统计建模与R软件假设检验习题答案

1、习题答案Ex5.1 x t.test(x,mu=225) One Sample t-testdata: xt = -3.4783, df = 19, p-value = 0.002516alternative hypothesis: true mean is not equal to 22595 percent confidence interval: 172.3827 211.9173sample estimates:mean of x 192.15原假设：油漆工人的血小板计数与正常成年男子无差异。备择假设：油漆工人的血小板计数与正常成年男子有差异。p值小于0.05，拒绝原假设，认为油漆工人

2、的血小板计数与正常成年男子有差异。上述检验是双边检验。也可采用单边检验。备择假设：油漆工人的血小板计数小于正常成年男子。 t.test(x,mu=225,alternative=less) One Sample t-testdata: xt = -3.4783, df = 19, p-value = 0.001258alternative hypothesis: true mean is less than 22595 percent confidence interval: -Inf 208.4806sample estimates:mean of x 192.15同样可得出油漆工人的血小板

3、计数小于正常成年男子的结论。Ex5.2 pnorm(1000,mean(x),sd(x)1 0.5087941 x 1 1067 919 1196 785 1126 936 918 1156 920 948 pnorm(1000,mean(x),sd(x)1 0.5087941x A B t.test(A,B,paired=TRUE) Paired t-testdata: A and Bt = -0.6513, df = 7, p-value = 0.5357alternative hypothesis: true difference in means is not equal to 095

4、 percent confidence interval: -15.62889 8.87889sample estimates:mean of the differences -3.375p值大于0.05，接受原假设，两种方法治疗无差异。Ex5.4（2）方差相同模型t检验： t.test(x,y,var.equal=TRUE) Two Sample t-testdata: x and yt = -0.6419, df = 38, p-value = 0.5248alternative hypothesis: true difference in means is not equal to 09

5、5 percent confidence interval: -2.326179 1.206179sample estimates:mean of x mean of y 2.065 2.625 方差不同模型t检验： t.test(x,y) Welch Two Sample t-testdata: x and yt = -0.6419, df = 36.086, p-value = 0.525alternative hypothesis: true difference in means is not equal to 095 percent confidence interval: -2.3

6、2926 1.20926sample estimates:mean of x mean of y 2.065 2.625 （3）方差检验： var.test(x,y) F test to compare two variancesdata: x and yF = 1.5984, num df = 19, denom df = 19, p-value = 0.3153alternative hypothesis: true ratio of variances is not equal to 195 percent confidence interval: 0.6326505 4.0381795

7、sample estimates:ratio of variances 1.598361接受原假设，两组数据方差相同。Ex5.5 a b t.test(a,b,var.equal=TRUE) Two Sample t-testdata: a and bt = -8.8148, df = 20, p-value = 2.524e-08alternative hypothesis: true difference in means is not equal to 095 percent confidence interval: -48.24975 -29.78358sample estimates

8、:mean of x mean of y 125.5833 164.6000可认为两者有差别。Ex5.6二项分布总体的假设检验： binom.test(57,400,p=0.147) Exact binomial testdata: 57 and 400number of successes = 57, number of trials = 400, p-value = 0.8876alternative hypothesis: true probability of success is not equal to 0.14795 percent confidence interval: 0.

9、1097477 0.1806511sample estimates:probability of success 0.1425P 值0.05，故接受原假设，表示调查结果支持该市老年人口的看法Ex5.7二项分布总体的假设检验： binom.test(178,328,p=0.5,alternative=greater) Exact binomial testdata: 178 and 328number of successes = 178, number of trials = 328, p-value = 0.06794alternative hypothesis: true probability of su