Department of Mathematics & Statistics Stat 2593 Final Examination 19 April 2001

Transcription

1 Department of Mathematics & Statistics Stat 2593 Final Examination 19 April 2001 TIME: 3 hours. Total Marks: 60. Indicate your answers clearly. Show all work. Remember to answer as a statistician should. 1. The average number of flaws in a bolt of a certain type of woven woolen cloth is marks The area of a bolt of this cloth is 500 square metres. Suppose that the number of flaws in any region of this cloth is independent of the number in any other nonoverlapping region. A single square metre of cloth is examined for flaws; the number of flaws in this square metre of cloth is denoted by X. Specify the name of the distribution of X, and the value of the mean and standard deviation of X. The probability that X is at most 3 is Use this fact to help you find the probability that X is at least Buck (from Deer Island) plays professional baseball. The probability that Buck will 6 marks get a hit against a right-handed pitcher is 0.35, and the probability of his getting a hit against a left-handed pitcher is Of the pitchers that Buck faces, 70% are right-handers, and 30% are left-handers. Find Buck s long-run batting average. (Note for non-baseball fans: batting average is just probability of getting a hit.) In a particular at-bat during the 2000 season, Buck was credited with a hit. Given this information, calculate the probability that the pitcher was right-handed. (i) Are the events Buck gets a hit and Buck faces a right-handed pitcher independent? Explain briefly. (ii) Are the events Buck gets a hit and Buck faces a right-handed pitcher mutually exclusive? Explain briefly.

2 3. A sample of 1/2 inch steel anchor bolts were measured for shear strength. A stem-leaf 6 marks plot of the resulting measurements, in kip, is shown below. Stem-and-leaf of s.stren N = 112 Leaf Unit = (16) Compute the value of the point estimate of the proportion of bolts (in the population from which the sample was drawn) having shear strength between 7.5 and 9.0 kip (inclusive). (Hint: From the stem-and-leaf plot, how many of the bolts in the sample have shear strength between 7.5 and 9.0 kip (inclusive)? What is the sample size?) Now estimate (answering as a statistician should) the proportion of bolts from this population having shear strength between 7.5 and 9.0 kip (inclusive). How many bolts should be sampled to estimate this proportion to within ±0.03, with 95% confidence? 4. Water samples were taken at four different locations in a river to determine whether 6 marks the quantity of dissolved oxygen, a measure of water pollution, varied from one location to another. Locations 1 and 2 were upstream from an industrial plant, one near shore, the other in midstream. Location 3 was adjacent to the industrial plant, and location 4 was slightly downriver, in midstream. The same number of water samples was taken from each location, but one was lost in the laboratory. The data were analyzed in Minitab. The results (parts of which have been obliterated) are shown on page 3....(continued over page) 2

3 MTB > Oneway oxy locn ; SUBC> Tukey 5. One-Way Analysis of Variance Analysis of Variance on oxy Source DF SS MS F p locn (i) (iv) (v) (vii) ***** Error (ii) (vi) Total (iii) Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev (--*---) (--*---) (---*--) (--*---) Pooled StDev = (viii) Tukey s pairwise comparisons Family error rate = Individual error rate = Critical value = 4.08 Intervals for (column level mean) - (row level mean) Supply the missing numbers labeled (i) through (viii) in the foregoing Minitab output. On the basis of your calculations complete (at the 0.05 significance level)the hypothesis test, being effected by the Minitab analysis. State your decision clearly and explain briefly in words what your decision means. Draw a diagram to summarize succinctly what the Tukey Multiple Comparisons output tells you about pairwise differences in means. (List the population means in the order implied by the corresponding sample means; indicate by underlining which means are evidently different, and which are not.) 3

4 5. Suppose that the length of time to failure, in thousands of hours, for a 6 marks transistor is a random variable Y with cumulative distribution function F(y) = { 1 e y 2 if y 0 0 if y<0. Find the probability that such a transistor operates for at least 1500 hours. Consider the sub-population of the transistors described in part that last for at least 1500 hours. Of this sub-population what fraction (or proportion ) last for at least 2000 hours? Now suppose that the lifetime Y of such transistors is modelled not by the cumulative distribution function specified in part, but by a Gamma distribution with parameters α and β. If the mean of Y is 0.84 and the variance of Y is 0.21, what are the values of α and β? 6. One of the operations in a plant consists of thread-grinding fittings for an aircraft 6 marks hydraulic system. The process is monitored every hour, by selecting a sample of five (5) fittings and measuring the pitch diameter of the threads in each fitting. From extensive sampling when the process was known to be in statistical control, it is know that the mean of the process is µ 0 = 35, and the standard deviation for the process is σ 0 = 4. (The units are inch in excess of inch.) Determine the upper and lower 3 sigma control limits for an x-chart. Suppose that the process goes out of control in that its mean changes to 39. (The standard deviation of the process does not change.) What is the probability that the problem will be detected (i.e. that x will fall outside the 3 sigma limits) the next time the process is monitored? (Make the usual assumption that goes with x-charts.) Now suppose that the process goes out of control in such a way that the probability that x falls outside of the 3 sigma limits (each time the process is monitored) is 0.6. What is the probability (assuming of course that the process is permitted to continue running) that exactly 4 of the next 7 values of x will fall outside of the 3 sigma limits? 7. Two methods for determining the percentage of iron in ore samples were compared 6 marks by splitting each of 87 ore samples into two parts. One part was then measured using method 1, and the other part was measured using method 2. The data were analyzed in Minitab in two ways, as shown on page 5. One of the analyses is correct and the other is incorrect....(continued over page) 4

