μ 1 = E ( X ) = μ = 1 n ∑ i = 1 n x i = 1 8 ( 74.001 + 74.005 + 74.003 + 74.001 + 74.000 + 73.998 + 74.006 + 74.002 ) = 74.002 \mu_1=E(X)=\mu\\=\frac{1}{n}\sum_{i=1}^nx_i\\=\frac{1}{8}(74.001+74.005+74.003+74.001+74.000+73.998+74.006+74.002)=74.002 μ1=E(X)=μ=n1∑i=1nxi=81(74.001+74.005+74.003+74.001+74.000+73.998+74.006+74.002)=74.002
σ 2 = 1 n ∑ i = 1 n ( x i − μ ) 2 = 1 8 [ ( μ − 74.001 ) 2 + ( μ − 74.005 ) 2 + ( μ − 74.003 ) 2 + ( μ − 74.001 ) 2 + ( μ − 74.000 ) 2 + ( μ − 73.998 ) 2 + ( μ − 74.006 ) 2 + ( μ − 74.002 ) 2 ] = 6 ∗ 1 0 − 6 \sigma^2=\frac{1}{n}\sum_{i=1}^n(x_i-\mu)^2\\=\frac{1}{8}[(\mu-74.001)^2+(\mu-74.005)^2+(\mu-74.003)^2+(\mu-74.001)^2+(\mu-74.000)^2+(\mu-73.998)^2+(\mu-74.006)^2+(\mu-74.002)^2]\\=6*10^{-6} σ2=n1∑i=1n(xi−μ)2=81[(μ−74.001)2+(μ−74.005)2+(μ−74.003)2+(μ−74.001)2+(μ−74.000)2+(μ−73.998)2+(μ−74.006)2+(μ−74.002)2]=6∗10−6
样本方差和总体方差的关系得到
S 2 = n n − 1 σ 2 = 6.86 ∗ 1 0 − 6 S^2=\frac{n}{n-1}\sigma^2=6.86*10^{-6} S2=n−1nσ2=6.86∗10−6
x<-c(74.001,74.005,74.003,74.001,74.000,73.998,74.006,74.002)
mean(x) # \mu
sum((x-mean(x))^2)/length(x) # \sigma^2
var(x) # S^2
由题意可得 X ∼ B ( 10 , p ) P ( X = k ) = C 10 k p k ( 1 − p ) 10 − k X\sim B(10,p)\ \ P(X=k)=C_{10}^kp^k(1-p)^{10-k} X∼B(10,p) P(X=k)=C10kpk(1−p)10−k
似然函数为 L ( p ) = ∏ i = 1 100 C 10 k p k ( 1 − p ) 1 − k L(p)=\prod_{i=1}^{100}C_{10}^kp^k(1-p)^{1-k} L(p)=∏i=1100C10kpk(1−p)1−k
l n L ( p ) = ( ∑ i = 1 100 x i ) l n p + ( 1000 − ∑ i = 1 100 x + i ) l n ( 1 − p ) + ∑ i = 1 100 l n C 10 x i lnL(p)=(\sum_{i=1}^{100}x_i)lnp+(1000-\sum_{i=1}^{100}x+i)ln(1-p)+\sum_{i=1}^{100}lnC_{10}^{x_i} lnL(p)=(∑i=1100xi)lnp+(1000−∑i=1100x+i)ln(1−p)+∑i=1100lnC10xi
d l n L d p = ∑ i = 1 100 x p − 1 1 − p ( 1000 − ∑ i = 1 100 x i ) = 0 \frac{dlnL}{dp}=\frac{\sum_{i=1}^{100}x}{p}-\frac{1}{1-p}(1000-\sum_{i=1}^{100}x_i)=0 dpdlnL=p∑i=1100x−1−p1(1000−∑i=1100xi)=0
解得: P ^ = ∑ i = 1 n x i n 2 = X ˉ n = 499 1000 = 0.499 \hat P=\frac{\sum_{i=1}^nx_i}{n^2}=\frac{\bar X}{n}=\frac{499}{1000}=0.499 P^=n2∑i=1nxi=nXˉ=1000499=0.499
由此可得
当总体 X X X服从二项分布 X ∼ B ( n , p ) X\sim B(n,p) X∼B(n,p)时,参数p的极大似然估计为 P ^ = X ˉ n \hat P=\frac{\bar X}{n} P^=nXˉ
sum(0*0,1*1,6*2,3*7,4*23,5*26,6*21,7*12,8*3,9*1,10*0)/1000
(1)
根据题意 σ \sigma σ已知,求 μ \mu μ的置信水平为0.95的置信区间
