Posted by Kim on .
A diabetes nurse educator believes that her patients have particularly bad control of their blood sugar. To test this she selected the fasting blood sugar values for 15 of the patients she saw this week. The national mean for patients with this level of severity of diabetes is 120. Test the research hypothesis that the scores from the nurses patients are higher than the average. Show all the components required.
Scores: 172,122,189,106,196,195,151,155,162,195,231,204,232,213,209

statistics 
PsyDAG,
Ho: mean = 120
Ha: mean > 120
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
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.