Fall 201T BE 2100 Honor stulent (Y/N7 You have 130 minutes to finish this exam H
ID: 3321751 • Letter: F
Question
Fall 201T BE 2100 Honor stulent (Y/N7 You have 130 minutes to finish this exam Honor student work for fiall markn s will have an lditional 10 minaten 1. A chenical plant is required to maistain ambsient salfur levels in the at an average level of no more than 12.50. The resalts of 3s randkosnly level produced a wurement" or the sealf sauple mean of 14.82 and a sample standard deviation of 3.91. saample mean of 14.82 and a sample standard devistion (5 point) what is the lower condence bound for the trun-w level of 95% average level (pa) with the confece test when -S96. Your answer man a. b. (5 point) What would be your conclusion for the following include the rejection region and the test statistic in order to receive full credit. Hai -12.5 (5 point) Find the sample size necessary to estimate a population mean to within 0.5 with 959 if the population standard deviation is 6.2.Explanation / Answer
1.
a)
b)
Test statistics, t = (14.82 - 12.5)/(3.91/sqrt(35)) = 3.5103
For alpha = 0.05, with a right tailed test
rejection region would be on the right of the normal curve. critical value of the test statistics = 1.6909
hence rejection region lies to the right of the critical value.
As value of test statistics lies in the rejection region, we reject null hypothesis.
This means there are significant evidence to conclude that sulphur levels are greater than 12.5
2.
CI for 95% n 35 mean 14.82 z-value of 95% CI 1.9600 std. dev. 3.91 SE = std.dev./sqrt(n) 0.66091 ME = z*SE 1.29536 Lower Limit = Mean - ME 13.52464 Upper Limit = Mean + ME 16.11536 95% CI (13.5246 , 16.1154 )Related Questions
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