P { − Z α 2 < X ˉ − μ σ / n < Z α 2 } = 1 − α P\{-Z_{\frac{\alpha}{2}}<\frac{\bar X-\mu}{\sigma/\sqrt{n}}
解得: ( X ˉ − σ n Z α 2 , X ˉ + σ n Z α 2 ) (\bar X - \frac{\sigma}{\sqrt{n}}Z_{\frac{\alpha}{2}},\bar X+ \frac{\sigma}{\sqrt{n}}Z_{\frac{\alpha}{2}}) (Xˉ−nσZ2α,Xˉ+nσZ2α)将 σ = 0.6 , 1 − α = 0.95 , Z α 2 = 1.96 , X ˉ = 6 \sigma = 0.6,1-\alpha =0.95,Z_{\frac{\alpha}{2}}=1.96,\bar X=6 σ=0.6,1−α=0.95,Z2α=1.96,Xˉ=6代入得置信区间为 ( 5.608 , 6.392 ) (5.608,6.392) (5.608,6.392)
x<-c(6.0,5.7,5.8,6.5,7.0,6.3,5.6,6.1,5.0)
m<-mean(x) # \bar X
sigma<-0.6 # \sigma
n<-length(x)
alpha<-0.05
m+sigma/sqrt(n)*qnorm(1-alpha/2)
m-sigma/sqrt(n)*qnorm(1-alpha/2)
σ \sigma σ未知,求 μ \mu μ的置信水平为0.95的置信区间
P { − t α 2 ( n − 1 ) < X ˉ − μ S / n < t α 2 ( n − 1 ) } = 1 − α P\{-t_{\frac{\alpha}{2}}(n-1)<\frac{\bar X-\mu}{S/\sqrt{n}}
( X ˉ − S n t α 2 ( n − 1 ) , X ˉ + S n t α 2 ( n − 1 ) ) (\bar X - \frac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1),\bar X+ \frac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1)) (Xˉ−nSt2α(n−1),Xˉ+nSt2α(n−1))
代入数据得置信区间为(5.558,6.442)
x<-c(6.0,5.7,5.8,6.5,7.0,6.3,5.6,6.1,5.0)
m<-mean(x) # \bar X
S<-sd(x) # S
n<-length(x)
alpha<-0.05
m+S/sqrt(n)*qt(1-alpha/2,n-1)
m-S/sqrt(n)*qt(1-alpha/2,n-1)
μ \mu μ未知,求 σ \sigma σ的置信水平为0.95的置信区间
P { χ 1 − α 2 2 ( n − 1 ) < ( n − 1 ) S 2 σ 2 < χ α 2 2 ( n − 1 ) } = 1 − α P\{ \chi^2_{1-\frac{\alpha}{2}}(n-1)<\frac{(n-1)S^2}{\sigma^2}< \chi^2_{\frac{\alpha}{2}}(n-1)\}=1-\alpha P{χ1−2α2(n−1)<σ2(n−1)S2<χ2α2(n−1)}=1−α
解得: ( ( n − 1 ) S 2 χ α 2 2 ( n − 1 ) , ( n − 1 ) S 2 χ 1 − α 2 2 ( n − 1 ) ) (\frac{(n-1)S^2}{ \chi^2_{\frac{\alpha}{2}}(n-1)},\frac{(n-1)S^2}{ \chi^2_{1-\frac{\alpha}{2}}(n-1)}) (χ2α2(n−1)(n−1)S2,χ1−2α2(n−1)(n−1)S2)
代入数据得: ( 7.4 , 21.1 ) (7.4,21.1) (7.4,21.1)
s<-11
alpha<-0.05
n<-9
df<-n-1
sqrt(df*s^2/qchisq(1-alpha/2,df))
sqrt(df*s^2/qchisq(alpha/2,df))
μ 1 , μ 2 , σ \mu_1,\mu_2,\sigma μ1,μ2,σ均未知,求 μ 1 − μ 2 \mu_1-\mu_2 μ1−μ2置信度为0.95的置信区间
直接套公式,置信区间为 ( X ˉ − Y ˉ − t α 2 ( n 1 + n 2 − 2 ) S w 1 n 1 + 1 n 2 , X ˉ − Y ˉ + t α 2 ( n 1 + n 2 − 2 ) S w 1 n 1 + 1 n 2 ) (\bar X-\bar Y-t_{\frac{\alpha}{2}}(n_1+n_2-2)S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}},\bar X-\bar Y+t_{\frac{\alpha}{2}}(n_1+n_2-2)S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}) (Xˉ−Yˉ−t2α(n1+n2−2)Swn11+n21,Xˉ−Yˉ+t2α(n1+n2−2)Swn11+n21)
