生物统计学课后答案

Chapter 4 4-1 For mean arterial pressure t = -1.969, and for total peripheral resistance t = -1.286. There are 23 degrees of freedom in each case. With 23 degrees of freedom 2.069 defines the most extreme 5% of the possible values of the t distribution when the treatment has no effect. Thus, these data do not provide sufficient evidence to reject the hypothesis that different anesthetic agents did not produce differences in mean arterial pressure or total peripheral resistance. 4-2 Yes. t = 3.14, = 20, P < .01. The blood pressure drops rather than increasing, however, so this change is not clinically desirable. 4-3 No. t = 1.33, = 20, P = .20. 4-4 Prob. 3-1: t= 3.967, = 40, P<.01; Prob. 3-5: t= -1.467, = 98, P<.01. In both these cases, we can reject the null hypothesis of no difference between the groups. t2=F. 4-5 The subgroups are nonsmokers, clean environment; nonsmokers, smoky environment and light smokers; moderate smokers and heavy smokers. Here are some of the resulting t values: nonsmokers, clean environment vs. nonsmokers, smoky environment, t = 6.249: light smokers vs. moderate smokers, t = 4.715; moderate smokers vs. heavy smokers, t = 2.358. Since there are a total of 10 pair-wise comparisons, to keep the overall error rate at 5%, these values of t must be compared with the critical value corresponding to P = .05/1 = .005 with 995 degrees of freedom, 2.807. These data support the hypothesis that chronic exposure to other peoples' smoke affects the lung function of nonsmokers. 4-6 All the groups have worse lung function than the nonsmokers breathing clean air (the control group). Nonsmokers in smoky office; q' = 6.249, p = 5; light smokers: q' = 7.499, p = 5; moderate smokers: q' = 12.220, p = 5; heavy smokers: q' = 14.558, p = 5. All exceed the critical values of q' for P < .01 with p = 5 and 995 degrees of freedom: 3.00. 4-7 Inactive vs. joggers, t = 5.616; inactive vs. runners, t = 8.214; joggers vs. runners, t = 2.598. To maintain the overall risk of erroneously rejecting the hypothesis of no difference at 5%, these values of t should be compared with the critical value for P = .05/3 = .017 with 207 degrees of freedom, 2.44. Therefore, the three samples are drawn from distinct populations. 4-8 Inactive vs. joggers, t = 5.616; inactive vs. runners, t = 8.214. To maintain the overall risk of erroneously rejecting the hypothesis of no difference at 5%, these values of t should be compared with the critical value for P = .05/2 = .025 with 207 degrees of freedom, 2.282. Therefore, the two
CHale Waihona Puke apter 22-1 Mean = 61,668, median = 13,956, standard deviation = 117,539, 25th percentile = 7861, 75th percentile = 70,133, mean – 0.67 standard deviations = 6623, mean + 0.67 standard deviations = 79,604. These data appear not to be drawn from a normally distributed population for several reasons. (1) The mean and median are very different. (2) All the observations are (and have to be, since you cannot have a negative viral load) greater than zero and the standard deviation is larger than the mean. If the population were normally distributed, it would have to include negative values of viral load, which is impossible. (3) The relationship between the percentiles and numbers of standards deviations about the mean are different from what you would expect if the data were drawn from a normally distributed population. 2-2 Mean = 4.30, median = 4.15, standard deviation = 0.67, 25th percentile = 3.89, 75th percentile = 4.84, mean – 0.67 standard deviations = 3.82, mean + 0.67 standard deviations = 4.90. These data appear to be drawn from a normally distributed population on the basis of the comparisons in the answer to Prob. 2-1. 2-3 Mean = 1709.3, median = 1750, standard deviation = 824.8, 25th percentile = 877.5, 75th percentile = 2350, mean – 0.67 standard deviations = 1156.7, mean + 0.67 standard deviations = 2261.9. These data appear to be drawn from a normally distributed population on the basis of the comparisons in the answer to Prob. 2-1. 2-4 There is 1 chance in 6 of getting each of the following values: 1, 2, 3, 4, 5, and 6. The mean of this population is 3.5. 2-5 The result is a sample drawn from the distribution of all means of samples of size 2 drawn from the population described in Prob. 2-4. Its mean is an estimate of the population mean, and its standard deviation is an estimate of the standard error of the mean of samples of size 2 drawn from the population in Prob. 2-4. 2-6 Given that the number of authors has to be an integer number and the size of the standard deviations compared with that of the means, the total number of authors is probably skewed toward larger numbers of authors, with a few papers containing many more authors than the rest of the population. Notice that the number of authors and spread of the population dramatically increased in 1976. The certainty with which you can estimate the mean number of authors is quantified with the standard error of the mean of each of the samples, the standard error of the mean decreasing as the precision with which you can estimate the true population means increases. The standard errors of the mean for the four different years are .11, .13, .10, .59.