Answer to Question #121048 in Statistics and Probability for William

Question

Plasma-glucose levels are used to determine the presence of diabetes. Suppose the mean ln (plasma-glucose) concentration (mg/dL) in 35- to 44-year-olds is 4.86 with standard deviation = 0.54. A study of 100 sedentary people in this age group is planned to test whether they have a higher or lower level of plasma glucose than the general population.
a1. If the expected difference is 0.10 ln units, then what is the power of such a study if a two-sided test is to be used with α = .05?
a2. How many people would need to be studied to have 80% power under the assumptions in Problem a1?

Answer 1

Expected difference: 0.10 units

$α = 0.05$

$= ϕ ( − z 1 − α / 2 + ( μ o − μ 1 ) √ n/ σ ) + ϕ ( − z 1 − α / 2 + ( μ o − μ 1 ) √ n/ σ ) ) = ϕ ( − z 1 − 0.05 / 2 + 0.10 √ 1 00/ 0.54 ) + ϕ ( − z 1 − 0.05 / 2 + − 0.10 √ 1 00/ 0.54 ) = ϕ ( − z 0.975 + 1.85 ) + ϕ ( − z 0.975 − 1.85 ) = ϕ ( 0.11 ) + ϕ ( − 3.81 ) = 0.1771 − 6.13 = 5.957$

The sample size required for the study to have 80% power is

$n = σ 2 ( Z 1 − β + Z 1 − α / 2 ) 2 ( μ o − μ 1 ) 2 = 0.54 2 ( Z 0.80 + Z 1 − 0.05 / 2 ) 0.10 2 = 0.54 2 ( 0.8416 + 1.96 ) 2 0.10 2 = 228.87$