network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 7 months |
| seen | Oct 29 '12 at 17:58 | |
| stats | profile views | 1 |
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Oct 29 |
asked | Statistics: normal distribution, finding truncated mean. |
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Oct 3 |
awarded | Scholar |
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Oct 3 |
accepted | How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$? |
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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 $k'$? |
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Oct 2 |
awarded | Editor |
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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 |
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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 $Z_\alpha = 1.645$ |
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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? |
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Oct 2 |
asked | How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$? |
