HW11

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ISyE 3030 Homework 11 Hannah McClure 11/27/2021 Exercise 11.2.3 Compressive strength x , intrinsic permeability y . n = 14 , ∑ y i = 572 , ∑ y 2 i = 23530 , ∑ x i = 43 , ∑ x 2 i = 157 . 42 , ∑ x i y i = 1697 . 80 a) ˆ β 1 = S xy S xx = ( ∑ i x i y i ) - n ¯ x ¯ y ( ∑ i x 2 i ) - n ¯ x 2 ¯ x = ∑ x i n = 43 14 = 3 . 0714 , ¯ y = ∑ y i n = 572 14 = 40 . 8571 ˆ β 1 = 1697 . 80 - 14 × 3 . 07 × 40 . 8571 157 . 42 - 14 × 3 . 0714 2 = - 59 . 05727 25 . 34853 = - 2 . 32981 ˆ β 0 = ¯ y - ˆ β 1 ¯ x = 40 . 8571 - ( - 2 . 32981 × 3 . 0714) = 48 . 01292 ˆ σ 2 = MSE = SSE n - 2 = SSTO - ˆ β 1 S xy n - 2 SSTO = i y 2 i - n ¯ y 2 = 23530 - 14 × 40 . 8571 2 = 159 . 7633 ˆ σ 2 = SSTO - ˆ β 1 S xy n - 2 = 159 . 7633 - ( - 2 . 32981 × (1697 . 80 - 14 × 3 . 0714 × 40 . 8571)) 14 - 2 = 1 . 847591 plot( 0 , 48.01292 , type = "n" , xlim = c( 0 , 10 ), xlab = "compressive strength (x)" , ylab = "intrinsic permeability (y)" ) abline( 48.01292 , - 2.32981 , col = "green" , lwd = 3 ) 1

0 2 4 6 8 10 30 40 50 60 compressive strength (x) intrinsic permeability (y) b) Compressive strength x = 4 . 3 . ˆ y i = ˆ β 0 + ˆ β 1 x i = 48 . 01292 - 2 . 32981 × 4 . 3 = 37 . 99474 c) Compressive strength x = 3 . 7 . μ y i | x i = E ( y i | x i ) = ˆ β 0 + ˆ β 1 x i = 48 . 01292 - 2 . 32981 × 3 . 7 = 39 . 39262 d) Observed value of permeability at x = 3 . 7 is y = 46 . 1 . e i = y i - ˆ y i ˆ y i = ˆ β 0 + ˆ β 1 x i = 48 . 01292 - 2 . 32981 × 3 . 7 = 39 . 39262 e i = 46 . 1 - 39 . 39262 = 6 . 70738 Exercise 11.2.4 Sound pressure level in decibels x , blood pressure rise in millimeters of mercury y . a) Yes, a simple linear regression model seems reasonable in this situation because it follows a linear pattern. input = read.table( "11-10.txt" , header = TRUE) attach(input) plot(x,y, main = "Relationship Between Noise Exposure and Hypertension" , xlab = "sound pressure level in decibels (x)" , ylab = "blood pressure rise in millimeters of mercury (y)" ) 2

60 70 80 90 100 0 2 4 6 8 Relationship Between Noise Exposure and Hypertension sound pressure level in decibels (x) blood pressure rise in millimeters of mercury (y) b) plot(x,y, main = "Relationship Between Noise Exposure and Hypertension" , xlab = "sound pressure level in decibels (x)" , ylab = "blood pressure rise in millimeters of mercury (y)" ) model = lm(y~x) x = x y = y n = length(x) barx = mean(x); print(barx) ## [1] 82.8 sxx = sum(xˆ 2 )-(sum(x))ˆ 2 /n; print(sxx) ## [1] 3059.2 sxy = sum(x*y)-sum(x)*sum(y)/n; print(sxy) ## [1] 533.2 b1 = sxy/sxx; print(b1) ## [1] 0.1742939 b0 = mean(y)-b1*mean(x); print(b0) ## [1] -10.13154 abline(model, col = "blue" , lwd = 3 ) 3

