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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.2012-10-03
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    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.2012-10-04
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    @MichaelChernick yes, thank you for answering, but it looks like you have discussed the topic in CV, but Im wondering what is CV...2012-10-04
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    @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...2012-10-04
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    @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.2012-10-04
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    Apparently he is arguing the the unpaired t test needs a covariance term for the variance of the difference in the denominator. I think there is a subtle issue here. When X and Y are paired and correlated taking average of the paired differences removes the variability due to the correlation of the paired variables and things reduce to an exact one sample t test. If you treat them unpaired that added variability is in both the mean for the Xs and the mean for the Ys. So although i think the Welch test may be valid it is far from the best.2012-10-04
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    The added variance makes it less powerful than the paired test. So typically you will see this when the correlation coefficient for the pair (X,Y) is large. I borrowed an example of temperature comaprisons in NYC vs DC over a two year cycle. The paired t tests detects the difference in average temperatures while the unpaired test will not. The p-value is very nonsignificant for the unpaired test and is very signifcant for the paired test. The summer peaks and winter troughs swamp out the smaller day to day differences.2012-10-04
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    @MichaelChernick to summarize, I cannot use the 2 sample independent t test because there's a strong correlation (or dependence) between the two variables. The experimental design is better suited with paried comparisons. correct?2012-10-05
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    yes better suited for the paired test, possible with unpaired but not a powerful test, use paired2012-10-05

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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?2012-10-04
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    Yes! an answer!!! Michael you're AWSOME!!!!2012-10-04
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    CV is short for Cross Validation. It is an SE site where statisticians, machine learning/data mining expert2012-10-04