Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
The human resources department of a consulting firm gives a standard creativity test to a randomly selected group of new hires every year. This year, 80 new hires took the test and scored a mean of 112.5 points with a standard deviation of 17.3. Last year, 55 new hires took the test and scored a mean of 117.5 points with a standard deviation of 17.3. Assume that the population standard deviations of the test scores of all new hires in the current year and the test scores of all new hires last year can be estimated by the sample standard deviations, as the samples used were quite large. Construct a 95% confidence interval for u1-u2, the difference between the mean test score u1 of new hires from the current year and the mean test score u2 of new hires from last year. Then complete the table below
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.
What is the lower limit of the 95% confidence interval
What is the upper limit of the 95% confidence interval
1 answer
0 answers