Question
Show algebraically that Equation 7.23 is equal to Equation 7.24. That is,$$\frac{N \sigma^2}{(N-1) \sigma_{\bar{X}}^2+\sigma^2}=\frac{n_0 N}{n_0+(N-1)}$$
Step 1
Step 1: Start with the left-hand side of the equation: \[ \frac{N \sigma^2}{(N-1) \sigma_{\bar{X}}^2+\sigma^2} \] Show more…
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