Suppose the distribution of cholesterol level in normal adul

Suppose the distribution of cholesterol level in normal adults is roughly symmetric with mean 219 mg/100mL and standard deviation 41 mg/100mL. A large number of physicians join a study. Each of them selects a random sample of size 25 from their regular visitors and measures their cholesterol level.

What is the expected value of the sample means?

What is the standard error of the sample means?

What proportion of the sample means do you expect to be greater than 223 mg/100mL?

If each physician increases the sample size from 25 to 50, what proportion of the sample means do you expect to be less than 218 mg/100mL?

One physician observes the following data on 25 individuals. She does not know the population mean or population variance of cholesterol levels. What is her estimate of the population mean and a 95% confidence interval?   

Solution

Expected value = 219 mg/100mL

standard error of the sample means = Sd/ sqrt(n)= 41/ sqrt(25) = 8.2

Proportion greater than 223 ----- z score= ( (223-219)/8.2) =0.4878

to z score to p value = 0.312846 or 31.28%

Proportion less than 219 ----- (218-219)/(41/ sqrt(50) )= -1/5.79827= -0.1724

to z score to p value =0.431719 or 43.17%

95%Confidence interval = 1.96* 8.2.......(1.96 is the z score of 95% ci or 0.05 alpha, two tailed)

= 16.072

219+- 16.072

Therefore Confidence interval is ( 202.928,235.072)