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# Thread: ASM 3L Manual Ed 7 Example 8G

1. ## ASM 3L Manual Ed 7 Example 8G

Just a quick question for those that have this manual. In this Hypothesis Testing example a null hypothesis H0 is said to be rejected if the sample mean Xbar > some value k. X here is a geometiric distribution with mean beta. And the sample is of 10 people. I understand that the sum of the 10 geometric distributions is a negative binomial with r = 10. So if we let Y = 10*Xbar be the negative binomial, why would we be finding a k such that Pr(Y > k) <= .05, our desired significance? Isn't that equivalent to Pr(10*Xbar > k)? How does that k equal the same k as in the Pr(Xbar > k) that is defined as the rejection criteria?

2. Originally Posted by jeanluc
Just a quick question for those that have this manual. In this Hypothesis Testing example a null hypothesis H0 is said to be rejected if the sample mean Xbar > some value k. X here is a geometiric distribution with mean beta. And the sample is of 10 people. I understand that the sum of the 10 geometric distributions is a negative binomial with r = 10. So if we let Y = 10*Xbar be the negative binomial, why would we be finding a k such that Pr(Y > k) <= .05, our desired significance? Isn't that equivalent to Pr(10*Xbar > k)? How does that k equal the same k as in the Pr(Xbar > k) that is defined as the rejection criteria?
I agree. I should've divided my k by 10 at the end and made it 0.3. However, this doesn't affect the answer to the question about the power of the test.