5 MTB > # Analysis number 1: MTB > let Diff = Meth.1 - Meth.2 MTB > TInterval 95.0 Diff. Variable N Mean StDev SE Mean 95.0 % C.I. Diff ( 0.015, 0.434) MTB > TTest 0.0 Diff ; SUBC> Alternative *. Test of mu = vs ***** Variable N Mean StDev SE Mean T P-Value Diff MTB > # Analysis number 2: MTB > TwoSample 95.0 Meth.1 Meth.2 ; SUBC> Alternative *. Twosample T for Meth.1 vs Meth.2 N Mean StDev SE Mean Meth Meth % C.I. for mu Meth.1 - mu Meth.2: ( -1.25, 1.70) T-Test mu Meth.1 = mu Meth.2 (vs *****): T= 0.30 P=0.76 DF= 171 State which of the two analyses is correct, giving a brief reason for your choice. State appropriate null and alternative hypotheses to be tested using these data. Explain briefly your choice of alternative hypothesis. On the basis of the correct analysis, state your decision about the hypothesis test, at the 0.05 significance level. Explain briefly in words what your decision means. 8. In an article in The Journal of the American Medical Association the relationship 6 marks between smoking habits and household income was studied for males aged 15 years and older in the Minhang district of China. The subjects of the study were classified as Current smokers, Former smokers, and Never Smokers, and were also classified into 5 income groups (1: less than 5000 yuan, 2: 5000 to 6999 yuan, 3: 7000 to 9999 yuan, 4: 10,000 to 14,999 yuan, and 5: 15,000 yuan or more). The counts were analyzed in Minitab; the results, parts of which have been obliterated, are given on page 6....(continued over page) 5

6 MTB > chis c1-c3 Expected counts are printed below observed counts Current Former Never Total (i) ****** 2 (ii) *** (iii) ***** ****** Total (iv) 3423 ChiSq = ***** (v) = df = (vi), p = ***** Supply the missing entries in the Minitab output, labeled (i), (ii),..., (vi). (Note that other entries, which you are NOT requested to supply, have been obliterated as well.) State clearly the null and alternative hypotheses being tested by this analysis. Summarize briefly in words, with reference to the standard significance levels (i.e. 0.10, 0.05, and 0.01) what you conclude from this analysis. 9. A stress analysis was conducted on epoxy-repaired truss joints in damaged timber 6 marks structures constructed from two species of wood, namely Southern Pine, and Ponderosa Pine. Repaired joints from forty (40) structures made of Southern Pine, and from another forty structures made of Ponderosa Pine, were selected for the study. The mean shear stress of each joint, in psi, was determined. The data were analyzed in Minitab in two ways, as shown on page 7. One of the analyses is correct and the other is incorrect....(continued over page) 6

7 MTB > # Analysis number 1: MTB > let Diff = s.pine - p.pine MTB > TInterval 95.0 Diff. Variable N Mean StDev SE Mean 95.0 % C.I. Diff ( , 159.2) MTB > TTest 0.0 Diff ; SUBC> Alternative 0. Test of mu = 0.0 vs mu not = 0.0 Variable N Mean StDev SE Mean T P-Value Diff MTB > # Analysis number 2: MTB > TwoSample 95.0 s.pine p.pine ; SUBC> Alternative 0. Twosample T for s.pine vs p.pine N Mean StDev SE Mean s.pine p.pine % C.I. for mu s.pine - mu p.pine: ( -170, 147) T-Test mu s.pine = mu p.pine (vs not =): T= P=0.88 DF= 62 State which of the two analyses is correct, giving a brief reason for your choice. Based upon the correct analysis, write down a 95% confidence interval for the mean difference, in strength of repaired joints, between the two species of wood. Does it appear that there is in fact a real difference between the two species of wood as regards joint strength? Explain, briefly. Suppose that your supervisor tells you that the Southern pine joints are on average 150 psi stronger than those of Ponderosa pine. Have you any reason to disbelieve her? Explain, briefly. 7

8 10. A study was conducted to examine the inhibiting properties of the sodium salts of 8 marks phosphoric acid on the corrosion of iron. A sample of fifty-five measurements of x = concentration of NaPO 4,andy = a measure of corrosion rate was collected. The data were analyzed in Minitab, with the following results: MTB > Regress corros 1 NaPO4 ; SUBC> Predict 22; SUBC> Predict 30; SUBC> Predict 35. The regression equation is corros = NaPO4 Predictor Coef Stdev t-ratio p Constant NaPO s = R-sq = 75.9% R-sq(adj) = 75.4% Analysis of Variance SOURCE DF SS MS F p Regression Error Total Fit Stdev.Fit 95.0% C.I. 95.0% P.I ( 2.859, 3.604) ( 0.461, 6.003) ***** ( *****, *****) ( *****, *****) ( 1.024, 1.899) ( , 4.242) Estimate the mean change in corrosion rate when the concentration of NaPO 4 increases by 1 unit. Calculate the point estimate of the mean corrosion rate when the concentration of NaPO 4 is 30 ppm. Now predict an observation of the corrosion rate when the concentration of NaPO 4 is 30 ppm. (d) (i) If you measured a very large number of samples of iron, exposed to water with a concentration of 22 ppm NaPO 4, would you be surprized if the mean of these observations was 1 unit? Why or why not? (ii) If you measured a single sample of iron, exposed to water with a concentration of 35 ppm NaPO 4, would you be surprized if your observation was 2 units? Why or why not? 8