posted by Jenn
Williams, Dunsiger, Jennings, and Marcus (2012) conducted a study to test how enhanced mood during exercise can increase the likelihood of physical activity in the future. They reported that at 6 months of physical activity, mood scores (rated on a scale from -5 to +5) after a walking exercise were 3.3 ± 1.2, and after 12 months, mood scores after a walking exercise were 3.3 ± 1.3 (mean ± standard error). Assuming these data are normally distributed, use the steps to conduct a z transformation to determine the following probabilities:
(a) What is the probability that a sample mean selected from this population would show no change (0) or a negative change (less than 0) at 6 months of physical activity?
(b) What is the probability that a sample mean selected from this population would show no change (0) or a negative change (less than 0) at 12 months of physical activity?
Unless there are typos, with the means the same, there was no difference.
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.