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60 70 80 90 100 0 2 4 6 8 Relationship Between Noise Exposure and Hypertension sound pressure level in decibels (x) blood pressure rise in millimeters of mercury (y) yhat = b0+b1*x e = y-yhat sse = sum(eˆ 2 ) sigma2 = sse/(n -2 ); print(sigma2) ## [1] 1.737026 summary(model) ## ## Call: ## lm(formula = y ~ x) ## ## Residuals: ## Min 1Q Median 3Q Max ## -1.8120 -0.9040 -0.1333 0.5023 2.9310 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -10.13154 1.99490 -5.079 7.83e-05 *** ## x 0.17429 0.02383 7.314 8.57e-07 *** ## --- ## Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 ## ## Residual standard error: 1.318 on 18 degrees of freedom ## Multiple R-squared: 0.7483, Adjusted R-squared: 0.7343 ## F-statistic: 53.5 on 1 and 18 DF, p-value: 8.567e-07 ˆ β 0 = - 10 . 13154 , ˆ β 1 = 0 . 17429 ˆ σ = 1 . 318 → ˆ σ 2 = 1 . 737 4

c) pmean = b0+b1* 85 ; print(pmean) ## [1] 4.683447 Predicted mean of rise in blood pressure level associated with a sound pressure level of 85 decibels is 4.683. Exercise 11.4.3 a) ˆ β 0 = 12 . 857 , ˆ β 1 = 2 . 3445 , se ( ˆ β 1 ) = 1 . 032 , se ( ˆ β 1 ) = 0 . 115 , SSE = 17 . 55 , MSR = 912 . 43 t-value for ˆ β 1 → t 0 = ˆ β 1 ˆ σ 2 S xx = 2 . 3445 0 . 115 = 20 . 387 pval = P ( t 8 > 20 . 287) × 2 = 3 . 6423 × 10 - 8 t-value for ˆ β 0 → t 0 = ˆ β 0 ˆ σ 2 1 n + ¯ x 2 S xx = 12 . 857 1 . 032 = 12 . 458 pval = P ( t 8 > 12 . 458) × 2 = 1 . 6107 × 10 - 6 MSE = SSE n - 2 = 17 . 55 8 = 2 . 1938 f 0 = MSR MSE = 912 . 43 2 . 1938 = 415 . 913 pval = P ( f 1 , 8 > 415 . 913) = 3 . 4948 × 10 - 8 ( 1 -pt( 20.287 , 8 ))* 2 ## [1] 3.642264e-08 ( 1 -pt( 12.458 , 8 ))* 2 ## [1] 1.610708e-06 1 -pf( 415.913 , 1 , 8 ) ## [1] 3.494749e-08 b) pval = 3 . 4948 × 10 - 8 ≈ 0 → Reject H 0 : β = 0 c) ˆ σ 2 = MSE = 2 . 1938 as calculated in (a). Exercise 11.4.5 a) α = 0 . 05 H 0 : β 1 = 0 , H 1 : β 1 = 0 f 0 = MSR MSE MSR = SSR/ 1 = SSR = ˆ β 1 S xy = - 2 . 32981 × - 59 . 05727 = 137 . 5922 MSE = ˆ σ 2 = 1 . 847 f 0 = MSR MSE = 137 . 5922 1 . 847591 = 74 . 47114 pval = P ( f 1 , 14 - 2 > 74 . 47114) = 1 . 71 × 10 - 6 (as calculated by R below) pval = 1 . 71 × 10 - 6 < α = 0 . 05 → Reject H 0 5

