SMAM 314 Computer Assignment 3 1.Suppose n = 100 lightbulbs are selected at random from a large population.. Assume that the light bulbs put on test until they fail. Assume that for the population of light bulbs the mean time to failure is 1100 hours with a standard deviation 110 hours, Use Minitab to do this part of the problem Simulate the results of this selection 100 times and in each case find a 90% confidence interval for the true mean time to failure. The following commands may be used: MTB > Random 100 c1-c100; SUBC> Normal 1100 110. MTB > OneZ c1-c100; SUBC> Sigma 110; SUBC> Confidence 90. cuments\ca3compspring10.mpj' MTB > Random 100 c1-c100; SUBC> Normal 1100 110. MTB > OneZ c1-c100; SUBC> Sigma 110; SUBC> Confidence 90. One-Sample Z: C1, C2, C3, C4, C5, C6, C7, C8,... The assumed standard deviation = 110 Variable N Mean StDev SE Mean 90% CI C1 100 1091.3 117.1 11.0 (1073.2, 1109.4) C2 100 1073.5 118.5 11.0 (1055.5, 1091.6) C3 100 1102.5 107.8 11.0 (1084.4, 1120.6) C4 100 1105.9 101.6 11.0 (1087.8, 1124.0) C5 100 1093.5 106.2 11.0 (1075.5, 1111.6) C6 100 1108.0 120.4 11.0 (1089.9, 1126.1) C7 100 1095.9 106.2 11.0 (1077.8, 1114.0) C8 100 1091.5 115.8 11.0 (1073.4, 1109.6) C9 100 1097.5 116.1 11.0 (1079.4, 1115.6) C10 100 1101.7 118.8 11.0 (1083.6, 1119.8) C11 100 1080.9 122.6 11.0 (1062.8, 1098.9) C12 100 1089.4 102.2 11.0 (1071.3, 1107.5) C13 100 1083.0 89.9 11.0 (1064.9, 1101.1) C14 100 1106.8 96.7 11.0 (1088.8, 1124.9) C15 100 1087.5 114.7 11.0 (1069.4, 1105.6) C16 100 1105.5 97.9 11.0 (1087.4, 1123.6) C17 100 1110.6 99.5 11.0 (1092.5, 1128.7) C18 100 1107.8 124.6 11.0 (1089.7, 1125.9) C19 100 1124.1 107.5 11.0 (1106.0, 1142.2) C20 100 1099.9 107.9 11.0 (1081.9, 1118.0) C21 100 1091.5 111.2 11.0 (1073.4, 1109.6) C22 100 1110.5 124.7 11.0 (1092.4, 1128.6) C23 100 1088.5 99.5 11.0 (1070.4, 1106.6) C24 100 1109.3 102.8 11.0 (1091.2, 1127.4) C25 100 1093.5 102.6 11.0 (1075.4, 1111.6) C26 100 1108.8 115.1 11.0 (1090.7, 1126.9) C27 100 1108.8 115.8 11.0 (1090.7, 1126.9) C28 100 1107.6 94.7 11.0 (1089.5, 1125.7) C29 100 1100.1 120.4 11.0 (1082.0, 1118.2) C30 100 1114.3 117.3 11.0 (1096.2, 1132.4) C31 100 1112.9 103.2 11.0 (1094.8, 1131.0) C32 100 1115.9 114.7 11.0 (1097.8, 1134.0) C33 100 1093.3 108.1 11.0 (1075.2, 1111.4)
C34 100 1080.7 109.4 11.0 (1062.6, 1098.8) C35 100 1119.9 100.3 11.0 (1101.8, 1138.0) C36 100 1104.7 101.3 11.0 (1086.6, 1122.8) C37 100 1103.9 105.7 11.0 (1085.8, 1122.0) C38 100 1096.9 118.7 11.0 (1078.8, 1115.0) C39 100 1097.9 127.0 11.0 (1079.8, 1116.0) C40 100 1097.0 90.3 11.0 (1078.9, 1115.1) C41 100 1101.6 105.2 11.0 (1083.5, 1119.7) C42 100 1105.4 101.9 11.0 (1087.3, 1123.5) C43 100 1110.4 111.8 11.0 (1092.4, 1128.5) C44 100 1099.2 124.4 11.0 (1081.1, 1117.3) C45 100 1111.6 111.2 11.0 (1093.5, 1129.7) C46 100 1089.1 100.9 11.0 (1071.0, 1107.2) C47 100 1103.9 101.1 11.0 (1085.8, 1122.0) C48 100 1112.7 98.6 11.0 (1094.6, 1130.8) C49 100 1117.8 118.0 11.0 (1099.7, 1135.9) C50 100 1113.1 119.8 11.0 (1095.0, 1131.2) C51 100 1089.1 110.2 11.0 (1071.0, 1107.2) C52 100 1100.3 111.3 11.0 (1082.2, 1118.4) C53 100 1114.1 