n 10 x N 30010 2 010 095 one sided From Appendix VIII K 2355 UTL x KS 300

N 10 x n 30010 2 010 095 one sided from appendix viii

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n = 10; x ~ N (300,10 2 ); 0.10; 0.95 ; one- sided From Appendix VIII: K = 2.355 UTL x KS 300 2.355(10) 323.55 8.46. Sample of 50 measurements on a normally distributed quality characteristic has a mean of 35 and a standard deviation of 2. Using a confidence probability of 0.95, find a value such that 90% of the future measurements on this quality characteristic will lie above it. n = 50; x ~ N (35,2 2 ); 0.10; 0.95 ; one- sided From Appendix VIII: K = 1.646 x KS 35 1.646(2) 31.708 8.47. A sample of 20 measurements on a normally distributed quality characteristic had x 350 and s 10 . Find an upper natural tolerance limit that has probability 0.90 of containing 95% of the distribution of this quality characteristic. n = 20; x ~ N (350,10 2 ); 0.05; 0.90 ; one- sided From Appendix VIII: K = 2.208 UTL x KS 350 2.208(10) 372.08 8.48. How large a sample is required to obtain a natural tolerance interval that has probability 0.90 of containing 95% of the distribution? After the data are collected, how would you construct the interval? 0.05; 0.90 2 2 7.779 1 ,4 0.10,4 1 2 2 1 2 0.05 7.779 n 1  ,4 77 2 4 2 0.05 4 After the data are collected, a natural tolerance interval would be the smallest to largest observations.
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8.49. A random sample of n = 40 pipe sections resulted in a mean wall thickness of 0.1264 in. and a standard deviation of 0.0003 in. We assume that wall thickness is normally distributed. (a) Between what limits can we say with 95% confidence that 95% of the wall thicknesses should fall? = 0.05; = 0.95; and two- sided From Appendix VII: K = 2.445 TI on x : x KS 0.1264 2.445(0.0003) [0.1257,0.1271] (b) Construct a 95% confidence interval on the true mean thickness. Explain the difference between this interval and the one constructed in part (a). 0.05; t /2, n 1 t 0.025,39 2.023 CI on x : x t /2, n 1 S n 0.1264 2.023 0.0003 40 [0.1263,0.1265] Part (a) is a tolerance interval on individual thickness observations; part (b) is a confidence interval on mean thickness. In part (a), the interval relates to individual observations (random variables), while in part (b) the interval refers to a parameter of a distribution (an unknown constant). 8.50. Find the sample size required to construct an upper nonparametric tolerance limit that contains at least 95% of the population with probability at least 0.80. How would this limit actually be computed from sample data? 0.05; 0.80 log(1 ) log(1 0.80) n 31.4 32 log(1 ) log(1 0.05) The largest observation would be the nonparametric upper tolerance limit.
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