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I have a following problem: Let's have two sport teams, X and Y. X's score at world tournaments is 1., 6. and 1. place. Y's: 2,7,5,3,3. I want to calculate with 95% probability who's better. I have chosen students t-test of a hypothesis

H0: µ1 = µ2

and it's alternative

H1: µ1 < µ2

where µ1 is the mean "place" of team X and µ2 of Y. I chose the alternative because the arithmetic mean of team X is better than Y's.

I have calculated the test statistic as -0.938 and from tables t 0.975 (6) = 2.447

What next? I know that if alternative hypothesis was µ1 != µ2 the rejection region would be R - {-2.447, 2.447} but with my more specific alternative hypothesis I'm not sure.

A'd be thankfull for any advice..

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up vote 1 down vote accepted

A one-sided alternative is really H0 µ1 >= µ2 vs H1 µ1 < µ2. The side is below zero so you should use -2.447 for the critical value or if you want alpha =0.05 use -t0.950(6)=t0.05(6). You reject if t is less than -2.447 or the critical value for alpha =0.05. In any event you can't reject. You might also think about the fact that the observations are very non-normal as they are discrete integer valued.

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