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Department of Mathematics & Statistics
STAT 2593 Final Examination
13 December, 1999
TIME: 3 hours. Total marks: 100.
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- Metal machine parts are often subject to micro-cracks which are difficult to detect 15 marks but make the part defective. Suppose that in a certain production process parts are automatically scanned for micro-cracks using an X-ray device and digital image processing. The probability that the system will correctly determine that a part has micro-cracks is 0.98. The probability that the system will incorrectly indicate that a part has micro-cracks when it is in fact defect free, is 0.03. The production system is of high quality so that the fraction of parts which actually have micro-cracks is 0.001. Let C be the event that a part has micro-cracks and D be the event that the automatic system indicates that a part has micro-cracks.
(a) What is the value of P (C ∩ D)? That is, what is the probability that a part has micro-cracks and the automatic detection system indicates that it has micro- cracks? (b) What is the value of P (C′^ ∩ D)? That is, what is the probability that a part has no micro-cracks and the automatic detection system (incorrectly) indicates that it has micro-cracks? (c) What is the value of P (D)? That is, what is the probability that the automatic detection system indicates that a part has micro-cracks? (d) Are the events C and D independent? Are they mutually exclusive? Explain briefly. (e) Suppose that the automatic detection system indicates that a part has micro- cracks. What is the probability that it does indeed have micro-cracks?
- Suppose that 30% of the balsam fir trees in a particular region are infested with spruce 6 marks budworm.
(a) Suppose that a random sample of 8 trees is chosen. Evaluate the probability that at least 2 of these trees are infested with spruce budworm. (b) Suppose that a random sample of 80 trees is chosen. Evaluate the probability that at least 20 of these trees are infested with spruce budworm.
- The MucKaines frozen food company processes vegetables into plastic pouches which 14 marks are designed to be taken directly from the freezer and placed into boiling water for cooking. At regular intervals, random samples of 25 pouches are taken and the pouches tested for integrity; the result of the test is either “pass” or “fail”. The following Minitab output displays the results for 30 such samples taken sequentially, as well as the total count of failing pouches.
MTB > print c
f.count 2 2 0 3 1 0 2 0 4 0 0 2 2 1 2 0 2 0 0 2 1 2 1 2 1 1 2 1 0 1
MTB > sum c
Sum of f.count = 37.
(a) Assuming the process to have been in control while these 30 samples were taken, use these results to construct an appropriate (3 SD) control chart. That is, determine the centre line and the upper and lower 3 SD control limits; you need not actually draw a chart. (b) Suppose that a sample of 25 pouches was taken at one of the regular sampling times. How many “fails” would there have to be in the sample to indicate that the process was out of control? (c) The Purple Dwarf company has a similar procedure with their frozen food pouches; they take samples of 20 pouches at regular intervals. Their control chart has a centre line at 0.07 with the UCL equal to 0.24, and LCL equal to 0. Suppose that the Purple Dwarf production process goes out of control in that the fraction of faulty pouches increases to 0.20. What is the probability that the next sample taken will indicate that the process is out of control? (Hint: You can save yourself a lot of tedious arithmetic by using the Binomial tables, Table A.1, pages 700 — 702 Devore, 4th ed., pages 670 — 671 Devore, 3rd ed.) (d) Yet another company (“Eye Gee Eh?”) has a similar procedure. At time t 0 their process shifts to a new state in such a manner that the probability that a sample will signal that a shift has occurred is 0.4. (i) What is the probability that process shift will remain undetected for the first 4 samples taken after time t 0? (ii) How many samples do you expect to be taken after time t 0 before the problem is detected?
- A study was conducted to determine if a certain treatment had any effect on the 5 marks amount of metal removed in a pickling operation. A random sample of pieces of metal was selected; each piece was cut into two parts. One part from each piece was treated; the other part was left untreated. All parts were then immersed in a pickling bath for 24 hours. The thickness of metal removed by the pickling bath was recorded for each part and the data were entered into a Minitab worksheet. Some descriptive statistics provided by Minitab are displayed below.
MTB > let c3 = c1-c MTB > name c3 ’diff’ MTB > desc c1-c
Variable N Mean Median TrMean StDev SE Mean untreat 38 14.945 14.850 14.721 5.679 0. treat 38 11.826 11.950 11.624 5.826 0. diff 38 3.118 3.200 3.150 1.008 0.
Variable Minimum Maximum Q1 Q untreat 4.100 29.400 11.300 18. treat 0.800 26.700 7.625 15. diff 0.100 5.200 2.525 3.
(a) Using appropriate information from the Minitab output, determine a 90% con- fidence interval for the the effect of the treatment upon the thickness of metal removed. (b) Does the treatment actually appear to have an effect? Explain briefly.
- In a study on graphic stress telethermometry (GST), a technique for detecting breast 9 marks cancer, a sample of 138 women known to have the disease was tested using GST. The number correctly diagnosed by GST to have breast cancer was 97.
(a) Construct a 95% confidence interval for the proportion of correct diagnoses by GST. (b) An oncologist claims that GST correctly diagnoses women with breast cancer to have the disease 75% of the time. Do the results of the study give you reason to doubt her claim? Explain briefly. (c) Suppose that a researcher says that in another study it was estimated that GST correctly diagnoses the presence of cancer in 74% of cases, and that this estimate is accurate “to within ±4% 19 times out of 20”. How large a sample was used in this study?
