容量分别是n1,n2.组成一个联合样本n1+n2,组合样本的方差是?从总体中抽取两组样本,其容量分别为n1与n2,设两组的样本均值分别为 X1与X2,样本方差分别为S1^2及S2^2,把这两组样本合并为一组容
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![容量分别是n1,n2.组成一个联合样本n1+n2,组合样本的方差是?从总体中抽取两组样本,其容量分别为n1与n2,设两组的样本均值分别为 X1与X2,样本方差分别为S1^2及S2^2,把这两组样本合并为一组容](/uploads/image/z/9490375-55-5.jpg?t=%E5%AE%B9%E9%87%8F%E5%88%86%E5%88%AB%E6%98%AFn1%2Cn2.%E7%BB%84%E6%88%90%E4%B8%80%E4%B8%AA%E8%81%94%E5%90%88%E6%A0%B7%E6%9C%ACn1%2Bn2%2C%E7%BB%84%E5%90%88%E6%A0%B7%E6%9C%AC%E7%9A%84%E6%96%B9%E5%B7%AE%E6%98%AF%3F%E4%BB%8E%E6%80%BB%E4%BD%93%E4%B8%AD%E6%8A%BD%E5%8F%96%E4%B8%A4%E7%BB%84%E6%A0%B7%E6%9C%AC%EF%BC%8C%E5%85%B6%E5%AE%B9%E9%87%8F%E5%88%86%E5%88%AB%E4%B8%BAn1%E4%B8%8En2%EF%BC%8C%E8%AE%BE%E4%B8%A4%E7%BB%84%E7%9A%84%E6%A0%B7%E6%9C%AC%E5%9D%87%E5%80%BC%E5%88%86%E5%88%AB%E4%B8%BA+X1%E4%B8%8EX2%2C%E6%A0%B7%E6%9C%AC%E6%96%B9%E5%B7%AE%E5%88%86%E5%88%AB%E4%B8%BAS1%5E2%E5%8F%8AS2%5E2%2C%E6%8A%8A%E8%BF%99%E4%B8%A4%E7%BB%84%E6%A0%B7%E6%9C%AC%E5%90%88%E5%B9%B6%E4%B8%BA%E4%B8%80%E7%BB%84%E5%AE%B9)
容量分别是n1,n2.组成一个联合样本n1+n2,组合样本的方差是?从总体中抽取两组样本,其容量分别为n1与n2,设两组的样本均值分别为 X1与X2,样本方差分别为S1^2及S2^2,把这两组样本合并为一组容
容量分别是n1,n2.组成一个联合样本n1+n2,组合样本的方差是?
从总体中抽取两组样本,其容量分别为n1与n2,设两组的样本均值分别为 X1与X2,样本方差分别为S1^2及S2^2,把这两组样本合并为一组容量为n1+n2的联合样本,
容量分别是n1,n2.组成一个联合样本n1+n2,组合样本的方差是?从总体中抽取两组样本,其容量分别为n1与n2,设两组的样本均值分别为 X1与X2,样本方差分别为S1^2及S2^2,把这两组样本合并为一组容
组合样本的方差是 S1^2 + S2^2
This problem is not as simple as the answer suggests here. When combining m random variables: n1, n2, ..., nm each with an average value of X1, X2, ..., Xm and a standard deviation S1, S2, ..., Sm, the average of the new random variable N = n1 + n2 + ... + nm will have an average value of X = X1 + X2 + ... + Xm, this is expected. The standard deviation of the new random variable, though, is S = (S1^2 + S2^2 + ... + Sm^2)^(1/2). So the variance of the new random variable or Var(N) = S1^2 + S2^2 + ... + Sm^2