n*a 发帖数: 12 | 1 for a given ANOVA study involving three treatments, with n1=n2=n3=9, it was
found the MSE=18. the appropriate estimeated standard deviation of the
sampling distributionof x1bar-x2bare to construct a confidence interval for
u1-u2 is 4
判断题,正确请给出理由
错误也请给出理由 |
c******a 发帖数: 6951 | 2 sqrt(18*2/9)=sqrt(4)=2
was
for
【在 n*a 的大作中提到】![](/moin_static193/solenoid/img/up.png) : for a given ANOVA study involving three treatments, with n1=n2=n3=9, it was : found the MSE=18. the appropriate estimeated standard deviation of the : sampling distributionof x1bar-x2bare to construct a confidence interval for : u1-u2 is 4 : 判断题,正确请给出理由 : 错误也请给出理由
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n*a 发帖数: 12 | 3
能不能在给解释的详细点啊
【在 c******a 的大作中提到】![](/moin_static193/solenoid/img/up.png) : sqrt(18*2/9)=sqrt(4)=2 : : was : for
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n*a 发帖数: 12 | |
n*a 发帖数: 12 | |
n*a 发帖数: 12 | |
n*a 发帖数: 12 | 7
我知道error的算法是squrt(18/9) 可是我不知道你上面乘以个2是什么意思.
【在 c******a 的大作中提到】![](/moin_static193/solenoid/img/up.png) : sqrt(18*2/9)=sqrt(4)=2 : : was : for
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d*********a 发帖数: 255 | 8 yes . there is a formula:
\sqrt{MSE*(1/ni+1/nj)}=\sqrt{18*(1/9+1/9)}=\sqrt{4}=2
【在 n*a 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 是有什么公式吗
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n*a 发帖数: 12 | 9 今天考完统计了 相当凄惨
唉 去复习第二科巴,下学期不能这么混了 |