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Topic: Why multiply t by sample std dev?
Replies: 6   Last Post: Oct 2, 2012 6:42 PM

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Bruce Weaver

Posts: 751
Registered: 12/18/04
Re: Why multiply t by sample std dev?
Posted: Oct 2, 2012 6:42 PM
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On 02/10/2012 2:02 PM, Paul wrote:
> On Oct 2, 9:39 am, Bruce Weaver <bwea...@lakeheadu.ca> wrote:
>> On 01/10/2012 9:54 PM, Rich Ulrich wrote:

>>> On Mon, 1 Oct 2012 12:45:45 -0700 (PDT), Paul
>>> <paul.domas...@gmail.com> wrote:

>>>> On Sun, 30 Sep 2012 12:55:11 -0700 (PDT), Paul
>>>> <paul.domas...@gmail.com> wrote:

>>> ... snip a bunch
>>>> Thanks, Dave, Rich,
>>>> I think I pin-pointed the source of my confusion. I didn't really
>>>> know what t was. It was just a curve and a table that we blindly used
>>>> according to a recipe. Well, Wikipedia

>>> (http://en.wikipedia.org/wiki/Student%27s_t-distribution) defined the
>>> t-value in a manner similar to

>>>> Rich:
>>>> [ (sample mean) - (population mean) ] / (sample standard deviation)
>>> No!
>>> It does not say that. It uses the sample standard ERROR.
>>> You are going to confuse yourself on into the future if you do
>>> not grasp the distinction and keep it clear.

>>> The Standard Error is the name used for the SD of a statistic.
>>> Yes, it is also a standard deviation. But the "SD of a sample"
>>> is distinct and larger than the "SD of a sample mean."

>> And to complete the thought, the (estimated) standard error (SE) of the
>> MEAN = the sample SD divided by the square root of the sample size.

> Thanks for the clarification of terminology. I also found
> http://en.wikipedia.org/wiki/Standard_error to be helpful. There
> should be square root of the sample size in the denominator:
> t = [ (sample mean) - (population mean) ] / [ (sample SD) (square root
> of sample size) ]
> where the denominator estimates the standard deviation of the
> estimated mean of the population.

The denominator for your single-sample t-test should be:

[ (sample SD) / (square root of sample size) ]

Bruce Weaver
"When all else fails, RTFM."

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