1 -pf( 74.47114 , 1 , 12 ) ## [1] 1.716851e-06 b) ˆ σ 2 = 1 . 847591 as found in part a) of 11.2.3 c) se ( ˆ β 0 ) = ˆ σ 2 1 n + ¯ x 2 S xx = 1 . 847591 1 14 + 3 . 071429 2 25 . 34853 = 0 . 9053 Exercise 11.4.6 a) α = 0 . 05 H 0 : β 1 = 0 , H 1 : β 1 = 0 f 0 = MSR MSE MSR = SSR/ 1 = SSR = ˆ β 1 S xy = 0 . 17429 × 533 . 2 = 92 . 93351 MSE = ˆ σ 2 = 1 . 737 f 0 = MSR MSE = 92 . 93351 1 . 737 = 53 . 50151 pval = P ( f 1 , 20 - 2 > 53 . 50151) = 8 . 567 × 10 - 7 (as calculated by R below) pval = 8 . 567 × 10 - 7 < α = 0 . 05 → Reject H 0 1 -pf( 53.50151 , 1 , 18 ) ## [1] 8.566753e-07 b) se ( ˆ β 0 ) = ˆ σ 2 1 n + ¯ x 2 S xx = 1 . 737 1 20 + 82 . 8 2 3059 . 2 = 1 . 9949 se ( ˆ β 1 ) = ˆ σ 2 S xx = 1 . 737 3059 . 2 = 0 . 0238 c) H 0 : ˆ β 0 = 0 , H 1 : ˆ β 0 = 0 , α = 0 . 05 t 0 = ˆ β 0 ˆ σ 2 1 n + ¯ x 2 S xx = - 10 . 132 1 . 737 ( 1 20 + 82 . 8 2 3059 . 1 ) = - 10 . 132 1 . 9949 = - 5 . 0787 t α 2 ,n - 2 = t 0 . 025 , 18 = 2 . 1009 (as calculated by R below) | t 0 | = 5 . 0787 > t 0 . 025 , 18 = 2 . 1009 → Reject H 0 pval = P ( t 18 > | - 5 . 0787 | ) = 7 . 835 × 10 - 5 < α = 0 . 05 → Reject H 0 qt( 0.975 , 18 ) ## [1] 2.100922 Exercise 11.6.4 Sound pressure level (x), blood pressure rise (y). Find 95% confidence interval on. . . a) ˆ β 1 6

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model2 = lm(y~x) confint(model2) ## 2.5 % 97.5 % ## (Intercept) -14.3226668 -5.940409 ## x 0.1242318 0.224356 The 95% confidence interval on ˆ β 1 is [0.1242,0.2244]. b) ˆ β 0 The 95% confidence interval on ˆ β 0 is [-14.3227,-5.9404]. c) Mean blood pressure rise when the sound pressure level is 85 decibels. ˆ μ Y | x 0 ± t α 2 ,n - 2 ˆ σ 2 1 n + ( x 0 - ¯ x ) 2 S xx ˆ μ Y | 85 = ˆ β 0 + ˆ β 1 × 85 = - 10 . 1315 + 0 . 17429 × 85 = 4 . 68315 4 . 68315 ± t 0 . 05 2 , 20 - 2 1 . 737 1 20 + (85 - 82 . 8) 2 3059 . 1 → 4 . 68315 ± 2 . 1009 1 . 737 1 20 + (85 - 82 . 8) 2 3059 . 1 → [4 . 054288 , 5 . 312012] d) 95% prediction interval on blood pressure rise when the sound pressure level is 85 decibels. predict(model2, newdata = data.frame( SQFT= 85 ), interval= "prediction" ) ## Warning: newdata had 1 row but variables found have 20 rows ## fit lwr upr ## 1 0.3260983 -2.7322013 3.384398 ## 2 0.8489801 -2.1564985 3.854459 ## 3 1.1975680 -1.7763914 4.171527 ## 4 2.0690377 -0.8397393 4.977815 ## 5 2.0690377 -0.8397393 4.977815 ## 6 2.0690377 -0.8397393 4.977815 ## 7 3.8119770 0.9712000 6.652754 ## 8 5.5549163 2.6947961 8.415037 ## 9 3.8119770 0.9712000 6.652754 ## 10 3.8119770 0.9712000 6.652754 ## 11 4.6834467 1.8439933 7.522900 ## 12 5.3806224 2.5263792 8.234866 ## 13 5.5549163 2.6947961 8.415037 ## 14 5.5549163 2.6947961 8.415037 ## 15 5.5549163 2.6947961 8.415037 ## 16 5.5549163 2.6947961 8.415037 ## 17 6.2520921 3.3599052 9.144279 ## 18 7.2978556 4.3327580 10.262953 ## 19 7.2978556 4.3327580 10.262953 ## 20 7.2978556 4.3327580 10.262953 7

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