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 Oct3 awarded Scholar Oct3 accepted How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$? Oct2 comment How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$? Before I do, have you checked to see if you do indeed get the required value for $k'$? Oct2 awarded Editor Oct2 revised How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$? added 2 characters in body; edited tags; edited title Oct2 comment How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$? @KennyTM $Z_\alpha = 1.645$ Oct2 comment How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$? @KennyTM My dad was reading from one of his books and he asked me to find how to get to that value of $k'$ from the equation in there. This is a little ahead of what I'm used to figuring out and I've not tried anything useful so I decided to ask here instead. The value for $k'$ is correct because the author write "But we know $k'=...$" after he writes down the exact equation above. Could it be that it's not possible to extract that value of $k'$ from the integral itself hence the word 'but' and it's taken from whatever was going on before? Oct2 asked How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$?