For All Significance Tests, Use a = 0.05 Significance Level
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STA 6167 – Exam 1 – Spring 2014 – PRINT Name ______
For all significance tests, use = 0.05 significance level.
Q.1. A simple linear regression was fit relating number of species of arctic flora observed (Y) and July mean temperature (X, in Celsius). The results of the regression model, based on n=19 temperature stations is given below.
ANOVA df SS MS F Significance F Regression 1 39858 39858 79.87 0.0000 Residual 17 8484 499 Total 18 48342
CoefficientsStandard Errort Stat P-value Lower 95%Upper 95% Intercept -34.49 16.56 -2.08 0.0527 -69.43 0.46 JulyTemp 24.60 2.75 8.94 0.0000 18.79 30.41
SS_XX X-bar 65.85 5.7 p.1.a. What proportion of the variation in number of species is “explained” by mean July temperature? R2 = SSR/TSS = 39858/48342 = .8245 p.1.b. Compute a 95% Confidence Interval for the population mean number of species, with mean July temperature of 6 degrees.
^ Y 6 = -34.49 + 24.60(6) = 113.11
2 2 2 骣 1(6-X) 1(6- 5.7) ^ 琪 1 ( 6 - X ) + = + =.0540SE Y6 = MSE + = 499(.0540) = 5.19 n S19 65.85 { } 琪 n S XX桫 XX t (.025,17) = 2.110 95% CI: 113.11焙 2.110(5.19) 113.11 焙 10.95 (102.16,124.06) p.1.c. Compute a 95% Prediction Interval for the number of species, at a single station with mean July temperature of 6 degrees.
^ Y 6,new = -34.49 + 24.60(6) = 113.11 骣 2 ^ 琪 1 (6 - X ) SE Y6,new = MSE 1 + + = 499(1.0540) = 22.93 { } 琪 n S 桫 XX t (.025,17) = 2.110 95% CI: 113.11焙 2.110(22.93) 113.11 焙 48.39 (64.72,161.50) Q.2. An experiment was conducted, relating the penetration depth of missiles (Y) to its impact factor (X). The results from the regression, and the residual versus fitted plot are given below (n=25).
Residuals vs Fitted Values ANOVA 0.4 df SS 0.3 Regression 1 1.585884 0.2 Residual 23 0.713406 0.1 Total 24 2.29929 0 Residuals -0.1 0 0.5 1 1.5 2 CoefficientsStandard Error-0.2 Intercept 0.633253 0.103076 -0.3 impact 0.06 0.008391 -0.4 p.2.a. Test H0: 1 = 0 (Penetration depth is not associated with impact factor) based on the t-test.
^ b 0.06 TS: t=1 = = 7.15 RR : t� t ( .025,23) 2.069 obs^ 0.008391 obs SE b { 1}
p.2.b. Test H0: 1 = 0 (Penetration depth is not associated with impact factor) based on the F-test. MSR (SSR 1) ( 0.713406 1) TS: F= = = = 51.13 RR : F� F ( .05,1,23) 4.279 obsMSE( SSE 23) ( 2.9929 23) obs p.2.c. The residual plot appears to display non-constant error variance. A regression of the squared residuals on the impact factors (X) is fit, and the ANOVA is given below. Conduct the Breusch-Pagan test to test whether the errors are related to X. Do you reject the null hypothesis of constant variance? Yes or No
ANOVA df SS Regression 1 0.007959 Residual 23 0.025824 Total 24 0.033783
SSR 2 2( e2 ) (0.007959 2) 0.0039795 2 2 TS: XBP= = = = 4.887 RR : X BP �c ( .05,1) 3.841 20.713406 25 2 0.00081432 (SSEy n) ( ) Q.3. An experiment was conducted to measure air permeability of fabric (Y) as a function of the following factors: warp density (X1), weft density (X2), and Mass per unit area (X3). There were n=30 observations, and 4 models are fit:
2 2 2 E( Y) =b0 + b 1 X 1 + b 2 X 2 + b 3 X 3 + b 12 X 1 X 2 + b 13 X 1 X 3 + b 23 X 2 X 3 + b 11 X 1 + b 22 X 2 + b 33 X 3 SSE = 72.4
E( Y) =b0 + b 1 X 1 + b 2 X 2 + b 3 X 3 + b 12 X 1 X 2 + b 13 X 1 X 3 + b 23 X 2 X 3 SSE = 86.5
E( Y) =b0 + b 1 X 1 + b 2 X 2 + b 3 X 3 SSE = 813.6
E( Y) =b0 + b 1 X 1 + b 2 X 2 + b 3 X 3 + b 23 X 2 X 3 SSE = 122.7 p.3.a. Use the first two models to test H0: .
Complete Model 1:SSE= 72.4 dfE = 30 - 10 = 20
Reduced Model 2:SSE= 86.5 dfE = 30 - 7 = 23
骣86.5- 72.4 琪 桫 23- 20 4.70 TS: F= = = 1.298 RR : F� F ( .05,3,20) 3.098 obs骣72.4 3.62 obs 琪 桫20 p.3.b. Use the 3rd and 4th models to test whether the weft-mass interaction is significant, controlling for all main effects.
