regressionConfIntervalInfo - 11.2 SIMPLE LINEAR REGRESSION...

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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:! (ll-12) 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 (11-13) | Computing SSAE using Equation 11-12 would be fairly tedious. A more convenient computing formula can be obtained by substituting if" I [30 + 81x, into Equation 11-12 and simplifying. The resulting computing formula is SSE : SST — an}... (1144) where SST = 2?:101 — i702 : 22-1 1y? — rifz is the total sum of squares of the response t-rariableltr. Formulas such as this are presented in Section E t-4. 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 1-2. EXERCISES. FOR HECT!O.\' l 12 11-1. 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 8531-1 (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] (11-15) 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 5-5 can be applied to show that V<Bl) : g3 (11-16) For the intercept, we can show in a similar manner that A A 2 1 1-2 Elfin) : Bu and V(Bn) : a t * (11-17) 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 11-92) that cov(f‘}0a 81) = —(T2)_C/SH. The estimate of (I2 could be used in Equations 11-16 and 1 1-17 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 11-13. ___ ___... . . iiiiiw ..__.____._.. . ., The Minitab computer output in Table 1 1-2 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 P-valuc 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 r-statistic 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 t-test 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 r-test (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 l1-3, using (I = 5.5 and Bi = 12.5, where the hypotheses are H0: BI 10 versus H1: B, as H]. Eir‘m. Consider the no-intereept model Y: Bx + e with the 5‘s NID(O, 0-2). 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 ll-20. 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.

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regressionConfIntervalInfo - 11.2 SIMPLE LINEAR REGRESSION...

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