統計檢驗是將抽樣結果和抽樣分布相對照而作出判斷的工作。主要分5個步驟:
- 建立假設
- 求抽樣分布
- 選擇顯著性水平和否定域
- 計算檢驗統計量
- 判定 —— 百度百科
假設檢驗(hypothesis test)亦稱顯著性檢驗(significant test),是統計推斷的另一重要內容,其目的是比較總體參數之間有無差別。假設檢驗的實質是判斷觀察到的“差別”是由抽樣誤差引起還是總體上的不同,目的是評價兩種不同處理引起效應不同的證據有多強,這種證據的強度用概率P來度量和表示。除t分布外,針對不同的資料還有其他各種檢驗統計量及分布,如F分布、X2分布等,應用這些分布對不同類型的數據進行假設檢驗的步驟相同,其差別僅僅是需要計算的檢驗統計量不同。
正態總體均值的假設檢驗
t檢驗
t.test() => Student's t-Test
require(graphics)
t.test(1:10, y = c(7:20)) # P = .00001855
t.test(1:10, y = c(7:20, 200)) # P = .1245 -- 不在顯著
## 經典案例: 學生犯困數據
plot(extra ~ group, data = sleep)

## 傳統表達式
with(sleep, t.test(extra[group == 1], extra[group == 2]))
Welch Two Sample t-test
data: extra[group == 1] and extra[group == 2]
t = -1.8608, df = 17.776, p-value = 0.07939
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-3.3654832 0.2054832
sample estimates:
mean of x mean of y
0.75 2.33
## 公式形式
t.test(extra ~ group, data = sleep)
Welch Two Sample t-test
data: extra by group
t = -1.8608, df = 17.776, p-value = 0.07939
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-3.3654832 0.2054832
sample estimates:
mean in group 1 mean in group 2
0.75 2.33
單個總體
- 某種元件的壽命X(小時)服從正態分布N(mu,sigma^2),其中mu、sigma^2均未知,16只元件的壽命如下;問是否有理由認為元件的平均壽命大於255小時。
X<-c(159, 280, 101, 212, 224, 379, 179, 264,
222, 362, 168, 250, 149, 260, 485, 170)
t.test(X, alternative = "greater", mu = 225)
One Sample t-test
data: X
t = 0.66852, df = 15, p-value = 0.257
alternative hypothesis: true mean is greater than 225
95 percent confidence interval:
198.2321 Inf
sample estimates:
mean of x
241.5
兩個總體
- X為舊煉鋼爐出爐率,Y為新煉鋼爐出爐率,問新的操作能否提高出爐率?
X<-c(78.1,72.4,76.2,74.3,77.4,78.4,76.0,75.5,76.7,77.3)
Y<-c(79.1,81.0,77.3,79.1,80.0,79.1,79.1,77.3,80.2,82.1)
t.test(X, Y, var.equal=TRUE, alternative = "less")
Two Sample t-test
data: X and Y
t = -4.2957, df = 18, p-value = 0.0002176
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
-Inf -1.908255
sample estimates:
mean of x mean of y
76.23 79.43
成對數據t檢驗
- 對每個高爐進行配對t檢驗
X<-c(78.1,72.4,76.2,74.3,77.4,78.4,76.0,75.5,76.7,77.3)
Y<-c(79.1,81.0,77.3,79.1,80.0,79.1,79.1,77.3,80.2,82.1)
t.test(X-Y, alternative = "less")
One Sample t-test
data: X - Y
t = -4.2018, df = 9, p-value = 0.00115
alternative hypothesis: true mean is less than 0
95 percent confidence interval:
-Inf -1.803943
sample estimates:
mean of x
-3.2
正態總體方差的假設檢驗
var.test() => F Test to Compare Two Variances
x <- rnorm(50, mean = 0, sd = 2)
y <- rnorm(30, mean = 1, sd = 1)
var.test(x, y) # x和y的方差是否相同?
var.test(lm(x ~ 1), lm(y ~ 1)) # 相同.
- 從小學5年級男生中抽取20名,測量其身高(厘米)如下;問:在0.05顯著性水平下,平均值是否等於149,sigma^2是否等於75?
