In questions 1-8, find the appropriate confidence interval given the following information: 1) n = 54, x = 13,0 = 2.5, a = 0.05 2) n = 12, x = 21.3,s = 6.3, CL = 99%, X~N 3) n = 84, x = 53,0 = 4.3, CL = 98%, Distribution Unknown 4) n = 15, x = 25,a = 9, CL = 90%, Distribution Unknown 5) n = 35, x = 85.4,s = 11.5, CL = 90%, Distribution Unknown 6) X1 = 69.8, n, = 30, $1 = 1.92, X2 = 63.8, n2 = 20, $2 = 2.18, a = 0.05, X~N, but variances are not equal. 7) X1 = 51.7, n, = 40, $1 = 2.1, x2 = 54.8, n2 = 35, $2 = 1.8, a = 0.05, X~N, variances are assumed to be equal 8) X1 = 122.05, m, = 55, 61 = 8.2, x2 = 111.65, n2 = 45, 62 = 7.8, a = 0.05, X~N 9) A study was conducted to estimate hospital costs for accident victims who wore seat belts. Forty randomly selected cases have an average of $9004 and a standard deviation of $5629. What can we infer, with 90% confidence, about the population mean? 10) When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A sample of 40 smokers has a mean cotinine level of 172.5. Assuming that o = 119.5, estimate the population mean for an a of 0.05? 11) Suppose that the average IQ score of 1200 randomly selected students that have successfully completed Stat 24000 is 105. Find a 90% confidence interval for the mean IQ score of Stat 24000 students. Based on an a of 0.1, can we conclude that Stat 24000 students are, on average, significantly smarter than the general population? Why or why not