Stat 401B Exam 2 Fall 2016

Transcription

1 Stat 40B Eam Fall 06 I have neither given nor received unauthorized assistance on this eam. Name Signed Date Name Printed ATTENTION! Incorrect numerical answers unaccompanied by supporting reasoning will receive NO partial credit. Correct numerical answers to difficult questions unaccompanied by supporting reasoning may not receive full credit. SHOW YOUR WORK/EXPLAIN YOURSELF! Completely absurd answers (that fail basic sanity checks but that you don't identify as clearly incorrect) may receive negative credit.

2 . A shop uses spot welding to join two parts of an assembly. Two different machines are used to make the same welds. Measured strengths for welds from the machines are summarized below. Machine # Machine # n = 5 n = 5 = 46.7 psi = 538. psi s = 34.5 psi s = 80.3 psi number above 800 psi = number above 800 psi = 7 number below 00 psi = 3 number below 00 psi = 0 8 pts a) Strengths below 00 psi or above 800 psi are undesirable/nonconforming. Is there clear evidence of a difference between the two machines in terms of the fractions of nonconforming welds they produce? (Show the whole 5-step format.) b) Give 90% two-sided confidence limits for comparing the standard deviations of weld strengths produced by the two machines. 3 pts c) On the basis of your limits from b) can you say with fair certainty which machine produces more consistent weld strengths? Why or why not?

3 Below are normal plots for the weld strengths for the two machines made on the same set of aes. 3 pts d) What does the plot indicate about the two distributions of weld strengths? e) Give (small sample) two-sided 95% confidence limits for the difference in mean weld strengths for the two machines. 5 pts f) If one judged the plot at the top of the page to fail to be approimately linear for one or both of the two machines, what basis might there nevertheless be for hoping that the nominal 95% confidence level used in e) is not ridiculously wrong as characterizing the reliability of the formula you used? 3 pts g) Would the same reasoning you applied in f) etend to use of the tolerance limits ± τ s? Eplain. 3

4 . At the end of this eam there is an R printout that concerns analysis of data from a study of the surface finish produced in a machining operation. A response variable, y, measuring a characteristic of surface finish related to ultimate part strength was studied as a function of machine settings. The first input variable, speed rate, was originally in rpm and has been coded by subtraction of 3500 and then division by 000 to produce values of the variable. The second input variable, feed rate, was originally in inches/revolution and has been coded by subtraction of.005 and then division by.004 to produce values of the variable. To begin, consider only the data with = (the lowest speed rate). Notice that from one perspective this amounts to r = 3 samples, each of size ( n = n = n 3 = ). a) Give two-sided 95% confidence limits for the standard deviation of y for any fied feed rate (at this lowest speed rate) under the one-way normal model. b) Give values of an F statistic and degrees of freedom for testing the hypothesis that (at this lowest speed rate) feed rate has no effect on average y. Is this hypothesis plausible? F = d.f. =, Plausible? 8 pts c) Because the values of the variable are equally spaced, the quantity ((mean y for = ) (mean y for = 0) ) ((mean y for = 0) (mean y for = ) ) is a measure of "curvature" or second derivative in mean response (at this lowest speed rate). Give and interpret 95% two-sided confidence limits for this quantity. 4

5 Continue to consider only the data with = (the lowest speed rate), but now treat a regression analysis with the predictor variable. 4 pts d) What fraction of the raw variation in y (at this lowest speed rate) can be "accounted for" using a model linear in? e) Give 95% two-sided confidence limits for the standard deviation of y at any fied value of under the SLR model (at this lowest speed rate). How does this compare to your answer to a)? Why, in retrospect, is this similarity or difference understandable? f) Give 95% two-sided confidence limits for the increase in mean y associated with a.00 inch/revolution increase in feed rate (according to the SLR analysis at this lowest speed rate). (Hint: look again at the top of the previous page and the description of how was coded. What would the answer here be if it was a " unit increase in " that was under discussion?) g) Give 95% prediction limits (based on the SLR model at this lowest at rate) the net y observed at =. (Plug in completely, but you need not simplify.) 5

6 Finally consider all the data and both and. There are several regressions reported on the output for the entire data set. Use them in answering the following. 5 pts h) Over the ranges represented by the data used here, which of the predictors ( or ) is most important in modeling y? Give quantitative rationale for your answer. Variable: Rationale: i) Give 95% two-sided confidence limits for the increase in mean y associated with a unit increase in when is held fied. 5 pts j) Give and interpret 95% two sided confidence limits for the intercept in the MLR model including both and for this particular data set. 4 pts k) What is the value of the sample correlation between the response and the predicted response for MLR including both and? l) Give 95% two-sided prediction limits for the net y when = 0 and = 0. (You don't need a special prediction call to get y ˆ or se y ˆ here. Consider again part j).) 6

7 R Code and Output > <-c(rep(-,6),rep(0,6),rep(,6)) > <-c(rep(c(-,-,0,0,,),3)) > treat<-c(,,,,3,3,4,4,5,5,6,6,7,7,8,8,9,9) > y<-c(7,9,77,77,93,90,7,9,75,85,9,9,9,8,79,80,9,90) > cbind(treat,,,y) treat y [,] [,] [3,] [4,] [5,] [6,] 3-90 [7,] [8,] [9,] [0,] [,] [,] [3,] [4,] [5,] [6,] [7,] [8,] 9 90 > aggregate(y,by=list(treat),mean) Group > aggregate(y,by=list(treat),sd) Group > summary(aov(y[:6] ~ as.factor([:6]))) Df Sum Sq Mean Sq F value Pr(>F) as.factor([:6]) 3436 Residuals e-06 *** Signif. codes: 0 *** 0.00 ** 0.0 *

8 > summary(lm(y[:6]~[:6])) Call: lm(formula = y[:6] ~ [:6]) Residuals: Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-05 *** [:6] *** Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error: 3. on 4 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: 93.4 on and 4 DF, p-value: > summary(aov(y[:6]~[:6])) Df Sum Sq Mean Sq F value Pr(>F) [:6] *** Residuals Signif. codes: 0 *** 0.00 ** 0.0 * > plot(,) > plot(,y) 8

9 > plot(,y) > summary(lm(y~)) Call: lm(formula = y ~ ) Residuals: Min Q Median 3Q Ma Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) *** Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error: 79.3 on 6 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on and 6 DF, p-value: > summary(lm(y~)) Call: lm(formula = y ~ ) Residuals: Min Q Median 3Q Ma Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) < e-6 *** e-5 *** Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error:.6 on 6 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on and 6 DF, p-value: 5.4e-5 9

10 > summary(lm(y~+)) Call: lm(formula = y ~ + ) Residuals: Min Q Median 3Q Ma Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-5 *** e-4 *** Signif. codes: 0 *** 0.00 ** 0.0 * Residual standard error:.58 on 5 degrees of freedom Multiple R-squared: 0.98, Adjusted R-squared: F-statistic: on and 5 DF, p-value:.797e-3 > summary(aov(y~+)) Df Sum Sq Mean Sq F value Pr(>F) 0.74 Residuals e-4 *** 34 Signif. codes: 0 *** 0.00 ** 0.0 *