AND CORRELATION IiVH. The type II error probability for the ttest for
H0: B] B”, can be computed in a similar manner to the t
tests of Chapter 9. if the true value of BI is B}, the value
at : LB”) * Bil/(0V(n — l)/S,.x is calculated and used as
the horizontal scale factor on the operating characteristic
curves for the rtest (Appendix Charts Vile through Vllh) and
the type II error probability is read from the vertical scale
using the curve for n — 2 degrees of freedom. Apply this pro
cedure to the football data of Exercise l13, using (I = 5.5
and Bi = 12.5, where the hypotheses are H0: BI 10 versus
H1: B, as H]. Eir‘m. Consider the nointereept model Y: Bx + e
with the 5‘s NID(O, 02). The estimate of 02 is s2 =
2:10? ’ SIX/(’1 ‘ 1) and : Uz/Ellzlxi‘z‘ (a) Devise a test statistic for H“: B = 0 versus Hp B 9'5 0.
(b) Apply the test in (a) to the model from Exercise ll20. i L‘in‘ttitienct li'itcrvals 0n the “Slope and Intercept. In addition to point estimates of the slope and intercept, it is possible to obtain confidence
interval estimates of these parameters. The width of these conﬁdence intervals is a measure of
the overall quality of the regression line. If the error terms, 6,, in the regression model are
normally and independently distributed, *2
(in e Ban/W9” and (a, _ M [all + a] ’1 Six are both distributed as trandom variables with n w 2 degrees of freedom. This leads to the following deﬁnition of 100(i * 0t) % conﬁdence intervals on the slope and intercept. Under the assumption that the observations are normally and independently distributed,
a 100“ — a)% conﬁdence interval on the slope B, in simple linear regression is A 62 
Bl * {warez _ 5 Bi 5 Bl + {man—2
\J St. A2 1 _
SH ( l 29) Similarly, a l00(l * 00% confidence intcrml 0n the intercept B0 is A l '7 £2 7
_ ‘2 _ _
Bo {oi/23772 U [n + SUE] ...
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This note was uploaded on 05/01/2011 for the course CHBE 2120 taught by Professor Gallivan during the Spring '07 term at Georgia Institute of Technology.
 Spring '07
 Gallivan

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