假设检验习题
囧囧有神
2016年1月11日
———
使用t.test()均值检验之前，需要对数据进行正态检验、方差齐性检验。本题原假设H0：油漆工人的血小板计数与正常男子无差异，即mu=225，H1:mu!=225。
x <- c(220, 188, 162, 230, 145, 160, 238, 188, 247, 113, 126, 245, 164, 231,
256, 183, 190, 158, 224, 175)
# 正态性检验：
shapiro.test(x) #p-value = 0.3768>0.05，无法拒绝原假设，因此认为X服从正态分布##
## Shapiro-Wilk normality test
##
## data: x
## W = 0.9506, p-value = 0.3768t.test(x, mu = 225)##
## One Sample t-test
##
## data: x
## t = -3.4783, df = 19, p-value = 0.002516
## alternative hypothesis: true mean is not equal to 225
## 95 percent confidence interval:
## 172.3827 211.9173
## sample estimates:
## mean of x
## 192.15p-value = 0.002516<0.05，拒绝原假设，即认为油漆工人的血小板计数与正常男子有差异。
———
x <- c(1067, 919, 1196, 785, 1126, 936, 918, 1156, 920, 948)
1 - pnorm(1000, mean(x), sd(x))## [1] 0.49120591000小时以上的概率为0.4912059
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成对数据的均值检验：
x <- c(113, 120, 138, 120, 100, 118, 138, 123)
y <- c(138, 116, 125, 136, 110, 132, 130, 110)
t.test(x, y, paired = TRUE)##
## Paired t-test
##
## data: x and y
## t = -0.6513, df = 7, p-value = 0.5357
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -15.628891 8.878891
## sample estimates:
## mean of the differences
## -3.375p-value = 0.5357，无法拒绝原假设，即认为两种方法疗效无差异。
———
x <- c(-0.7, -5.6, 2, 2.8, 0.7, 3.5, 4, 5.8, 7.1, -0.5, 2.5, -1.6, 1.7, 3, 0.4,
4.5, 4.6, 2.5, 6, -1.4)
y <- c(3.7, 6.5, 5, 5.2, 0.8, 0.2, 0.6, 3.4, 6.6, -1.1, 6, 3.8, 2, 1.6, 2, 2.2,
1.2, 3.1, 1.7, -2)（1）检验数据是否服从正态分布：
shapiro.test(x) #p-value = 0.7527>0.5，故无法拒绝原假设，认为服从正态分布##
## Shapiro-Wilk normality test
##
## data: x
## W = 0.9699, p-value = 0.7527shapiro.test(y) #p-value = 0.7754，服从正态分布##
## Shapiro-Wilk normality test
##
## data: y
## W = 0.971, p-value = 0.7754ks.test(x, "pnorm", mean(x), sd(x))## Warning in ks.test(x, "pnorm", mean(x), sd(x)): ties should not be present
## for the Kolmogorov-Smirnov test##
## One-sample Kolmogorov-Smirnov test
##
## data: x
## D = 0.1065, p-value = 0.9771
## alternative hypothesis: two-sidedks.test(y, "pnorm", mean(y), sd(y))## Warning in ks.test(y, "pnorm", mean(y), sd(y)): ties should not be present
## for the Kolmogorov-Smirnov test##
## One-sample Kolmogorov-Smirnov test
##
## data: y
## D = 0.1197, p-value = 0.9368
## alternative hypothesis: two-sidedsort(x)## [1] -5.6 -1.6 -1.4 -0.7 -0.5 0.4 0.7 1.7 2.0 2.5 2.5 2.8 3.0 3.5
## [15] 4.0 4.5 4.6 5.8 6.0 7.1x1 <- table(cut(x, breaks = c(-6, -3, 0, 3, 6, 9)))
p <- pnorm(c(-3, 0, 3, 6, 9), mean(x), sd(x))
p <- c(p[1], p[2] - p[1], p[3] - p[2], p[4] - p[3], 1 - p[4])
chisq.test(x1, p)## Warning in chisq.test(x1, p): Chi-squared approximation may be incorrect##
## Pearson's Chi-squared test
##
## data: x1 and p
## X-squared = 15, df = 12, p-value = 0.2414（2）均值检验：
t.test(x, y, var.equal = TRUE) #方差相等##
## Two Sample t-test
##
## data: x and y
## t = -0.6419, df = 38, p-value = 0.5248
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -2.326179 1.206179
## sample estimates:
## mean of x mean of y
## 2.065 2.625t.test(x, y, var.equal = FALSE) #方差不等##
## Welch Two Sample t-test
##
## data: x and y
## t = -0.6419, df = 36.086, p-value = 0.525
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -2.32926 1.20926
## sample estimates:
## mean of x mean of y
## 2.065 2.625t.test(x, y, paired = TRUE) #成对数据##
## Paired t-test
##
## data: x and y
## t = -0.6464, df = 19, p-value = 0.5257
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -2.373146 1.253146
