hw2fall10sol

# hw2fall10sol - 0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 8 10 12...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 route-number 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 Max-2.2-1.2 0.3 0.8 1.8 Coefficients: Estimate Std. Error t value Pr(&gt;|t|) (Intercept) 10.2000 0.6633 15.377 3.18e-07 *** x 4.0000 0.4690 8.528 2.75e-05 ***--- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 1.483 on 8 degrees of freedom Multiple R-Squared: 0.9009, Adjusted R-squared: 0.8885 F-statistic: 72.73 on 1 and 8 DF, p-value: 2.749e-05 (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 2.75e-05 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: &gt; Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max-2.2-1.2 0.3 0.8 1.8 Coefficients: Estimate Std. Error t value Pr(&gt;|t|) (Intercept) 10.2000 0.6633 15.377 3.18e-07 *** x 4.0000 0.4690 8.528 2.75e-05 ***--- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 1.483 on 8 degrees of freedom Multiple R-Squared: 0.9009, Adjusted R-squared: 0.8885 F-statistic: 72.73 on 1 and 8 DF, p-value: 2.749e-05 &gt; 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...
View Full Document

## 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.

### Page1 / 6

hw2fall10sol - 0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 8 10 12...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online