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Question

A sample of n=28 individuals is randomly selected from a population with a mean of µ= 63, and a treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M=68.

a. If the sample variance = 96, are the data sufficient to reject the null and conclude that the treatment has a significant effect using a two-tailed test with alpha = .05? Be sure to show all formulas with symbols (and plug in numbers), steps, processes and calculations for all parts of all answers above (including how you arrived at the critical value)

Answer 1

Ho: mu = 63

Ha: mu != 63

M = 68

sample variance = 68

sample standard deviation (SD) = 8.25

z = (M - mu)/(SD/sqrt(n) = (68 - 63)/(8.25/sqrt(28)) = 3.2

critical values for alpha = 0.05 are +/-1.96

since computed z = 3.2 > 1.96 we can reject Ho and conclude the treatment has a significant effect at a 5% level of significance.