## sample estimates:
## mean of the differences
## -0.56p值均大于0.05，故无法拒绝原假设，即认为两组数据均值无差异。
（3）方差检验检验：
var.test(x, y)##
## F test to compare two variances
##
## data: x and y
## F = 1.5984, num df = 19, denom df = 19, p-value = 0.3153
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
## 0.6326505 4.0381795
## sample estimates:
## ratio of variances
## 1.598361p-value = 0.3153>0.05，故无法拒绝原假设，即认为两组数据方差无差异。
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（1）检验数据是否服从正态分布：
x <- c(126, 125, 136, 128, 123, 138, 142, 116, 110, 108, 115, 140)
y <- c(162, 172, 177, 170, 175, 152, 157, 159, 160, 162)
shapiro.test(x)##
## Shapiro-Wilk normality test
##
## data: x
## W = 0.9396, p-value = 0.4934shapiro.test(y)##
## Shapiro-Wilk normality test
##
## data: y
## W = 0.938, p-value = 0.5313ks.test(x, "pnorm", mean(x), sd(x))##
## One-sample Kolmogorov-Smirnov test
##
## data: x
## D = 0.1464, p-value = 0.9266
## alternative hypothesis: two-sidedks.test(y, "pnorm", mean(y), sd(y))##
## One-sample Kolmogorov-Smirnov test
##
## data: y
## D = 0.2222, p-value = 0.707
## alternative hypothesis: two-sidedp值均大于0.05，无法拒绝原假设，数据服从正态分布。
（2）方差齐性检验：
var.test(x, y)##
## F test to compare two variances
##
## data: x and y
## F = 1.9646, num df = 11, denom df = 9, p-value = 0.32
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
## 0.5021943 7.0488630
## sample estimates:
## ratio of variances
## 1.964622p-value = 0.32>0.05，无法拒绝原假设，即认为两组数据方差是相同的。
（3）两样本均值检验
t.test(x, y, var.equal = TRUE)##
## Two Sample t-test
##
## data: x and y
## t = -8.8148, df = 20, p-value = 2.524e-08
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -48.24975 -29.78358
## sample estimates:
## mean of x mean of y
## 125.5833 164.6000p-value = 2.524e-08<0.05，拒绝原假设，认为新药组和对照组病人的抗凝血酶活力是有差异的
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根据题意，应用二项分布总体的假设检验，原假设H0:老年人口比重为14.7%。
binom.test(57, 400, p = 0.147)##
## Exact binomial test
##
## data: 57 and 400
## number of successes = 57, number of trials = 400, p-value = 0.8876
## alternative hypothesis: true probability of success is not equal to 0.147
## 95 percent confidence interval:
## 0.1097477 0.1806511
## sample estimates:
## probability of success
## 0.1425p-value = 0.8876>0.05，无法拒绝原假设，认为调查结果支持老年人口比重为14.7%的看法。
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原假设H0:母雏比例<=0.5,H1:母雏比例>0.5。
binom.test(178, 328, p = 0.5, alternative = "greater")##
## Exact binomial test
##
## data: 178 and 328
## number of successes = 178, number of trials = 328, p-value =
## 0.06794
## alternative hypothesis: true probability of success is greater than 0.5
## 95 percent confidence interval:
## 0.4957616 1.0000000
## sample estimates:
## probability of success
## 0.5426829p-value = 0.06794>0.05，无法拒绝原假设，即认为这种处理未能增加母雏的比例。
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检验是否服从理论分布，应用chisq.test()：
chisq.test(c(315, 101, 108, 32), p = c(9, 3, 3, 1)/16)##
## Chi-squared test for given probabilities
##
## data: c(315, 101, 108, 32)
## X-squared = 0.47, df = 3, p-value = 0.9254p-value = 0.9254>0.05，无法拒绝原假设，即认为结果服从自由组合规律。
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应用pearson卡方检验
y <- c(92, 68, 28, 11, 1, 0)
x <- 0:5
q <- ppois(x, mean(rep(x, y)))
n <- length(y)
p[1] <- q[1]
p[n] <- 1 - q[n - 1]
