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Unformatted text preview: Stat 231 Final Exam December 16, 1997 Professor Vardeman 1. The first page of printout attached to this exam summarizes some data (collected by a student group) on the diameters of holes bored in certain steel housing covers. A member of the group measured the diameters of < oe * parts 7 oe 3 times each. Data values were recorded in units of "!$ mm above the nominal (or ideal) diameter. Consider first only the measurements made on housing number 1. (a) Give a 90% twosided confidence interval for the long run mean measured diameter for housing 1. (No need to simplify.) (b) Give a 90% twosided prediction interval for the next measured diameter for housing number 1. (No need to simplify.) (c) Give a 90% upper confidence bound for the long run standard deviation of measured diameters for housing number 1. Now consider the measurements made on both housings 1 and 2. (d) Test the hypothesis that there is no difference between the long run mean measured diameters for housings 1 and 2 using ! oe !". (Show all 7 steps.) Finally, consider the measurements made on all 9 housings. 1 (e) What model assumptions must be made in order to do inference for comparing these 9 mean long run measured diameters? (f) Under the assumptions of part (e), what :value is associated with a test of H! ." oe .# oe oe .* ? With ! oe !& should this hypothesis be rejected? Explain. :value oe __________ Explanation: Reject? yes no (circle one) (g) Based on the model assumptions of part (e) give a 90% twosided confidence interval for the difference in long run mean measured diameters for housings 1 and 2. (No need to simplify.) (h) Suppose that the parts measured by these students were in fact a random sample from a large lot of such housings and there is interest in estimating how much the holes bored in these housings actually vary in diameter (where this is to be assessed without the corruption of measurement error). Give an ANOVAbased estimate of the long run standard deviation among housing mean measurements. 2. Stat 231 concerns both PROBABILITY and STATISTICS. Define and contrast these two subjects. 3. Pages 2 through 4 of the printout attached to this exam concern a study of the effectiveness of armor plating. Armorpiercing bullets were fired at an angle of 40 against armor plate of 2 thickness B" (in !!" inches) and Brinell hardness number B# , and the resulting "ballistic limit," C (in feet/second), was measured. (a) There are two simple linear regressions included on the printout. Which of these seems to be the most effective summarization of the data? Explain. Find the value of G: for this model. Best SLR Model: ______________ Explanation: G: oe _____________________ (b) Using the second of the two simple linear regressions, give a 90% twosided confidence interval for the increase in mean ballistic limit that accompanies a 1 unit increase in Brinell hardness number. (No need to simplify.) Henceforth use the MLR analysis beginning in the middle of page 3 of the printout. (c) Give and interpret a :value for testing whether B" and B# together provide some important ability to explain or predict C. :value oe ___________ Interpretation: (d) Give the value of and degrees of freedom for an F statistic used to test the hypothesis that after accounting for B# , B" provides no important additional ability to predict C. .0 oe ______ , ______ 0 oe __________ (e) Tomorrow I plan to test a piece of armor plating #&$ inch thick and with Brinell hardness number %!(. Give me an interval that is in some sense "90% sure" to contain tomorrow's ballistic limit. 4. In the manufacture of a certain glass product, defects of three different types (A, B and C) can occur on the product. These differ in both importance and frequency of occurrence. Suppose 3 that on one item, \A , \B and \C are the numbers of occurrences of these defects and that these can be modeled as independent Poisson random variables. (a) Suppose that A oe E\A oe # Find the probability that \A 2. (b) I decide to assign "demerits" to an inspected item according to the expression H oe \A +#\B %\C . Suppose that A oe E\A oe #, B oe E\B oe 1 and that C oe E\C oe 05. Find the mean and variance of H. (c) For a different product (having a different product grading scheme) items have a mean number of demerits of & with a corresponding standard deviation of demerits of "!. Approximate the probability that a sample of "!! of these items has a total of at least 48 demerits. (Hint: >9>+6 ./7/<3>= %) is the same as B %).) 