Posted by **Kim** on Thursday, March 14, 2013 at 8:09pm.

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**, Friday, March 15, 2013 at 12:33pm
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.

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