and standard deviation of 8. After treatment is administered to the individuals in the sample the sample mean is found to be 33. If n=64 what conclusion would you make (2-tailed test, a=0.05.)?
z = (33-30)/(8/sqrt(64))= 3/(8/8) = 3
Since z > 1.96 we reject the null hypothesis
1.96 is the 5% value for Normal distribution ( P(z < -1.96) =0.025 and p(z> 1.96) =0.025
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