D. A 90% confidence interval for the ratio of two variances is (.023,1.99). Based on the confidence interval you will fail to reject H 0 =!

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1 SMAM 314 Review for Exam 3 1. Mark the following statements true (T) or false(f) A. A null hypothesis that is rejected at α=.01 will always be rejected at α=.05. Β. One hundred 90% confidence intervals are obtained for the mean of a population with an unknown mean. About 90 of these confidence intervals contain the true mean. It is possible to tell which of the 90 confidence intervals actually contain the true mean. C. An algebra test is administered before and after students take a course in remedial algebra. It is appropriate to use a t test on independent random samples for such data. D. A 90% confidence interval for the ratio of two variances is (.023,1.99). Based on the confidence interval you will fail to reject H 0 1 = 2 in favor of H 1 1 " 2. E. Two independent random samples are compared. A F test rejects the hypothesis of equal variances. Sample 1 has 10 observations. Sample 2 has 12 observations. It is always appropriate to test the hypothesis of equality of means vs. one mean being larger than the other using a student t distribution with 20 degrees of freedom. For F and G indicate the correct choice F. For the same sample size a 95% confidence interval is a. wider than a 90% confidence interval but narrower than a 99% confidence interval b. narrower than both a 90% and a 99% confidence interval c. wider than both a 90% and a 99% confidence interval d. the average of the widths of the 90% and 95% confidence interval e. none of the choices a-d. G. You have a real data set. You do not know the true population mean. You find % two sided confidence intervals. Which of the following is not true. a. About 95 of the 100 confidence intervals will contain the true population mean. b. You cannot tell which of the 100 confidence intervals contain the true mean. c. You can tell which of the 100 confidence intervals contain the true mean. d. For values inside a confidence interval you fail to reject the null hypothesis that those values are the true mean. e. If a sample size greater than 100 were used the width of the confidence intervals would be less. 2. A machine produces metal rods used in an automobile suspension system. A random sample of 6 rods is selected and the diameter is measured. The resulting data in millimeters is shown A. Find and interpret a 95% two sided confidence interval for the true mean diameter of the rods. B. Based only on the confidence interval (do not perform a formal test of hypothesis) would H 0 µ = 8.20 be rejected in favor of H 1 µ 8.20 at α =.05? Explain. (5 points)

2 3. A study is to be conducted of the proportion of homeowners who have a high-speed Internet connection. A sample of 1000 people shows that 632 have a high speed internet connection. A. Find a 90% confidence interval on the true proportion of people that have a high speed internet connection. B.How large a sample is needed to be 95% confident that the error in estimating this quantity is less than 0.02? 4.A random sample of 500 registered voters in Phoenix, Arizona is asked whether they favor the use of oxygenated fuels year round to reduce air pollution. Suppose 325 voters respond positively. A. Perform an appropriate test of hypothesis to determine whether more than 60% of the voters favor the use of these fuels.use α=.05. Write the complete report. B. Find the p value. C. Would you reject the null hypothesis at α =.01.? Explain. 5.To study the effectiveness of wall insulation in saving energy for home heating the energy consumption in MWH for 6 houses in Bristol England was recorded for two winters. The first winter was before insulation and the second winter was after insulation House Before Insulation After Insulation A. Find a 95% two sided confidence interval on the difference in the energy consumption before and after insulation. B. Based on the confidence interval is there a statistically significant difference at α =.05 in the energy consumption before and after insulation. Explain.(5 points) 6.The breaking strengths of n = 20 bundles of wool fibers have a sample mean x = and a sample standard deviation s x = In addition the breaking strengths of m = 25 bundles of synthetic fibers have a sample mean of y = and a sample standard deviation s y = Consider the output below.

3 A. For a test of hypothesis H 0 1 = 2 H 1 1 " 2 what would be your conclusion at (1) α=.05? Explain (3 points) (2) α=.01? Explain (3 points) B. Based on the Minitab output below write the hypothesis test report. No computations are needed. Chose a level of significance and specify it. (15 points) Two-Sample T-Test and CI Sample N Mean StDev SE Mean Difference = mu (1) - mu (2) Estimate for difference: % CI for difference: ( , ) T-Test of difference = 0 (vs not =): T-Value = P-Value = DF = 23

4 This is a test for equality of proportions. We have not covered this in class but you will be able to do it by following instructions. The test is based on the normal distribution for large samples. 7. A company is performing a failure analysis for two of its products. It is found that for the first product 76 out of 243 failures were due to operator misuse while for the second product 122 of the 320 failures were due to operator misuse. A. Perform an appropriate test of hypothesis to determine whether there is a significant difference at α=.05 in the proportion of products that failed due to operator misuse for the two products. Write the complete report. (15 points) H 0 = p 2 H 1 p 2 Assumptions Large sample Independent random samples Region of rejection What would be the region of rejection based on percentiles of the normal distribution. Calculation Let x i,i = 1,2 be the number of failures due to operator misuse for the ith product. Let n i,i = 1,2 be the number of failures for the ith product. p i = x i,i = 1,2, p = x + x 1 2 n i n 1 + n 2 p Z = 1 p 2 p(1 p) " % $ # n 1 n ' 2 & Reject or do not reject H 0 Conclusion B. Obtain the p value. C. Find a 95% confidence interval and tell whether the confidence interval supports the results in part A. p p1 (1 p 2 ± z 1 ) "/2 + n 1 p 2 (1 p 2 ) n 2

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