X<-scan()
136 144 143 157 137 159 135 158 147 165
158 142 159 150 156 152 140 149 148 155
var.test(X,Y)
F test to compare two variances
data: X and Y
F = 34.945, num df = 19, denom df = 9, p-value = 6.721e-06
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
9.487287 100.643093
sample estimates:
ratio of variances
34.94489
- 對煉鋼爐的數據進行分析
X<-c(78.1,72.4,76.2,74.3,77.4,78.4,76.0,75.5,76.7,77.3)
Y<-c(79.1,81.0,77.3,79.1,80.0,79.1,79.1,77.3,80.2,82.1)
var.test(X,Y)
F test to compare two variances
data: X and Y
F = 1.4945, num df = 9, denom df = 9, p-value = 0.559
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.3712079 6.0167710
sample estimates:
ratio of variances
1.494481
二項分布的總體檢驗
- 有一批蔬菜種子的平均發芽率為P=0.85,現在隨機抽取500粒,用種衣劑進行浸種處理,結果有445粒發芽,問種衣劑有無效果。
binom.test(445,500,p=0.85)
Exact binomial test
data: 445 and 500
number of successes = 445, number of trials = 500, p-value = 0.01207
alternative hypothesis: true probability of success is not equal to 0.85
95 percent confidence interval:
0.8592342 0.9160509
sample estimates:
probability of success
0.89
- 按照以往經驗,新生兒染色體異常率一般為1%,某醫院觀察了當地400名新生兒,有一例染色體異常,問該地區新生兒染色體是否低於一般水平?
binom.test(1,400,p=0.01,alternative="less")
Exact binomial test
data: 1 and 400
number of successes = 1, number of trials = 400, p-value = 0.09048
alternative hypothesis: true probability of success is less than 0.01
95 percent confidence interval:
0.0000000 0.0118043
sample estimates:
probability of success
0.0025
非參數檢驗
數據是否正態分布的Neyman-Pearson 擬合優度檢驗-chisq
- 5種品牌啤酒愛好者的人數如下
A 210
B 312
C 170
D 85
E 223
問不同品牌啤酒愛好者人數之間有沒有差異?
X<-c(210, 312, 170, 85, 223)
chisq.test(X)
Chi-squared test for given probabilities
data: X
X-squared = 136.49, df = 4, p-value < 2.2e-16
- 檢驗學生成績是否符合正態分布
X<-scan()
25 45 50 54 55 61 64 68 72 75 75
78 79 81 83 84 84 84 85 86 86 86
87 89 89 89 90 91 91 92 100
A<-table(cut(X, br=c(0,69,79,89,100)))
#cut 將變量區域划分為若干區間
#table 計算因子合並后的個數
p<-pnorm(c(70,80,90,100), mean(X), sd(X))
p<-c(p[1], p[2]-p[1], p[3]-p[2], 1-p[3])
chisq.test(A,p=p)
Chi-squared test for given probabilities
data: A
X-squared = 8.334, df = 3, p-value = 0.03959
#均值之間有無顯著區別
大麥的雜交后代芒性狀的比例 無芒:長芒: 短芒=9:3:4,而實際觀測值為335:125:160 ,檢驗觀測值是否符合理論假設?
chisq.test(c(335, 125, 160), p=c(9,3,4)/16)
Chi-squared test for given probabilities
data: c(335, 125, 160)
X-squared = 1.362, df = 2, p-value = 0.5061
- 現有42個數據,分別表示某一時間段內電話總機借到呼叫的次數,
接到呼叫的次數 0 1 2 3 4 5 6
出現的頻率 7 10 12 8 3 2 0
問:某個時間段內接到的呼叫次數是否符合Possion分布?
x<-0:6
y<-c(7,10,12,8,3,2,0)
mean<-mean(rep(x,y))
q<-ppois(x,mean)
n<-length(y)
p[1]<-q[1]
p[n]<-1-q[n-1]
for(i in 2:(n-1))
p[i]<-1-q[i-1]
chisq.test(y, p= rep(1/length(y), length(y)) )
Chi-squared test for given probabilities
data: y
X-squared = 19.667, df = 6, p-value = 0.003174
Z<-c(7, 10, 12, 8)
n<-length(Z); p<-p[1:n-1]; p[n]<-1-q[n-1]
chisq.test(Z, p= rep(1/length(Z), length(Z)))
Chi-squared test for given probabilities
data: Z
X-squared = 1.5946, df = 3, p-value = 0.6606
P值越小越有理由拒絕無效假設,認為總體之間有差別的統計學證據越充分。需要注意:不拒絕H0不等於支持H0成立,僅表示現有樣本信息不足以拒絕H0。
傳統上,通常將P>0.05稱為“不顯著”,0.0l<P≤0.05稱為“顯著”,P≤0.0l稱為“非常顯著”。
反饋與建議
- 作者:ShangFR
- 郵箱:shangfr@foxmail.com
