& standard devaition of 8. After treatment is administered the sample mean is 33. If sample is n=16 scores whats the standard error? If n=16what conclusion would you make (2-tailed test)? If sample consists n=64 scores thats the standard error?

If n= 16 the standard error is ; mean /sqrt (n) = 30/sqrt (16) = 30/4 = 7.5

If n= 16, z score = x-mean/std error = 33 - 30/7.5 = 0.4

As z score o.4 is less than z crit= 1.96, the null hypothesis cannot be rejected. This means that the sample score is compatible with the mean of 30.

If n= 64, the standard error is: mean/sqrt(n) = 30/sqrt(64) = 30/8 = 3.75

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