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Paul
Posts:
208
Registered:
2/23/10
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Why multiply t by sample std dev?
Posted:
Sep 30, 2012 3:55 PM
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A stats book I'm using describes confidence intervals for estimating a
population mean assuming: (i) normally distributed population, (ii)
small sample size, and (iii) unknown population standard deviation.
They use the t-distribution. Depending on whether they are doing a 2-
tail or 1-tail test, they find the proper spot under the t-
distribution, take the t-value and multiply by the sample standard
deviation to get the offset from the sample mean that defines the
confidence interval.
I am confused by the part about multiplying the t-value by the sample
standard deviation. If it was the *normal* distribution, we multiply
the z-value by the sample standard deviation because the normal
distribution has a standard deviation of 1. So multiplying by the
sample standard deviation is simply rescaling the horizontal axes.
However, the standard deviation for the t-distribution is sqrt[df/
(df-2)]. To do similar rescaling, shouldn't the t-value be divided by
sqrt[df/(df-2)] to get it in terms of standard deviations in t (after
all, the z-value is basically in terms of the standard deviation under
the normal distribution), *then* multipled by the sample standard
deviation to get it in terms of the units of measure for the random
variable?
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