4、
临界值法
在显著水平α=0.05\alpha=0.05α=0.05下检验假设 H0:μ≤10,H1:μ>10H_0:\mu\le 10,H_1:\mu>10H0:μ≤10,H1:μ>10
n=20,Xˉ=10.2,S=0.5099n=20,\bar X=10.2,S=0.5099n=20,Xˉ=10.2,S=0.5099
t0.05(n−1)=t0.05(19)=1.7291t_{0.05}(n-1)=t_{0.05}(19)=1.7291t0.05(n−1)=t0.05(19)=1.7291
拒绝域为t=Xˉ−μ0S/n≥tα(n−1)=1.7291t=\frac{\bar X-\mu_0}{S/\sqrt{n}}\geq t_{\alpha}(n-1)=1.7291t=S/nXˉ−μ0≥tα(n−1)=1.7291
ttt的观察值为10.2−100.5099/2=1.754>1.7291\frac{10.2-10}{0.5099/\sqrt{2}}=1.754>1.72910.5099/210.2−10=1.754>1.7291
即：落在拒绝域内，在α=0.05\alpha=0.05α=0.05下拒绝H0H_0H0
P值法
P{t≥t0}=P{t≥1.754}=0.04776<0.05P\{t\ge t_0\}=P\{t\ge 1.754\}=0.04776<0.05P{t≥t0}=P{t≥1.754}=0.04776<0.05
x=c(9.8,10.4,10.6,9.6,9.7,9.9,10.9,11.1,9.6,10.2,10.3,9.6,9.9,11.2,10.6,9.8,10.5,10.1,10.5,9.7)m=mean(x)# 临界值法alpha=0.05(m-10)/sd(x)*sqrt(20) > qt(alpha,19,lower.tail = FALSE)# p值法1-pt((m-10)/sd(x)*sqrt(20),19)
7、
临界值法
检验假设 H0:μ1−μ2≤2,H1:μ1−μ2>2H_0:\mu_1-\mu_2\le 2,H_1:\mu_1-\mu_2>2H0:μ1−μ2≤2,H1:μ1−μ2>2
X∼N(μ1,σ2),Y∼N(μ2,σ2)X\sim N(\mu_1,\sigma^2),Y\sim N(\mu_2,\sigma^2)X∼N(μ1,σ2),Y∼N(μ2,σ2)
Xˉ=5.25,Yˉ=1.5,S12=0.9318,S22=1,SW=0.9828\bar X=5.25,\bar Y=1.5,S_1^2=0.9318,S_2^2=1,S_W=0.9828Xˉ=5.25,Yˉ=1.5,S12=0.9318,S22=1,SW=0.9828
ttt检验,α=0.05\alpha=0.05α=0.05
拒绝域为W={Xˉ−YˉSw1n1+1n2>tα(n1+n2−2}W=\{\frac{\bar X-\bar Y}{S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}>t_\alpha(n_1+n_2-2\}W={Swn11+n21Xˉ−Yˉ>tα(n1+n2−2}
t0.05(22)=1.7171t_0.05(22)=1.7171t0.05(22)=1.7171
样本值为5.25−1.5−20.99828112+112=4.3616>1.7171\frac{5.25-1.5-2}{0.99828\sqrt{\frac{1}{12}+\frac{1}{12}}}=4.3616>1.71710.99828121+1215.25−1.5−2=4.3616>1.7171
拒绝原假设
p值法
P{t≥t0}=P{t≥4.3616}=0.000125<0.05P\{t\ge t_0\}=P\{t\ge 4.3616\}=0.000125<0.05P{t≥t0}=P{t≥4.3616}=0.000125<0.05
拒绝原假设
x=c(6,4,5,5,6,5,5,6,4,6,7,4)y=c(2,1,2,2,1,0,3,2,1,0,1,3)# 临界值法alpha=0.05(mean(x)-mean(y)-2)/0.9828/sqrt(1/12+1/12) > qt(alpha,19,lower.tail = FALSE)# p值法1-pt(4.3616,19)
17、
（1）
H0:σ12=σ22,H1:σ12≠σ22H_0:\sigma_1^2=\sigma_2^2,H_1:\sigma_1^2\neq\sigma_2^2H0:σ12=σ22,H1:σ12=σ22
S12S22=1.09\frac{S_1^2}{S_2^2}=1.09S22S12=1.09
F0.025(9,9)=4.03,F0.975(9,9)=0.248F_{0.025}(9,9)=4.03,F_{0.975}(9,9)=0.248F0.025(9,9)=4.03,F0.975(9,9)=0.248
0.248<1.9<4.030.248<1.9<4.030.248<1.9<4.03 接受原假设
拒绝域为S12S22≥F0.025(99)∣∣S12S22≤F0.975(9.9)\frac{S_1^2}{S_2^2}\ge F_{0.025}(99) || \frac{S_1^2}{S_2^2}\le F_{0.975}(9.9)S22S12≥F0.025(99)∣∣S22S12≤F0.975(9.9)
PPP值为P0=0.55>0.05P_0=0.55>0.05P0=0.55>0.05
接受原假设
x=c(101,100,99,99,98,100,98,99,99,99)y=c(100,98,100,99,98,99,98,98,99,100)# 临界值法v1=var(x)/var(y)v1 >= qf(0.025,9,9) && v1 = qf(1-0.025,9,9)# p值法pf(v1,9,9)
（2）
σ12=σ22=σ2σ\sigma_1^2=\sigma_2^2=\sigma^2\quad\sigmaσ12=σ22=σ2σ未知
Xˉ=99.2,Yˉ=98.9,S12=0.805,S22=0.77,Sw2=0.805,∣t∣=0.748\bar X=99.2,\bar Y=98.9, S_1^2=0.805,S_2^2=0.77,S_w^2=0.805,|t|=0.748Xˉ=99.2,Yˉ=98.9,S12=0.805,S22=0.77,Sw2=0.805,∣t∣=0.748
拒绝域为∣t∣≥tα2(n1+n2−2)=t0.025(18)=2.1009>0.748|t|\ge t_{\frac{\alpha}{2}(n_1+n_2-2)}=t_{0.025}(18)=2.1009>0.748∣t∣≥t2α(n1+n2−2)=t0.025(18)=2.1009>0.748
接受原假设
P0=P{∣t∣≥t0}=0.4645<0.05P_0=P\{|t|\ge t_0\}=0.4645<0.05P0=P{∣t∣≥t0}=0.4645<0.05
接受原假设
x=c(101,100,99,99,98,100,98,99,99,99)y=c(100,98,100,99,98,99,98,98,99,100)# 临界值法alpha=0.05(mean(x)-mean(y))/((9*var(x)+9*var(y))/18)/sqrt(1/9+1/9) > qt(alpha/2,18,lower.tail = FALSE)# p值法1-pt(0.748,18)
