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Answers to selected homework problems (Ch 16-21) Chapter 16 4a. 2.064 (use df = 24) b. 0.685 6. 3.939 to 7.061 mean percent changes in level 8a. df = 14 b. t is between 1.761 and 2.145. The probabilities for these entries are p = .05 and p = 0.025 c. P-value is between p = .025 and .05. d. Significant at the = 0.05 level, not significant at the = 0.01 level. 10. t is approx –8.87 (from table), so P < .001. Conclusion: We would expect a sample mean of 59.59 percent of nitrogen less than 0.1% of the time if the null hypothesis were true. Thus there is very strong evidence for Ha, that Cretaceous air is quite different from modern air. 14a. H0: = 0; Ha: > 0. Use a one-sided alternative because researchers believe that CO2 will increase the growth rate. b. Using x = 1.916, and S = 1.05, we find P = .0436. Thus, about 4% of the time we would have a sample mean such as this when the actual mean was really 0. c. Because n = 3 is such a small sample, we can’t tell if this is Normally distributed. 16a. H0: = $0, Ha: > $0; t = 43.47 and P = 0. Thus, we reject H0. The new policy would increase credit card usage. b. $312 to $352 c. The sample size is large, and we have a SRS. c. Choose another SRS of size 200 without the offer as a control group and repeat the study comparing the mean increase. 18. The mean number of unexcused absences for all workers is 8.10 to 11.66 days. 28. Since 2-sided use p = .0025. The critical value is 3.174. So we would reject H0 for t < -3.174 or t > 3.174. Chapter 17 6a. t = 7.36 means that if df = k is reasonable large, the t(k) distribution is similar to the Normal distribution, wo we can estimate the P values by this. Based on the 68-95-99.7 rule, we would almost never get a value as large as 7.36 from the N(0,1) distribution, so we can say P is essentially 0. Thus, the result is significant. b. df = 32. 24a. H0: 1 = 2 (where 1 = men’s rage, and 2 = women’s rage) Ha: 1 > 2 t = 6.129, P is essentially 0. Thus, there is extremely strong evidence for Ha, that the male mean is higher than the female mean. b. They contacted only people with telephones, so the entire population may include poorer people, who may have more rage. 26. H0: 1 = 2; Ha: 1 < 2; p = .02956. Thus there is strong evidence at the = 0.03 level for Ha . Note: this probably should be 2-sided, Ha: 1 2, since researchers wouldn’t know beforehand. Then p = 0.06. Chapter 17 on-line quiz: (If you use your calculator, your answers are more precise than the options given in the multiple choice for numbers 2, 3, and 4) 2. 34.6 38.59 3. P .0487 4. 4 5.85 yards Chapter 18 4a. The mean is p = .14 and s = .0155 (approx) b1. probability is 0.02% --not likely b2. about 26% 6. The population must be at least 10 times as large as the sample, and hers is not. 16a. m = .0460 (approx) b. n 635 18. H0: p = .5; Ha: p .5 So z = -2.40 (approx), and P = .0162 (approx). Thus, strong evidence for Ha. 22. Approx. 0.4918 to 0.6706 28a. m = 0.0306 –Yes b. We don’t have sample sizes for each gender, but we can figure it out. c. Greater than 0.03 since the sample size will be smaller. Chapter 19 2. Approx 0.09152 to 0.29081. 6. ˆ ˆ ˆ ˆ H0: p t = p c ; Ha: p t > p c ; z = 4.82 (could be negative if you set this up as control – treatment). Z is off the chart, so p will be tiny, and we declare that the drug is successful. 8a. approx 0.60933 to 0.65071. b. 0.36933 to 0.45064. 12. ˆ ˆ ˆ ˆ H0: p b = p w ; Ha: p b < p w ; P = .0124. Since the probability that pb = pw given this ample proportion is only 1.24%, we conclude that black parents are less likely to rate schools favorably. Chapter 20 2a. C or better with < 2 hours: approx 55% 2-12 hours: 74.7% > 12 hours: 37.5% graph should have % on y-axis and three bars. b. Some, but not too much time in extra curricular activities seems to be beneficial. 4a. <2 2-12 > 12 TOTAL hours hours hours GRADE C or up 13.8 62.71 5.51 82 D or F 6.22 28.29 2.49 37 TOTAL 20 91 8 119 b. Compare the expected counts (above) with the observed counts in problem 20.2. We expected more in the < 2 and > 12 for passing than what was observed and less in the 2-12 hours category. So again we see that moderate activity seems beneficial. 10a. 2 = 10.827, P = .013, df = 3. Thus, there is strong evidence for a relationship between gender and major. b. one expected count is less than 5, which is only 1/8, so the chi-square procedure is acceptable. c. Chart of percents: Female Male Accounting 30.2 34.8 Admin 40.4 24.8 Econ 2.2 3.7 Finance 27.1 36.7 The biggest difference is in admin. A higher percentage of women chose this major. Greater proportions of men chose other fields, especially Finance. d. The largest chi-square components are the two from the “Administration” row. Many more women than we expect (91 actual, 76.36 expected) chose this major, whereas only 40 men chose this (54.64 expected) e. 46.5% did not respond. 16. 2 = 41.08, P = 0.0000000258. Thus, we reject H0 and claim that the differences between US and Canada are statistically significant. We can also see this because how far away 2 is from the mean of the distribution, which is df = 4. Chapter 21 2a. r2 = 0.9915 (rounded). Thus time explains almost all of the change in length. b. estimate of = -2.3948 cm = 0.15848 cm/min = 0.80588 cm c. The regression equation is y = -2.3948 + 0.15848 x 4. The Minitab output give b = 0.158483 and SEb = 0.003661. There were n = 18 observations, so df = 16. For a 95% confidence interval, we use t* = 2.120. So 0.158483 2.120 SEb = 0.1507 to 0.1662 cm/min. The slope represents the average rate in cm/min at which icicles grow under the conditions of this experiment. 10. Find the LinReg is y = -22.97x + 260.56; r = -.8428 with n = 19. Table F (p. 661) tells us P < 0.0005. So there is strong evidence for Ha: < 0 (note: one-sided). 18a. The 95% CI for the slope is seen in cells F12 and G12 in Fig 21.9. So the interval is .0033 to .0138. This can be verified by: b t*SEb = .0086 (2.145)(.0025). (Note: Use Table C to find t* for confidence level 95% and df = 16-2 = 14.) b. A 90% CI for is .0086 (1.761)(.0025) = .0042 to .0130. Chapter 22 2a. Ho: all age groups have the same mean road-rage measurement. Ha: at least one group has a different mean. b. F = 34.96, P < .01 implies the F test is quite significant, giving strong evidence that the means are different. The sample means suggest that road rage decreases with age. 8a. To use ANOVA make sure the largest sample standard deviation is no more than 33 .1944 twice the smallest. For 22.3 we calculate 1.16 –ok 24 .8106 b. 9.970/8.341 1.20 –ok 22a. The means for each of the four groups: Weeks xi Si 2 123.8 4.6043 4 123.6 6.5422 8 134.4 9.5289 16 116.4 16.0873 The mean does not consistently decrease over time. b. Considering that 16.0873/4.6043 is greater than 2, ANOVA should not be used. c. The 2-sample t test does not require that the standard deviations be equal, ANOVA does.