TABLE 12-14TABLE 10-3TABLE 10-4TABLE 10-911/14/03 252x0323 ECO252 QBA2 Name (Page layout view!) SECOND HOUR EXAM Hour of Class Registered October 30, 2003 Circle 11am 12:30pmI. (53 points) Do all the following? (2points each unless noted otherwise).1. A manufacturer revises a manufacturing process and finds a fall in the defect rate of%.4%5 a) The fall in defects is statistically significant because 5% is larger than 4%.b) The fall in defects is statistically significant because the confidence interval supports H0.c) The fall in defects is not statistically significant because 4% is smaller than 5%.d) The fall in defects is not statistically significant because the confidence interval would lead us to reject H0.2. If we wish to determine whether there is evidence that the proportion of successes is higher in group 1 than in group 2, the appropriate test to use isa) the z test.b) the 2test.c) both of the aboved) none of the aboveTABLE 12-14Recent studies have found that American children are more obese than in the past. The amount of time children spend watching television has received much of the blame. A survey of 100 ten-year-olds revealed the following with regards to weights and average number of hours a day spent watching television. We are interested in testing whether the average number of hours spent watching TV and weights are independent at 1% level of significance.WeightsTV HoursTotal0-3 3-6 6+More than 10 lbs. overweight 1 9 20 30Within 10 lbs. of normal weight 20 15 15 50More than 10 lbs. underweight 10 5 5 20Total 31 29 40 1003. Referring to Table 12-14, if there is no connection between weights and average number of hours spent watching TV, we should expect how many children to be spending 3-6 hours, on average, watching TV and are more than 10 lbs. underweight?a) 5b) 5.8c) 6.2d) 8Note the following:1. You will be penalized if you do not compute the sample variance of the d column in question 20, so you might want to do it now.2. This test is normed on 50 points, but there are 74 points possible including the take-home. You may not finish the exam and might want to skip some questions.3. A table identifying methods for comparing 2 samples is at the end of the exam.10/24/03 252x03234. Turn in your computer output from computer problem 1 only tucked inside this exam paper. (3 points - 2 point penalty for not handing this in.)MTB > TwoT 90.0 'educ' 'sex';SUBC> Alternative -1.Two-Sample T-Test and CI: EDUC, SEXTwo-sample T for EDUCSEX N Mean StDev SE MeanFemale 788 13.19 3.03 0.11Male 651 13.28 2.85 0.11Difference = mu (Female) - mu (Male )Estimate for difference: -0.09190% upper bound for difference: 0.108T-Test of difference = 0 (vs <): T-Value = -0.58 P-Value = 0.280 DF = 1412The computer output above refers to a test very much like the Minitab test you ran of two independent samples. The major difference is that 1439 numbers appear in column 1 (labeled EDUC) which give number of years of education completed and the computer sorted them by gender using the words ‘female’ and ‘male’ in column 5 (labeled SEX). The variable Fx can thus refer to an imaginary column of female education figures and Mx to in imaginary column of male education figures. Call this the GSSEduc output. 5. Referring to the GSSEduc output, and using the rules taught in class, the null hypothesis that wastested is .a) H0: 0– MFb) H0: 0– MFc) H0: 0– MFd) H0: 0– MF6. Referring to the GSSEduc output, we can conclude, (doing no more calculations) that, for the particular population that was sampleda) At the 10. level, there is sufficient evidence that women had fewer years of education than men.b) At the 10. level, there is a difference between the years of education gotten by men and women.c) At the 10. there is insufficient evidence that the average men’s education level is higher than the women’s.d) At the 10. level, there is sufficient evidence to conclude that there is no differencebetween men’s and women’s education level.7. Referring to the GSSEduc output, the most commonly used methods to find degrees of freedom are (i) to calculate 14372651788221 nndf, or (ii) to say that since we have large sample to use z, which is equivalent to saying that the degrees of freedom are infinite, yet the computer claims 1412df. Explain, briefly, what the computer probably did (and assumed) to get that number. 210/24/03 252x03238. (Wonnacott and Wonnacott) A small piece of hose in the cooling system of a new engine has a lifetime that varies normally (following the Normal distribution) around a mean of 18 months with a standard deviation of 4 months. The first maintenance check occurs at 12 months. What isthe probability that the hose will wear out before the maintenance check? (This is the same as theper cent of hoses that will wear out before the first maintenance check!) Make a diagram!9. In problem 8 above, the manufacturer decides that too much money is being spent on maintenance checks. If the manufacturer is willing to accept having 20% of hoses wear out before the fist maintenance check, how many months (to the nearest 100th of a month) can the manufacturer wait until the check? (This is the same as finding the 20th percentile of the distribution)10. The t test for the difference between the means of 2 independent populations assumes that the respectivea) sample sizes are equal.b) sample variances are equal.c) populations are approximately normal.d) all of the aboveTABLE 10-3The use of preservatives by food processors has become a controversial issue. Suppose 2 preservatives are extensively tested and determined safe for use in meats. A processor wants to compare the preservatives for their effects on retarding spoilage. Suppose 15 cuts of fresh meat are treated with preservative A and 15 are treated with preservative B, and the number of hours until spoilage begins is recorded for each of the 30 cuts of meat. The results are summarized in the table below.Preservative A Preservative BAx = 106.4 hoursAx = 96.54 hoursAs= 10.3 hours Bs = 13.4 hours11. Referring to Table 10-3, state the test statistic for determining if the population variance for preservative B is larger than the population variance for preservative A.a) F = 3.100b) F = 1.300c) F = 1.693d) F = 0.591310/24/03
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