for (i in 2:(n - 1)) p[i] <- q[i] - q[i - 1]
p## [1] 0.447087927 0.359905781 0.144862077 0.038871324 0.007822854 0.001450038chisq.test(y, p = p)##
## Chi-squared test for given probabilities
##
## data: y
## X-squared = 2.1596, df = 5, p-value = 0.8267警告由于pearson卡方检验要求分组后的每组频数大于等于5。
应用pearson卡方检验
z <- c(92, 68, 28, 12)
n <- length(z)
p <- p[1:n - 1]
p[n] <- 1 - q[n - 1]
chisq.test(z, p = p)##
## Chi-squared test for given probabilities
##
## data: z
## X-squared = 0.9113, df = 3, p-value = 0.8227p-value = 0.8227>0.05，无法拒绝原假设，即认为服从泊松分布。
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应用ks.test()，检验两组数据是否来自同一总体分布
x <- c(2.36, 3.14, 7.52, 3.48, 2.76, 5.43, 6.54, 7.41)
y <- c(4.38, 4.25, 6.53, 3.28, 7.21, 6.55)
ks.test(x, y)##
## Two-sample Kolmogorov-Smirnov test
##
## data: x and y
## D = 0.375, p-value = 0.6374
## alternative hypothesis: two-sidedp-value = 0.6374>0.05，无法拒绝原假设，即认为两组数据来自同一总体分布。
列联表的独立性检验，原假设：分娩过程中使用胎儿电子检测仪对剖腹产率无影响。
x <- c(358, 2492, 229, 2745)
dim(x) <- c(2, 2)
x## [,1] [,2]
## [1,] 358 229
## [2,] 2492 2745chisq.test(x)##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: x
## X-squared = 37.4143, df = 1, p-value = 9.552e-10p-value = 9.552e-10<0.05，拒绝原假设，认为分娩过程中使用胎儿电子检测仪对剖腹产率有影响。
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列联表的独立性检验，原假设：锻炼时间和1500米长跑记录是相互独立的。
x <- c(45, 46, 28, 11, 12, 20, 23, 12, 10, 28, 30, 35)
dim(x) <- c(4, 3)
x## [,1] [,2] [,3]
## [1,] 45 12 10
## [2,] 46 20 28
## [3,] 28 23 30
## [4,] 11 12 35chisq.test(x)##
## Pearson's Chi-squared test
##
## data: x
## X-squared = 40.401, df = 6, p-value = 3.799e-07p-value = 3.799e-07<0.05，拒绝原假设，认为锻炼时间对1500米长跑记录是有影响的。
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由于有频率小于5，因此使用fisher精确检验
x <- c(3, 6, 4, 4)
dim(x) <- c(2, 2)
fisher.test(x)##
## Fisher's Exact Test for Count Data
##
## data: x
## p-value = 0.6372
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.04624382 5.13272210
## sample estimates:
## odds ratio
## 0.521271p-value = 0.6372>0.05，无法拒绝原假设，即认为两变量是独立的，两种工艺对产品质量没有影响。
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在相同个体上的两次测定，因此使用McNeMar检验：
x <- c(58, 1, 8, 2, 42, 9, 3, 7, 17)
dim(x) <- c(3, 3)
mcnemar.test(x)##
## McNemar's Chi-squared test
##
## data: x
## McNemar's chi-squared = 2.8561, df = 3, p-value = 0.4144p-value = 0.4144>0.05，无法拒绝原假设，即认为两次测定结果是相同的。
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（1）使用符号检验分析，原假设为：中位数>=14.6，备择假设：中位数<14.6。
x <- c(13.32, 13.06, 14.02, 11.86, 13.58, 13.77, 13.51, 14.42, 14.44, 15.43)
binom.test(sum(x > 14.6), length(x), alternative = "less")##
## Exact binomial test
##
## data: sum(x > 14.6) and length(x)
## number of successes = 1, number of trials = 10, p-value = 0.01074
## alternative hypothesis: true probability of success is less than 0.5
## 95 percent confidence interval:
## 0.0000000 0.3941633
## sample estimates:
## probability of success
## 0.1p-value = 0.01074<0.05，拒绝原假设，认为该鱼塘的鱼的长度在14.6之下。
（2）使用Wilcoxon符号秩检验分析，原假设为：中位数>=14.6，备择假设：中位数<14.6。
x <- c(13.32, 13.06, 14.02, 11.86, 13.58, 13.77, 13.51, 14.42, 14.44, 15.43)
wilcox.test(x, mu = 14.6, alternative = "less", exact = FALSE)##
## Wilcoxon signed rank test with continuity correction
##
## data: x
## V = 4.5, p-value = 0.01087
## alternative hypothesis: true location is less than 14.6p-value = 0.01087<0.05，拒绝原假设，认为该鱼塘的鱼的长度在14.6之下。
———
