One-Sample Z: C1, C2, C3, C4, C5, C6, C7, C8,... The assumed standard deviation = 110

Transcription

1 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 MTB > OneZ c1-c100; SUBC> Sigma 110; SUBC> Confidence 90. cuments\ca3compspring10.mpj' MTB > Random 100 c1-c100; SUBC> Normal 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 C (1073.2, ) C (1055.5, ) C (1084.4, ) C (1087.8, ) C (1075.5, ) C (1089.9, ) C (1077.8, ) C (1073.4, ) C (1079.4, ) C (1083.6, ) C (1062.8, ) C (1071.3, ) C (1064.9, ) C (1088.8, ) C (1069.4, ) C (1087.4, ) C (1092.5, ) C (1089.7, ) C (1106.0, ) C (1081.9, ) C (1073.4, ) C (1092.4, ) C (1070.4, ) C (1091.2, ) C (1075.4, ) C (1090.7, ) C (1090.7, ) C (1089.5, ) C (1082.0, ) C (1096.2, ) C (1094.8, ) C (1097.8, ) C (1075.2, )

2 C (1062.6, ) C (1101.8, ) C (1086.6, ) C (1085.8, ) C (1078.8, ) C (1079.8, ) C (1078.9, ) C (1083.5, ) C (1087.3, ) C (1092.4, ) C (1081.1, ) C (1093.5, ) C (1071.0, ) C (1085.8, ) C (1094.6, ) C (1099.7, ) C (1095.0, ) C (1071.0, ) C (1082.2, ) C (1096.0, ) C (1058.4, ) C (1095.5, ) C (1078.3, ) C (1080.9, ) C (1082.8, ) C (1096.2, ) C (1080.7, ) C (1095.4, ) C (1076.2, ) C (1083.6, ) C (1069.3, ) C (1088.1, ) C (1076.6, ) C (1086.4, ) C (1086.6, ) C (1082.8, ) C (1065.4, ) C (1097.1, ) C (1089.7, ) C (1080.3, ) C (1081.7, ) C (1078.7, ) C (1069.9, ) C (1069.8, ) C (1069.2, ) C (1090.3, ) C (1087.9, ) C (1071.5, ) C (1065.3, ) C (1075.3, ) C (1095.8, ) C (1072.6, ) C (1075.4, ) C (1078.9, ) C (1084.0, ) C (1091.0, ) C (1082.9, ) C (1089.8, ) C (1087.3, ) C (1069.5, ) C (1072.6, ) C (1076.2, ) C (1075.8, )

3 C (1083.4, ) C (1063.5, ) C (1078.5, ) C (1089.2, ) 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 = 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 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 Ninety five intervals cover Forty four intervals cover 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 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.

4 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 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 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.

5 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> (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.