- The breaking strength (in thousands of kilograms) of samples of steel cable is a 9 marks random variable Y which has cumulative distribution function
F (y) =
1 − e−^0.^05 y 2 y ≥ 0 0 y < 0
(a) Suppose that a piece of cable meets requirements if its breaking strength is at least 1200 kilograms. What is the probability that a sample of the cable meets requirements? (b) What is the median value of the breaking strength? That is, what is the number m so that exactly half of all the samples of cable in the population have breaking strength less than m? (c) Suppose that a piece of cable is tested at 2 thousand kilograms, and does not break. What is then the probability that it will not break when tested at 3 thousand kilograms?
- A manufacturer of cloth is interested in the underlying distribution of the number X 6 marks of defects in the bolts of cloth that he produces.
(a) A proposed model for the underlying frequencies of the categories: {X = 0}, {X = 1}, {X = 2}, {X = 3}, {X ≥ 4 } is:
H 0 : p 0 = 0. 30 , p 1 = 0. 25 , p 2 = 0. 20 , p 3 = 0. 10 , p 4 = 0. 15.
If a chi-squared goodness of fit test is planned based on a random sample of size n = 175, what would be the expected number of sample values, under H 0 , that fall in the category {X = 3}? (b) Repeat part (a) if the null hypothesis stipulates that X follows a Poisson distri- bution with mean number of defects equal to 2.
- In a study conducted in northern Maine on the performance of a Timberjack 550 18 marks Cable Skidder data were collected on a number of variables. These included machine production (in solid cubic metres of wood per machine hour) and on the distance from the area being logged to the road (i.e. the distance that the logs had to be skid- ded). Since one would expect machine production to decrease as distance increases, a regression model was fitted to predict machine production from the reciprocal of distance, which we shall refer to as “proximity”. Below are shown a graph of machine production versus proximity and the results of fitting the regression model in Minitab. ...... (continued over page)
Analysis of Variance
Source DF SS MS F P Regression 1 611.50 611.50 21.96 0. Residual Error 37 1030.29 27. Total 38 1641.
Predicted Values
Fit StDev Fit 90.0% CI 90.0% PI 26.747 0.886 ( 25.252, 28.241) ( 17.720, 35.774) 28.770 1.096 ( 26.921, 30.618) ( 19.677, 37.862) ****** 1.410 ( ******, ******) ( ******, ******)
(a) Test the hypothesis that the slope of the regression line is 0.75 (against a two- sided alternative). Give your conclusions at the “standard” significance levels, i.e. 0.10, 0.05, 0.01. (b) Determine the fitted values of machine production when the proximity is (i) 6, and (ii) 15. Plot the two corresponding points on the Minitab graph, and hence draw the fitted regression line. (c) Find a 90% prediction interval for machine production when the proximity is 12. (d) A professor of Forestry says that when the proximity is 8, the mean machine production is 26. Do the data give you any reason to doubt his assertion? Explain briefly. (e) A Forestry student says that when she was working a summer job, a machine production of 39 was achieved at a site whose proximity was 10. Should you believe her? Explain briefly. (f) One point in particular on the graph of machine production versus proximity stands out as being an outlier. Circle this point. If this point were removed from the data set, and the regression line were re-estimated, which of the following would occur: (i) both the slope and intercept would increase, (ii) both the slope and intercept would decrease, (iii) the slope would increase and the intercept would decrease, (iv) the slope would decrease and the intercept would increase? Explain briefly.
- A Social Engineer (!) conducted a survey to determine whether the incidence of certain 9 marks types of crime varied amongst a number of large American cities (1 = Boston, 2 = New York, 3 = Chicago, 4 = Detroit). The particular crimes of interest were assault, burglary, larceny, and homicide. The results were tabulated and analyzed using Minitab with the following results (parts of which have been obliterated):
MTB > chis c1-c
Expected counts are printed below observed counts
assault burglary larceny homicide Total 1 16 12 45 (i) 91 19.47 13.30 44.34 13.
2 31 20 100 25 176 37.66 25.73 85.76 xxxxx
3 26 19 46 10 101 21.61 14.76 (ii) 15.
4 28 18 39 19 104 22.25 15.20 50.68 15.
Total (iii) 69 230 xxxxx 472
Chi-Sq = 0.619 + 0.128 + 0.010 + xxxxx + 1.178 + 1.276 + 2.364 + 0.127 + 0.891 + 1.215 + (iv) + 1.897 + 1.483 + 0.514 + 2.691 + 0.620 = 16. DF = (v), P-Value = xxxxx
(a) State the appropriate null and alternative hypotheses clearly. (b) Supply the missing entries in the Minitab output, labeled (i), (ii), ..., (v). (Note that other entries, which you are NOT requested to supply, have been obliterated as well.) (c) On the basis of the Minitab output (including the bits supplied by you) state your decision regarding the hypothesis being tested, with respect to each of the “standard” significance levels (i.e. 0.10, 0.05 and 0.01). For each significance level, indicate briefly “in words” what your decision means.
