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.
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To determine the probabilities, we will first conduct a z-transformation using the formula:
z = (x - μ) / σ
where:
- x is the value we want to transform,
- μ is the population mean, and
- σ is the population standard deviation.
Step 1: Calculate the population mean (μ) and standard deviation (σ) for each time point.
For 6 months of physical activity:
μ = 3.3
σ = 1.2
For 12 months of physical activity:
μ = 3.3
σ = 1.3
Step 2: Calculate the z-scores for the desired values.
(a) At 6 months:
z = (0 - 3.3) / 1.2
z = -2.75
(b) At 12 months:
z = (0 - 3.3) / 1.3
z = -2.54
Step 3: Determine the probabilities using the z-table or a statistical software.
(a) Using the z-table, find the probability corresponding to the z-score of -2.75. This probability represents the area under the standard normal distribution curve to the left of -2.75.
(b) Similarly, find the probability corresponding to the z-score of -2.54 for 12 months.
Please note that in order to find the exact probabilities, you would need to use a z-table or a statistical software that provides accurate values for the area under the standard normal distribution curve.