代入数据得: ( − 0.002 , 0.006 ) (-0.002,0.006) (−0.002,0.006)
x1<-c(0.143,0.142,0.143,0.137)
x2<-c(0.140,0.142,0.136,0.138,0.140)
m1<-mean(x1)
m2<-mean(x2)
alpha=0.05
n1<-length(x1)
n2<-length(x2)
sw<-sqrt(((n1-1)*var(x1)+(n2-1)*var(x2))/(n1+n2-2))
m1-m2-qt(1-alpha/2,n1+n2-2)*sw*(sqrt(1/n1+1/n2))
m1-m2+qt(1-alpha/2,n1+n2-2)*sw*(sqrt(1/n1+1/n2))
μ 1 , μ 2 \mu_1,\mu_2 μ1,μ2均未知,求 σ A 2 σ B 2 \frac{\sigma_A^2}{\sigma_B^2} σB2σA2置信度为0.95的置信区间
置信区间为
( S 1 2 S 2 2 1 F α 2 ( n 1 − 1 , n 2 − 1 ) , S 1 2 S 2 2 1 F 1 − α 2 ( n 1 − 1 , n 2 − 1 ) ) (\frac{S_1^2}{S_2^2}\frac{1}{F_{\frac{\alpha}{2}}(n_1-1,n_2-1)},\frac{S_1^2}{S_2^2}\frac{1}{F_{1-{\frac{\alpha}{2}}}(n_1-1,n_2-1)}) (S22S12F2α(n1−1,n2−1)1,S22S12F1−2α(n1−1,n2−1)1)
代入数据解得: ( 0.222 , 3.601 ) (0.222,3.601) (0.222,3.601)
sa=0.5419
sb=0.6065
alpha=0.05
n=10
del=sa/sb
del/qf(1-alpha/2,n-1,n-1)
del/qf(alpha/2,n-1,n-1)
(1)
依题意求 σ \sigma σ未知 μ \mu μ的单侧置信上限
μ ˉ = X ˉ + S n t α ( n − 1 ) \bar \mu=\bar X+\frac{S}{\sqrt{n}t_\alpha(n-1)} μˉ=Xˉ+ntα(n−1)S
代入数据解得: ( − ∞ , 6.108 ) (-\infty,6.108) (−∞,6.108)
x<-c(6.0,5.7,5.8,6.5,7.0,6.3,5.6,6.1,5.0)
m<-mean(x) # \bar X
sigma<-0.6 # \sigma
n<-length(x)
alpha<-0.05
m+sigma/sqrt(n)/qt(1-alpha,n-1)
(2)
依题意求 σ \sigma σ未知 μ 1 − μ 2 \mu_1-\mu_2 μ1−μ2的单侧置信下限
μ 1 − μ 2 ‾ = X ˉ − Y ˉ − t α ( n 1 + n 2 − 2 ) S w 1 n 1 + 1 n 2 \underline{\mu_1-\mu_2}=\bar X-\bar Y-t_\alpha(n_1+n_2-2)S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} μ1−μ2=Xˉ−Yˉ−tα(n1+n2−2)Swn11+n21
代入数据解得: ( − 0.001 , + ∞ ) (-0.001,+\infty) (−0.001,+∞)
x1<-c(0.143,0.142,0.143,0.137)
x2<-c(0.140,0.142,0.136,0.138,0.140)
m1<-mean(x1)
m2<-mean(x2)
alpha=0.05
n1<-length(x1)
n2<-length(x2)
sw<-sqrt(((n1-1)*var(x1)+(n2-1)*var(x2))/(n1+n2-2))
m1-m2-qt(1-alpha,n1+n2-2)*sw*(sqrt(1/n1+1/n2))
依题意求 μ 1 , μ 2 \mu_1,\mu2 μ1,μ2未知 σ 1 2 σ 2 2 \frac{\sigma_1^2}{\sigma_2^2} σ22σ12的单侧置信上限
σ 1 2 σ 2 2 ‾ = S 1 2 S 2 2 1 F 1 − α ( n 1 − 1 , n 2 − 1 ) \overline {\frac{\sigma_1^2}{\sigma_2^2}}=\frac{S_1^2}{S_2^2}\frac{1}{F_{1-{\alpha}}(n_1-1,n_2-1)} σ22σ12=S22S12F1−α(n1−1,n2−1)1
代入数据解得: ( − ∞ , 0.281 ) (-\infty,0.281) (−∞,0.281)
sa=0.5419
sb=0.6065
alpha=0.05
n=10
del=sa/sb
del/qf(1-alpha,n-1,n-1)