（1）使用符号检验法检验两测定无显著差异，原假设为两总体无差异。
x <- c(48, 33, 37.5, 48, 42.5, 40, 42, 36, 11.3, 22, 36, 27.3, 14.2, 32.1, 52,
38, 17.3, 20, 21, 46.1)
y <- c(37, 41, 23.4, 17, 31.5, 40, 31, 36, 5.7, 11.5, 21, 6.1, 26.5, 21.3, 44.5,
28, 22.6, 20, 11, 22.3)
binom.test(sum(x > y), length(x))##
## Exact binomial test
##
## data: sum(x > y) and length(x)
## number of successes = 14, number of trials = 20, p-value = 0.1153
## alternative hypothesis: true probability of success is not equal to 0.5
## 95 percent confidence interval:
## 0.4572108 0.8810684
## sample estimates:
## probability of success
## 0.7p-value = 0.1153>0.05，无法拒绝原假设，即认为两侧定是无差异的。
（2）Wilcoxon符号秩检验法检验两测定无显著差异，原假设为两总体无差异。
wilcox.test(x, y, paired = TRUE, exact = FALSE)##
## Wilcoxon signed rank test with continuity correction
##
## data: x and y
## V = 136, p-value = 0.005191
## alternative hypothesis: true location shift is not equal to 0p-value = 0.005191<0.05，拒绝原假设，即认为两侧定是有差异的。
（3）使用Wilcoxon秩和检验法检验两测定无显著差异，原假设为两总体无差异。
wilcox.test(x, y, exact = FALSE)##
## Wilcoxon rank sum test with continuity correction
##
## data: x and y
## W = 274.5, p-value = 0.04524
## alternative hypothesis: true location shift is not equal to 0p-value = 0.04524<0.05，拒绝原假设，即认为两侧定是有差异的。
（4）对数据作正态检验和方差齐性检验
正态性检验：
ks.test(x, "pnorm", mean(x), sd(x))##
## One-sample Kolmogorov-Smirnov test
##
## data: x
## D = 0.1407, p-value = 0.8235
## alternative hypothesis: two-sidedshapiro.test(x)##
## Shapiro-Wilk normality test
##
## data: x
## W = 0.9507, p-value = 0.3773ks.test(y, "pnorm", mean(y), sd(y))##
## One-sample Kolmogorov-Smirnov test
##
## data: y
## D = 0.1014, p-value = 0.973
## alternative hypothesis: two-sidedshapiro.test(y)##
## Shapiro-Wilk normality test
##
## data: y
## W = 0.9667, p-value = 0.6848p-value均大于0.05，无法拒绝原假设，即认为x、y是正态的。
方差齐性检验：
var.test(x, y)##
## F test to compare two variances
##
## data: x and y
## F = 1.1406, num df = 19, denom df = 19, p-value = 0.7772
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
## 0.4514788 2.8817689
## sample estimates:
## ratio of variances
## 1.140639p-value = 0.7772>0.05，无法拒绝原假设，即认为x、y方差是相同的。
t检验：
t.test(x, y, var.equal = TRUE)##
## Two Sample t-test
##
## data: x and y
## t = 2.2428, df = 38, p-value = 0.03082
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 0.8125529 15.8774471
## sample estimates:
## mean of x mean of y
## 33.215 24.870p-value = 0.03082<0.05，拒绝原假设，即认为x、y是有差异的。
———
spearman秩相关检验：
x <- c(24, 17, 20, 41, 52, 23, 46, 18, 15, 29)
y <- c(8, 1, 4, 7, 9, 5, 10, 3, 2, 6)
cor.test(x, y, method = "spearman", exact = F)##
## Spearman's rank correlation rho
##
## data: x and y
## S = 10, p-value = 5.484e-05
## alternative hypothesis: true rho is not equal to 0
## sample estimates:
## rho
## 0.9393939p-value = 5.484e-05<0.05，拒绝原假设，认为相关的，且为正相关。
kendall秩相关检验：
cor.test(x, y, method = "kendall", exact = F)##
## Kendall's rank correlation tau
##
## data: x and y
## z = 3.3094, p-value = 0.000935
## alternative hypothesis: true tau is not equal to 0
## sample estimates:
## tau
## 0.8222222p-value = 0.000935<0.05，拒绝原假设，认为相关的，且为正相关。
———
使用Wilcoxon检验
x <- rep(1:5, c(0, 1, 9, 7, 3))
y <- rep(1:5, c(2, 2, 11, 4, 1))
wilcox.test(x, y, exact = F)##
## Wilcoxon rank sum test with continuity correction
##
## data: x and y
## W = 266, p-value = 0.05509
## alternative hypothesis: true location shift is not equal to 0