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Unformatted text preview: 0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 8 10 12 14 16 18 20 22 x y Figure 1: The fitted line using the shipment routenumber of ampules data STAT5044: Regression and ANOVA The Solution of Homework #2 Inyoung Kim Problem# 1. The shipment route (X) and the number of ampules to be broken upon arrival (Y). The summary of simple linear regression fit is following: Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max2.21.2 0.3 0.8 1.8 Coefficients: Estimate Std. Error t value Pr(>t) (Intercept) 10.2000 0.6633 15.377 3.18e07 *** x 4.0000 0.4690 8.528 2.75e05 *** Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 1.483 on 8 degrees of freedom Multiple RSquared: 0.9009, Adjusted Rsquared: 0.8885 Fstatistic: 72.73 on 1 and 8 DF, pvalue: 2.749e05 (a) The estimated function is Y = 10 . 2 + 4 * x . Linear regression function apprears to give a good fit here because the data are around the linear regression function and the pvalue of testing 1 = 0 is 2.75e05 supporting that linear fit is statistically 1 significant, R 2 = 0 . 9, and pearson correlation is 0.949158. However, the number of unique values of X are only four. If we have more various values of X, the fitted line would be more informative. (b) A point estimate of the expected number of broken ampules when X = 1 transfer is made is Y = 10 . 2 + 4 * 1 = 14 . 2 (c) The expected number of ampules broken when there are 2 transfer is Y = 10 . 2 + 4 * 2 = 18 . 2. Therefore, the estimation of the increase is Y (2) Y (1) = 4(2 1) = 4. (d) Since x = 1 and y = 14 . 2, y = 10 . 2 + 4 * x . Problem# 2. Refer to Airfreight breakage (Problem 1). (a) Estimate 1 with a 95% confidence interval. Interpret your interval estimate. A.(a) Summary of simple linear regression: > Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max2.21.2 0.3 0.8 1.8 Coefficients: Estimate Std. Error t value Pr(>t) (Intercept) 10.2000 0.6633 15.377 3.18e07 *** x 4.0000 0.4690 8.528 2.75e05 *** Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 1.483 on 8 degrees of freedom Multiple RSquared: 0.9009, Adjusted Rsquared: 0.8885 Fstatistic: 72.73 on 1 and 8 DF, pvalue: 2.749e05 > confint(lmfit) 2.5 % 97.5 % (Intercept) 8.670370 11.729630 x 2.918388 5.081612 With confidence coefficient .95, we estimate that the mean number of broken ampules increases by somewhere between 2.92 and 5.08 for each additional unit time that a carton is transfered. NOTE: In 95% confidence interval, 95% is NOT a probability that interval includes means of number of broken ampules...
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This note was uploaded on 01/02/2012 for the course STAT 5044` taught by Professor Staff during the Fall '11 term at Virginia Tech.
 Fall '11
 Staff

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