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seen Oct 29 '12 at 17:58

Oct
3
awarded  Scholar
Oct
3
accepted How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$?
Oct
2
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 $k'$?
Oct
2
awarded  Editor
Oct
2
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
Oct
2
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 $Z_\alpha = 1.645$
Oct
2
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?
Oct
2
asked How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$?