112.3 11.0 (1096.0, 1132.2) C54 100 1076.5 115.7 11.0 (1058.4, 1094.5) C55 100 1113.6 106.8 11.0 (1095.5, 1131.7) C56 100 1096.3 115.9 11.0 (1078.3, 1114.4) C57 100 1099.0 121.1 11.0 (1080.9, 1117.1) C58 100 1100.9 120.8 11.0 (1082.8, 1119.0) C59 100 1114.3 120.0 11.0 (1096.2, 1132.4) C60 100 1098.8 109.2 11.0 (1080.7, 1116.9) C61 100 1113.5 117.1 11.0 (1095.4, 1131.6) C62 100 1094.3 103.9 11.0 (1076.2, 1112.4) C63 100 1101.7 106.7 11.0 (1083.6, 1119.7) C64 100 1087.4 115.8 11.0 (1069.3, 1105.5) C65 100 1106.2 115.2 11.0 (1088.1, 1124.3) C66 100 1094.7 111.4 11.0 (1076.6, 1112.8) C67 100 1104.5 109.3 11.0 (1086.4, 1122.6) C68 100 1104.7 115.6 11.0 (1086.6, 1122.8) C69 100 1100.9 111.1 11.0 (1082.8, 1119.0) C70 100 1083.5 109.4 11.0 (1065.4, 1101.6) C71 100 1115.2 112.6 11.0 (1097.1, 1133.3) C72 100 1107.8 114.4 11.0 (1089.7, 1125.9) C73 100 1098.4 122.0 11.0 (1080.3, 1116.5) C74 100 1099.8 104.1 11.0 (1081.7, 1117.9) C75 100 1096.8 115.9 11.0 (1078.7, 1114.8) C76 100 1088.0 104.4 11.0 (1069.9, 1106.1) C77 100 1087.9 100.5 11.0 (1069.8, 1106.0) C78 100 1087.3 123.5 11.0 (1069.2, 1105.4) C79 100 1108.4 100.8 11.0 (1090.3, 1126.5) C80 100 1106.0 119.6 11.0 (1087.9, 1124.1) C81 100 1089.6 105.3 11.0 (1071.5, 1107.7) C82 100 1083.4 111.9 11.0 (1065.3, 1101.5) C83 100 1093.4 109.9 11.0 (1075.3, 1111.5) C84 100 1113.9 116.6 11.0 (1095.8, 1132.0) C85 100 1090.7 114.9 11.0 (1072.6, 1108.8) C86 100 1093.5 113.0 11.0 (1075.4, 1111.6) C87 100 1096.9 123.2 11.0 (1078.9, 1115.0) C88 100 1102.1 95.8 11.0 (1084.0, 1120.2) C89 100 1109.1 92.1 11.0 (1091.0, 1127.2) C90 100 1100.9 108.0 11.0 (1082.9, 1119.0) C91 100 1107.9 110.9 11.0 (1089.8, 1126.0) C92 100 1105.3 112.3 11.0 (1087.3, 1123.4) C93 100 1087.6 108.2 11.0 (1069.5, 1105.7) C94 100 1090.7 126.0 11.0 (1072.6, 1108.8) C95 100 1094.3 104.4 11.0 (1076.2, 1112.4) C96 100 1093.9 117.1 11.0 (1075.8, 1112.0)
C97 100 1101.5 112.7 11.0 (1083.4, 1119.5) C98 100 1081.6 106.8 11.0 (1063.5, 1099.7) C99 100 1096.6 106.9 11.0 (1078.5, 1114.7) C100 100 1107.3 101.4 11.0 (1089.2, 1125.3) Answer the following questions in complete sentences below. Staple your computer output to the assignment. A. How many of the intervals contain the true mean time to failure? Ninety five intervals contain the true mean to failure. For this class n = 26,x = 91,s = 2.742 B. Would you expect all 100 of the intervals to contain the true mean time to failure? Explain. I would expect about 90 of the intervals to contain the true mean time to failure because the probability the intervals contain the true mean is 0.90. C. Do all of the intervals have the same width? Why or why not? All of the intervals have the same width. The width is 2z.05 σ / sample to sample. n. All the quantities in this formula are fixed and do not change from D. Suppose you took 80% intervals instead of 90%. Would they be narrower or wider? What about 95% intervals? The 80% intervals would be narrower. The 95% intervals would be wider. A wider interval has a higher probability of containing the true mean. E. How many intervals contain 1080? 1100? 