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This is a paired comparison questionenter image description here:

The answer is: enter image description here

However, this is my thought of using a t test on difference of population means

enter image description here

$$t=\frac{583.125 - 415 - 0}{\sqrt{\frac{262.5900866^2}{8}+\frac{261.5721698^2}{8}}}$$

$t= 1.2829$

enter image description here

$df=13.99978881$

critital value $t_{0.05}(df=14)$= 1.76131

$1.2829 < 1.76131$

therefore no difference between the mean daily sales from the two stores. Why am I getting a different conclusion? Is it reasonable to think about it this way?

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I suppose $s_1^2$ is the sample variance of $\bar{x}_1$ and $s_2^2$ is the sample variance of $\bar{x}_2$. In that case, the sample variance of $\bar{x}_{1}-\bar{x}_{2}$ is not equal to the sum of variance unless covariance is equal to zero. Hence, the denominator of your statistics does not seem to be correct. Moreover, the degrees of freedom are 7 because you only lose 1 observation when computing the var of the difference of the means. –  Cristian Oct 3 '12 at 19:03
    
Sorry I didn't look at the formula used for the two sample t test. I assumed it was correct. Maybe that is another mistake but the key point is that it is the sample variance of the paired difference that needs to be used. If you like my answer so much why not give me an upvote. It seems to be awfully difficult on this site to get upvotes. –  Michael Chernick Oct 4 '12 at 1:40
    
@MichaelChernick yes, thank you for answering, but it looks like you have discussed the topic in CV, but Im wondering what is CV... –  user133466 Oct 4 '12 at 15:35
    
@MichaelChernick why do you think using the formula for 2 sample t test is incorrect in this scenario? I think we can use it because has been used to determine difference of population means. my thoughts are, if anything, this and the paired t test should come to the same conclusion... –  user133466 Oct 4 '12 at 17:59
    
@user133466 If you have a big seasonal effect the variance of the two sample t test can be much higher than the paired test and that will hold even though one has more dfs than the other. Also you are using the unequal variance assumption for the two sample test. So it does not have a t distribution under the null hypothesis. It becomes the Behrens-Fisher problem and has Welch's distribution which is approximated by a t test. I thought Cristian was implying that you were using the wrong formula for the test because there was an algebraic eror in the formula you used. –  Michael Chernick Oct 4 '12 at 18:55
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1 Answer

up vote 1 down vote accepted

You are using the formula for the unpaired two sample t test. The paired t test is like a one-sample test of the null hypothesis that the paired difference is 0 vs alternative that it is different from 0. The test statistic and the degrees of freedom are different. It is a more appropriate and more powerful test when the pairs are positively correlated. I have discussed this in posts on CV.

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thanks for answering michael, what exactly is a CV? –  user133466 Oct 4 '12 at 0:10
    
Yes! an answer!!! Michael you're AWSOME!!!! –  user133466 Oct 4 '12 at 0:28
    
CV is short for Cross Validation. It is an SE site where statisticians, machine learning/data mining expert –  Michael Chernick Oct 4 '12 at 17:41
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