Question

A clinical trial is being conducted in order to determine the efficacy of a new drug that will be used to treat type II diabetes. The efficacy of the medication will be determined by the blood glucose readings obtained from the patients. If the researcher wants a margin of error less than or equal to 5 mg/dL and the standard deviation for blood glucose readings among type II diabetics was previous documented as 15 mg/dL, how many patients should be recruited for each group of individuals in the study assuming a 95% confidence interval will be used to quantify the mean difference in blood glucose scores between the control group and the treatment group? [1] n for the treatment group = 32 and n for the control group = 13 [2] n for the treatment group = 33 and n for the control group = 33 [3] n for the treatment group = 65 and n for the control group = 65 [4] n for the treatment group = 1729 and n for the control group = 1729

Answer 1

z value for 95% confidence interval = 1.96

Margin of error, E = z * Std Error

5 = 1.96 * Std Error

Std Error = 5 / 1.96 = 2.55

For margin of error less than or equal to 5 mg/dL, the standard error should be less than 2.55.

Std Error =

For [1] n for the treatment group = 32 and n for the control group = 13

Std Error = = 4.933451

[2] n for the treatment group = 33 and n for the control group = 33

Std Error = = 3.692745

[3] n for the treatment group = 65 and n for the control group = 65

Std Error = = 2.631174

[4] n for the treatment group = 1729 and n for the control group = 1729

Std Error = = 0.5101628

The standard error is less than 2.55 for n for the treatment group = 1729 and n for the control group = 1729

Thus, the answer is,

[4] n for the treatment group = 1729 and n for the control group = 1729