A tool manufacturer tests three types of cutters used in a lathe operations. One type is a laminated steel cutter consisting of a very hard high carbon steel sandwiched between two softer steels. Another is a special high-speed tool steel cutter developed using powder metallurgy. The final one is made from cryogenically treated steel. Several cutters of each type are tested to see how long they will last (in hours) in continuous operation until they need to be sharpened. The times are recorded for each cutter used in the study, and the results are summarized below. n is the number of cutters used of the type indicated. Type of cutter laminated steel high-speed steel cryogenically treated steel Average time until sharpening needed 29.84 28.58 33.83 Standard deviation 7.08 8.07 6.93 n 32 27 38 Assume the data are three independent SRSs, one from each of the three populations of cutters, and that the distribution of the times until sharpening is needed is normal. A partial ANOVA table is given below. Source Cutter type Error Total df Sums of squares 505.26 Mean square 53.45 1. A) B) C) D) The degrees of freedom for cutter type (group) are 2. 3. 94. 97. 2. A) B) C) D) The degrees of freedom for error are 2. 3. 94. 97. Page 1 F-ratio 3. For this example, we notice A) this is a randomized, designed experiment. B) the data show no strong evidence of a violation of the assumption that the three populations have the same standard deviation. C) ANOVA cannot be used on these data because the sample sizes are different. D) the data show very strong evidence of a violation of the assumption that the three populations have the same standard deviation. 4. The null hypothesis for the ANOVA F test is that the population mean time until sharpening is needed is the same for all three cutter types. The alternative hypothesis is A) that the population mean time until sharpening is needed is larger for cryogenically treated steel cutters than for the other two types of cutters. B) that the population mean time until sharpening is needed is larger for high-speed steel cutters than for the other two types of cutters. C) that the population mean time until sharpening is needed is larger for laminated steel cutters than for the other two types of cutters. D) none of the above. 5. The value of the ANOVA F-statistic for testing equality of the population means of the three cutter types is A) 3.15. B) 4.73. C) 6.30. D) 9.45. 6. The P-value for the ANOVA F test for testing equality of the population means of the three cutter types is A) less than 0.001. B) between 0.001 and 0.010. C) between 0.010 and 0.025. D) greater than 0.025. A company runs a three-day workshop on strategies for working effectively in teams. On each day a different strategy is presented. Forty-eight employees of the company attend the workshop. At the outset, all 48 are divided into twelve teams of four. The teams remain the same for the entire workshop. Strategies are presented in the morning. In the afternoon, the teams are presented with a series of small tasks, and the number completed successfully using the strategy taught that morning is recorded for each team. The mean number of tasks completed successfully by all teams each day and the standard deviation is computed. The results are given below. Page 2 Day (strategy) 1 2 3 Means 15.25 15.64 15.25 StdDevs 7.10 14.14 14.03 The researchers did an ANOVA F test of the data and obtained the following results. Source Day Error Total Sums of squares 1.36 5321.71 5323.08 Mean square 0.68 152.05 F-ratio 0.0045 7. A) B) C) D) The degrees of freedom in the numerator for this test are 36. 33. 2. 1. 8. A) B) C) D) The degrees of freedom in the denominator of the F test are 33. 12. 3. 2. 9. A) B) C) D) The P-value of the ANOVA F test is larger than 0.10. between 0.10 and 0.05. between 0.05 and 0.01. below 0.01. 10. In this example, we notice A) there is clear evidence of bias in the results. This is undoubtedly due to the lack of blinding on the part of the subjects. B) the data show very strong evidence of a violation of the assumption that the three populations have the same standard deviation. C) ANOVA cannot be used on these data because the sample sizes are less than 20. D) the assumption that the data are independent for the three days is unreasonable, because the same teams were observed each day. Page 3 11. For this example, which of the following conclusions is most reasonable? A) There is moderate evidence that the strategies taught are effective in increasing the number of tasks completed successfully for the first two days, but then the effect appears to wear off.. B) An ANOVA F test is not appropriate for these data. Instead, the researchers should have done several tests to see if the number of tasks completed successfully differed for the three days. This analysis would have shown the treatment was effective. C) The data provide strong evidence that the mean number of tasks completed successfully differ for the three strategies taught. D) The data appear to provide little or no evidence that the strategies taught differ in their effectiveness in helping teams complete tasks successfully. A researcher wished to compare the effect of the frequency of the rate of stepping on heart rate in a step-aerobics workout. A collection of 30 adult volunteers, 15 women and 15 men, was selected from a local gym. The men were randomly divided into three groups of five subjects each. Each group did a standard step-aerobics workout with group 1 at a low rate of stepping, group 2 at a medium rate of stepping, and group 3 at a rapid rate. The women were also randomly divided into three groups of five subjects each. As with the men, each group did a standard step-aerobics workout with group 1 at a low rate of stepping, group 2 at a medium rate of stepping, and group 3 at a rapid rate. The mean heart rate at the end of the workout for all subjects was determined in beats per minute. A partial ANOVA table for these data is given below. Analysis of variance for heart rate: Source step rate gender group*gender Error Total 12. A) B) C) D) DF SS 2553.2 70.7 231.7 5762.0 8617.6 MS Page 4 P 0.48 The factors in the experiment are rate of stepping and heart rate. rate of stepping and gender. heart rate and gender. gym membership and gender. F 0.623 13. The advantages of studying the two factors, gender and stepping rate, in the same experiment are: A) it is more efficient to study two factors simulaneously rather than separately. B) we can reduce the residual variation in a model by including a second factor thought to influence the response. C) we can investigate interaction between factors. D) all of the above. 14. A) B) C) D) The degrees of freedom for error are 5. 11. 24. 29. 15. A) B) C) D) The pooled standard error is 15.5. 75.9. 240.1. 5762.0. 16. Using the information in the ANOVA table and the example description, the degrees of freedom for interaction are A) 1. B) 2. C) 3. D) 6. 17. The plots and the P-value for the test for interaction show little evidence of interaction. This means that A) there is little difference in the heart rates of men and women. B) the change in heart rate due to the different stepping rates are similar for men and women. C) changes in stepping rate are positively associated with heart rate. D) step exercise is equally beneficial to men and women. 18. A) B) C) D) The numerical value of the F statistic for the test for gender is 0.01. 0.29. 5.32. 81.49. Page 5 19. A) B) C) D) The numerical value of SSM, the sum of squares for the model, is 2553.2. 2866.6. 5762.0. 8617.6. 20. Using the information in the ANOVA table and the example description, the P-value for the test for the main effect of step rate is A) less than 0.01. B) between 0.01 and 0.05. C) between 0.05 and 0.10. D) greater than 0.10. 21. If the data were analyzed as a one-way ANOVA with six \"treatments\" corresponding to the six combinations of stepping rate and gender, the sum of squares for error would have been A) 2623.9. B) 2855.6. C) 5762.0. D) 8617.6. Page 6