1120? Thirty eight intervals cover 1080. Ninety five intervals cover 1100. Forty four intervals cover 1120. F. Suppose you took samples of size 225 instead of size 100. Would you expect more or fewer of the intervals to cover 1080? 1100? 1120? Would these intervals be narrower or wider? (Answer the question by thinking about it not by doing another computer simulation) You would expect fewer intervals to contain 1080 and 1120. You would expect about the same number of intervals to contain 1100 because it is the true mean. The confidence intervals would be narrower because n is in the denominator. As n increases the width of the confidence interval decreases.
G. Suppose you calculated 90% confidence intervals for 100 sets of real data. About how many of these intervals would you expect to contain the true mean time to failure? Could you tell which intervals were successful and which were not? Why or why not? You would expect about 90 of the intervals to contain the true mean. You could not tell which intervals are successful because you do not know the true mean. 2. The times of first sprinkler activation for a series of tests with fire prevention sprinkler systems using an aqueous film forming foam were in sec 27 41 22 27 23 35 30 33 24 27 28 22 24 The system has been designed so that the true average activation time is at most 25 sec under such conditions. Use Minitab to do the following (1) Make a normal probability plot. (2) Perform a test of hypothesis to determine whether the average activation time is significantly greater than 25 sec. One-Sample T: Times Test of mu = 25 vs > 25 95% Lower Variable N Mean StDev SE Mean Bound T P Times 13 27.92 5.62 1.56 25.15 1.88 0.043 MT Answer the following questions in the space provided below. Staple your computer output to the back of the assignment. A. Based on the normal probability plot is it reasonable to assume that the data is normally distributed? Explain.
The points on the graph lie reasonably close to a straight line so one may assume the data is normally distributed. B. For the test of hypothesis (1) What is the null and the alternative hypothesis? H 0 µ = 25 H 1 µ > 25 (2) What other important assumption is needed besides normality? Unknown standard deviation and a small sample. (3)What is the region of rejection at α=.05? T>1.782 (4)What is the numerical value of the test statistic and what is the p value? T = 1.88 pvalue =.043 (5) Would you reject H 0 at α=.05? Explain. Yes because T=1.88>1.7982 (6) At α=.05 would you conclude that there is sufficient evidence that the average activation time is greater than 25 sec?explain. Yes because I rejected H 0 at α=.05 (7) Is the lower confidence bound consistent with your conclusion in (5)? Explain. LCB =25.51 Concidence interval (25.51, ) does not contain 25. (8)At α=.01 would you conclude that there is sufficient evidence you conclude that there is sufficient evidence that the average activation time is greater than 25 sec?explain. I would not conclude there is sufficient evidence that the activation time is greater than 25 because the p value is.043 greater than.01.