A study is trying to determine the mean fasting (total) cholesterol of teenage b
ID: 3232407 • Letter: A
Question
A study is trying to determine the mean fasting (total) cholesterol of teenage boys in the US. The study collects data on 49 boys and finds a mean cholesterol of 195 mg/dL and a standard deviation of 50 mg/dL.
(3 pts) Which of the following is correct? Explain why.
xbar = 195 mg/dL ii. = 195 mg/dL
(3 pts) Which of the following is correct? Explain why.
s = 50 mg/dL ii. = 50 mg.dL
(4 pts) Compute a 95% confidence interval for the mean fasting cholesterol in the population, based on the study results. Note: use the sample standard deviation as an estimate of the population standard deviation.
Explanation / Answer
The sample size is given by, n = 49
mean chol. levels = 195mg/dL
Standard deviation of 50mg/dL
1. This is a statistic not a parameter. Thus we say xbar and not
Mu. Mu is a parameter. parameter ties to a population and statistic
to a sample
So, xbar = 195mg/dL is right
2. Again, for the above reasons s is the sample deviation and sigma is the
population deviation. Thus, s = 50mg/dL
A 95% Ci is given by:
xbar +/- Z*s/sqrt(N)
=195+/- 1.96*50/sqrt(49)
=181 to 209
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