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Question : A sample of n=25 individuals is randomly selected from a population with a mean of = 65, and a treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M=69.

a. If the sample variance = 100, 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?   (show all steps and formulas including how you arrived at the critical value)

Explanation / Answer

Given,

Sample of size =n=25

Degree of freedom = n-1 = 25 - 1 = 24

Population mean = m = 65

Sample mean = M =69

Sample variance = s^2 =10

Standard error =S. E=sqrt (s^2/n)=sqrt (100/25)= 2

S.E=2

Use t - test (sample size n<=30, unknown population variance)

t=(M - m) /S. E.

t = (69 - 65) / 2

t = 2

For critical value use t - test, two tailed.

t(alpha, df) = t(0.05, 24) = 2.064 (using t-table)

Conclusion :

The calculated value t = 2 is less than critical value t(alpha, df) = 2.064.

So, we accept null hypothesis.

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