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From: Bacle on 21 Nov 2009 23:00
Hi, everyone.
I would appreciate your help with the following:

I wonder if I am missing something obvious re the idea
of hypothesis-testing for regression; specifically, we are testing whether y and x are linearly-related.

My layout always assumes the null hypothesis Ho to be that there is no linear relation, i.e., that the slope of the regression line y=bo+b1x is zero.

Question: why isn't it enough to just calculate the correlation coefficient r from the data.?. I just don't see what we gain by using the test statistic:

t=b1/s_b1


(Note that we are assuming b=0 in Ho, so that we

are actually using t=(b1-0)/s_b1

Where y=bo+b1x is the regression line of y by x, i.e., b1 is the slope of the least-squares line , and s_b1 is the sample standard deviation of b1.


Doesn't the correlation coefficient r give us the necessary information to decide whether there is a linear relation between y and x.?

Thanks.
From: Ray Koopman on 23 Nov 2009 00:58
On Nov 22, 9:40 pm, Bacle <ba...(a)yahoo.com> wrote:
> Anyone.?

The two tests are equivalent.
From: Ray Koopman on 23 Nov 2009 04:21
On Nov 22, 11:08 pm, Bacle <ba...(a)yahoo.com> wrote:
>> On Nov 22, 9:40 pm, Bacle <ba...(a)yahoo.com> wrote:
>>> Anyone.?
>>
>> The two tests are equivalent.
>
> Thanks, Prof. Koopman.
> Wouldn't it be easier to calculate r than to do a hypothesis test.?.
> On top of that, a hypothesis test only returns a confidence interval,
> while calculating r would give us a definitive answer; a non-zero
> coefficient would be a yes, there is a linear relation .
>
> Is there a reason to do one tewst over the other, or are the tests
> basically interchangeable.?

Neither b1 nor r is a *test* statistic. They are closely related
*descriptive* statistics: b1 = r*s_y/s_x, where s_y and s_x are the
sample standard deviations of the y and x variables. The only
difference between b1 and r is their units: b1 is in units of y/x,
whereas r is unitless. They both need testing. The test statistic is
t = b1/s_b1 = r*sqrt[(n-2)/(1-r^2)], with n-2 degrees of freedom.

A warning: some people erroneously interpret the form of the t as
implying that s_r = sqrt[(1-r^2)/(n-2)]. It is much more complicated
than that.

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