In: Statistics and Probability
A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 4 points with 99% confidence assuming s=13.7 based on earlier studies? Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required?
A) 99% confidence level requires_____ Subject. (round up to the nearest subject)
B) 90% confidence level requires_____ Subject. (round up to the nearest subject)
a)
Standard Deviation , σ =
13.7
sampling error , E = 4
Confidence Level , CL= 99%
alpha = 1-CL = 1%
Z value = Zα/2 = 2.576 [excel
formula =normsinv(α/2)]
Sample Size,n = (Z * σ / E )² = ( 2.576
* 13.7 / 4 ) ²
= 77.831
So,Sample Size needed=
78
b)
Standard Deviation , σ =
13.7
sampling error , E = 4
Confidence Level , CL= 90%
alpha = 1-CL = 10%
Z value = Zα/2 = 1.645 [excel
formula =normsinv(α/2)]
Sample Size,n = (Z * σ / E )² = ( 1.645
* 13.7 / 4 ) ²
= 31.738
So,Sample Size needed=
32