The document discusses a study that measured cholesterol levels in a subpopulation of hypertensive female smokers. A random sample of 20 subjects found a mean cholesterol level of 221 mg/dL. Given the original population had a mean of 211 mg/dL and standard deviation of 39, a 90% confidence interval for the subpopulation's mean is calculated as [203.48, 238.52] mg/dL. A second 90% confidence interval is calculated without knowing the subpopulation's standard deviation, using the sample's standard deviation of 31, as [205.78, 236.22] mg/dL.

1. Suppose the mean cholesterol levele for all 40 to 74 -year-old females livine in Cinalin in 11 .
211 me/at, and the sasdard dectation is of = 39 mid / all . A stikfy was conducted to meawe the
cholesterod fevels in a subpopulation of hrypertensive female imoker Sipgpose an tandomi
sample of size & 20 hypertensive female smoker provided a mean ehalesterol lovet of Fi. = 21 k
mg / dt . Assume that the distribution of cholesterollevels in this subpopalation is approximately
nocmal. t. Find a 90% contidence interval for the mean cholesterol level of this subpopalation, if
the subpopalation has the same stindard deviation = 39 as the original poputation. (Note: Round
the Jower fimit and upper limit to 2 decimal places of accuracy, and enter in the answer box.) 90
%. Confidence Intervali b. Find a 90% confidence interval for the mean cholesterol level of this
subpopulation, if the standard deviation of the subpopulation is not known, but the sample
standard deviation from the study is s = 31 mg / dL . (Note: Round the lower limit and upper
limit to 2 decimal places of accuracy, and enter in the answer box.)