5. Endoftheline inspection is done on 100 items manufactured on a particular production line with the results below (in terms of actual Good/Defective character and inspection result). Inspection Result Pass Fail Actual Part Good 72 8 Condition Defective 4 16 Consider first the random selection of a single item from the lot. (a) What it the probability that the item is actually "Good" or has a "Pass" inspection result? (b) What is the conditional probability that it is "Good" given that it "Passes" final inspection? 4 Consider now the random selection (without replacement) of two items from the lot. Define random variables \ oe the number of "Good&Pass" items in the sample, and ] oe the number of "Defective&Fail" items in the sample. (c) Below is a partially complete table giving the joint probability function for \ and ] . Finish filling it in. Boe! Coe# Coe" Coe! !$)) !"$$ "(%& &"'% Boe" Boe# (d) ^ oe \ ] is the number of items in the sample that are correctly classified by the inspection process. Using your answer to (c) find T ^ oe #. 6. A commonly used model in reliability work is that with [ oe the life of a device of interest , ] oe ln[ has a normal distribution with some mean . and some standard deviation 5. Suppose that the guarantee period for a device is .5 year and that its actual lifetime follows a model like this where . oe ! and 5 oe (. Find the probability that the device needs to be replaced during its warranty period. 5 MTB > Oneway 'diameter' 'part'. OneWay Analysis of Variance Analysis Source part Error Total of Variance on diameter OF SS MS 8 211.33 26.42 18 70.67 3.93 26 282.00 F 6.73 p
0.000 Individual 95% CIs For Mean Based on Pooled StDev Level 1 2 3 4 5 6 7 8 9 Pooled stOev
N 3 3 3 3 3 3 3 3 3 Mean 4.333 4.667 2.000 3.667 0.000 2.667 2.667 2.333 10.667 1.981 stOev 4.509 1.155 2.000 2.082 1.000 1.155 1.155 1.155 0.577 ++++(*) (*) (*) (*) (*) (*) (*) (*) (*) = ++++ 1 MTB > print Data Display c1c3 Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 y 927 978 1028 906 1159 1055 1335 1392 1362 1374 1393 1401 1436 1327 950 998 1144 1080 1276 1062 xl 253 258 259 247 256 246 257 262 255 258 253 252 246 250 242 243 239 242 244 234 x2 317 321 341 350 352 363 365 375 373 391 407 426 432 469 257 302 331 355 385 426 MTB > Regress 'y' 1 'xl'; SUBC> Constant. Regression Analysis The regression equation y =  886 + 8.27 xl Predictor Constant xl s Coef 886 8.267 R~q of Variance OF 1 18 19 is Stdev 1348 5.395 11.5% tratio 0.66 1.53 Rsq(adj) p 0.519 0.143 = 180.2 = 6.6% Analysis SOURCE Regression Error Total SS 76220 584368 660589 MS 76220 32465 F 2.35 P
0.143 2 MTB > Regress 'y' 1 'x2'; SUBC> Constant. Regression Analysis The regression equation y = 211 + 2.64 x2 Predictor Constant x2 s Coef 211.0 2.6388 Rsq of variance OF 1 18 19 is Stdev 228.5 0.6174 50.4% tratio 0.92 4.27 Rsq(adj) p 0.368 0.000 = 135.0 = 47.6% Analysis SOURCE Regression Error Total Unusual Obs. 15 20 SS 332705 327883 660589 MS 332705 18216 F 18.26 p 0.000 Observatir':1s x2 Y 950.0 257 1062.0 426 Fit 889.2 1335.1 Stdev.Fit 74.3 47.4 Residual 60.8 273.1 St.Resid 0.54 X 2.16R R denotes X denotes an obs. with a large st. resid. an obs. whose X value gives it large influence. MTB > Name c4 = 'SRES1' c5 = 'FITS1' MTB > Regress 'y' 2 'xl' 'x2'; SUBC> SResiduals 'SRES1'; Fits 'FITS1'; SUBC> SUBC> Constant; SUBC> Pure; Predict 'xl' 'x2'. SUBC> Regression Analysis The regression equation is y =  1674 + 7.61 xl + 2.59 x2 Predictor Constant xl x2 s Coef 1674.1 7.612 2.5939 Rsq of Variance OF 2 17 19 SS 397230 263358 660589 MS 198615 15492 F 12.82 p 0.000 Stdev 947.4 3.730 0.5698 60.1% tratio 1. 77 2.04 4.55 Rsq(adj)
p 0.095 0.057 0.000 = 124.5 = 55.4% Analysis SOURCE Regression Error Total 3 SOURCE OF
1 1 xl
x2 SEQ SS 76220 321010 Fit 1074.1 1122.5 1182.0 1114.0 1187.7 1140.1 1229.0 1293.0 1234.6 1304.1 1307.5 1349.2 1319.1 1445.5 834.7 959.0 1003.8 1088.9 1181.9 1212.2 Cannot Stdev.Fit 41.9 49.6 47.0 31.1 37.4 31.3 38.7 53.4 34.0 43.2 37.7 44.2 48.9 64.5 73.5 52.1 52.5 40.6 37.0 74.4 ( ( ( ( ( ( ( ( (
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( 95.0% C.!. 985.7, 1162.4) 1017.9, 1227.1) 1281. 2) 1082.8, 1048.3, 1179.7) 1108.8, 1266.6) 1074.1, 1206.1) 1147.3, 1310.7) 1180.4, 1405.7) 1162.8, 1306.4) 1212.9, 1395.2) 1228.1, 1387.0) 1255.9, 1442.5) 1215.9, 1422.3) 1581. 6) 1309.5, 679.6, 989.8) 849.1, 1069.0) 893.1, 1114.5) 1003.1, 1174.7) 1103.9, 1260.0) 1055.2, 1369.2) ( ( ( ( ( ( ( ( (
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( ( ( 95.0% P.!. 1351. 2) 797.0, 839.8, 1405.2) 901. 2, 1462.8) 843.2, 1384.7) 1461. 9) 913.4, 1410.9) 869.3, 953.9, 1504.1) 1578.8) 1007.2, 962.3, 1506.9) 1582.1) 1026.0, 1581. 9) 1033.1, 1070.5, 1627.9) 1601. 3) 1036.9, 1741.3) 1149.7, 529.7, 1139.7) 674.3, 1243.8) 1288.9) 718.8, 812.6, 1365.2) 1455.9) 907.9, 906.2, 1518.2) do pure error test 4 ...
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This note was uploaded on 02/11/2012 for the course STAT 231 taught by Professor Staff during the Fall '08 term at Iowa State.
 Fall '08
 Staff

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