Complete Model 4:SSE= 122.7 dfE = 30 - 5 = 25
Reduced Model 3:SSE= 813.6 dfE = 30 - 4 = 26
骣813.6- 122.7 琪 桫 26- 25 690.9 TS: F= = = 140.77 RR : F� F ( .05,1,25) 4.242 obs骣122.7 4.91 obs 琪 桫 25 Q.4. A regression model was fit, relating the share of big 3 television network prime-time market share (Y, %) to household penetration of cable/satellite dish providers (X = MVPD) for the years 1980-2004 (n=25). The regression results and residual versus time plot are given below.
ANOVA Residuals df SS MS F 6 Regression 1 7073.7 7073.7 685.7 4
Residual 23 237.3 10.3 2
Total 24 7311.0 0 Residuals 1 3 5 7 9 11 13 15 17 19 21 23 25 -2 CoefficientsStandard Error t Stat P-value -4 Intercept 112.029 2.090 53.61 0.0000 mvpd -0.863 0.033 -26.19 0.0000 -6 p.4.a. Compute the correlation between big 3 market share and MVPD.
SSR 7073.7 ^ R2= = =.9675 R = sgnb R 2 = - .9836 TSS 7311.0 { 1} p.4.b. The residual plot appears to display serial autocorrelation over time. Conduct the Durbin-Watson test, with null hypothesis that errors are not autocorrelated.
25 2 (et- e t-1 ) =161.4 d L(a = 0.05, n = 25, p = 1) = 1.29 d U ( a = 0.05, n = 25, p = 1) = 1.45 t=2
25 2 (et- e t-1 ) t=2 161.4 DW=n = =0.6802� dL (a = 0.05, n = 25, p = 1) 1.29 Reject H0 2 237.3 et t=1 p.4.c. Data were transformed to conduct estimated generalized least squares (EGLS), to account for the auto-correlation.
The parameter estimates and standard errors are given below. Obtain 95% confidence intervals for 1, based on Ordinary Least Squares (OLS) and EGLS. Note that the error degrees’ of freedom are 23 for OLS and 22 for EGLS (estimated the autocorrelation coefficient). beta-egls SE(b-egls) 110.577 3.469 -0.845 0.055 OLS: t ( .025,23) = 2.069 : - 0.863焙 2.069(0.033) - 0.863 焙 0.068( - 0.931, - 0.795) EGLS: t ( .025,22) = 2.074 : - 0.845焙 2.074(0.055) - 0.845 焙 0.114( - 0.959, - 0.731) Q.5. Regression analyses were fit, relating various chemical levels to age for stranded bottlenose dolphins in South Carolina and Florida. This plot gives the quadratic fit, relating mercury/selenium molar ratio (Y) to age (X) for the Florida dolphins. Complete 2 the following parts. Note: The data were NOT centered. The model fit was: E(Y) = 0 + 1X + 2X
n = 14 Predicted value when age = 15: 0.1295 + 0.1479(15) – 0.0046(152) = 1.313
Test H0: 1 = 2 = 0
2 臌轾R p [.6151 2] TS: Fobs= = = 8.79 RR : F obs � F ( .05,2,11) 3.982 轾1-R2 n - p ' 轾.3849 14- 3 臌( ) ( ) 臌( ) ( ) Q.6. A study was conducted to determine which factors were associated with percent release (Y) of hydroxypropyl methylcellulose (HPMC) tablets. The factors were: X1 = Carr’s compressibility index, X2 = angle of repose, X3 = solubility, X4 =molecular weight, X5 = compression force X6 = apparent viscosity of 4% (w/v) HPMC.
The sample size was n=18, and the authors reported the fit of the following models. p.6.a. Complete the table in terms of AIC and SBC (BIC).
Predictors p' SSE AIC SBC X1,X2,X3,X4,X5,X6 7 42.62 29.5151 35.7477 X1,X2,X3,X4,X5 6 42.62 27.5151 32.8574 X1,X2,X3,X4 5 48.58 27.8711 32.3230 X1,X3,X4,X6 5 48.58 27.8711 32.3230 X1,X3,X4 4 52.86 27.3910 30.9524 X2,X3,X4 4 75.31 33.7623 37.3238 X3,X4,X6 4 48.85 25.9709 29.5324
Models 3 and 4: AIC =18( ln(48.58)) + 2(5) - 18( ln(18)) = 69.8978 + 10 - 52.0267 = 27.8711 SBC =18( ln(48.58)) +( ln(18)) (5) - 18( ln(18)) = 69.8978 + 14.4519 - 52.0267 = 32.3230 Model 7: AIC =18( ln(48.85)) + 2(4) - 18( ln(18)) = 69.9976 + 8 - 52.0267 = 25.9709 SBC =18( ln(48.85)) +( ln(18)) (4) - 18( ln(18)) = 69.9976 + 11.5615 - 52.0267 = 29.5324 p.6.b. Which model is “best” based on AIC: Model 7 BIC: Model 7 p.6.c. R2 for the complete model was 0.9278. Compute the total (corrected) sum of squares (TSS):
SSE SSE SSE 42.62 42.62 R2=1 -� - 1 � R 2 = TSS = = 590.305 TSS TSS1- R2 1 - .9278 .0722