统计建模与R软件

Step: AIC=35.07 y ~ x1 + x4

Df Deviance AIC 29.073 35.073 - x4 1 33.535 37.535 - x1 1 39.131 43.131 只留下x1和x4两个变量。

> summary(glm.new)

Call:

glm(formula = y ~ x1 + x4, family = binomial, data = cancer)

Deviance Residuals:

Min 1Q Median 3Q Max

-1.4825 -0.6617 -0.1877 0.1227 2.2844

Coefficients:

Estimate Std. Error z value Pr(>|z|)

(Intercept) -6.13755 2.73844 -2.241 0.0250 * x1 0.09759 0.04079 2.393 0.0167 * x4 -1.12524 0.60239 -1.868 0.0618 . ---

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 44.987 on 39 degrees of freedom Residual deviance: 29.073 on 37 degrees of freedom AIC: 35.073

Number of Fisher Scoring iterations: 6 回归方程为

P=exp(-6.13755+0.09759x1-1.12524x4)/(1+exp(-6.13755+0.09759x1-1.12524x4)) 概率估计略。

Ex6.9

我表示不想做了...

我没弄明白nls()函数里所说的start即初始值怎么设置。好像可以随便设置，只要保证函数收敛即可？？

Ex7.1 (1)

>lamp<-data.frame(X=c(115,116,98,83,103,107,118,116,73,89,85,97),A=factor(rep(1:3,c(4,4,4))))

> lamp.aov<-aov(X~A,data=lamp);summary(lamp.aov) Df Sum Sq Mean Sq F value Pr(>F) A 2 1304 652.0 4.923 0.0359 * Residuals 9 1192 132.4 P值小于0.05，有显著差异。 (2)

对甲的区间估计：

> a<-c(115,116,98,83) > t.test(a)

One Sample t-test

data: a

t = 13.1341, df = 3, p-value = 0.0009534

alternative hypothesis: true mean is not equal to 0 95 percent confidence interval: 78.04264 127.95736 sample estimates: mean of x 103

或者用这个命令更简单：

>attach(lamp) > t.test(X[A==1])

乙的均值估计为111，95%置信区间为99.59932, 122.40068。 丙的均值估计为86，95%置信区间为70.08777, 101.91223。 (3)多重检验： > attach(lamp) P值不做调整：

> pairwise.t.test(X,A,p.adjust.method = \

Pairwise comparisons using t tests with pooled SD

data: X and A

1 2 2 0.351 -

3 0.066 0.013

P值进行Holm调整：

P value adjustment method: none

> pairwise.t.test(X,A,p.adjust.method = \

Pairwise comparisons using t tests with pooled SD

data: X and A

1 2 2 0.35 - 3 0.13 0.04

P value adjustment method: holm

不论采取哪种方法，都可看出乙和丙有显著差异。

Ex7.2 (1)

>lamp<-data.frame(X=c(20,18,18,17,15,16,13,18,22,17,26,19,26,28,23,25,24,25,18,22,27,24,12,14),A=factor(rep(1:4,c(10,6,6,2)))) > lamp.aov<-aov(X ~ A, data=lamp);summary(lamp.aov) Df Sum Sq Mean Sq F value Pr(>F)

A 3 351.7 117.24 15.11 2.28e-05 *** Residuals 20 155.2 7.76

P值小于0.05，可认为四个厂生产的产品的变化率有显著差异。 (2)

> attach(lamp) P值不做调整：

> pairwise.t.test(X,A,p.adjust.method = \

Pairwise comparisons using t tests with pooled SD

data: X and A

1 2 3 2 8.0e-05 - - 3 0.00053 0.47666 -

4 0.05490 6.1e-05 0.00020

P value adjustment method: none P值进行Holm调整：

> pairwise.t.test(X,A,p.adjust.method = \

Pairwise comparisons using t tests with pooled SD

data: X and A

1 2 3 2 0.00040 - - 3 0.00158 0.47666 -

4 0.10979 0.00036 0.00079

P value adjustment method: holm

由此可得，除了A1和A4，A2和A3这两组的差异不显著外，其他组合的差异都很显著。

Ex7.3

>lamp1<-data.frame(X=c(30,27,35,35,29,33,32,36,26,41,33,31,43,45,53,44,51,53,54,37,47,57,48,42,82,66,66,86,56,52 ,76,83,72,73,59,53),A=factor(rep(1:3,c(12,12,12)))) >attach(lamp1) 正态性检验：

> shapiro.test(X[A==1])

Shapiro-Wilk normality test

data: X[A == 1]

W = 0.9731, p-value = 0.9407

> shapiro.test(X[A==2])

Shapiro-Wilk normality test

data: X[A == 2]

W = 0.9708, p-value = 0.9193

> shapiro.test(X[A==3])

Shapiro-Wilk normality test

data: X[A == 3]

W = 0.9371, p-value = 0.4613 数据在三种水平下均是正态的。 方差齐性检验：

> bartlett.test(X~A,data=lamp1)

Bartlett test of homogeneity of variances

data: X by A

Bartlett's K-squared = 12.139, df = 2, p-value = 0.002312 P值小于0.05，认为各组方差不等。

Ex7.4

>lamp<-data.frame(X=c(2.79,2.69,3.11,3.47,1.77,2.44,2.83,2.52,3.83,3.15,4.70,3.97,2.03,2.87,3.65,5.09,5.41,3.47,4.92,4.07,2.18,3.13,3.77,4.26),