Question

Suppose the average cholesterol level of children ages 2-14 is 175 mg/dL with a standard deviation = 30 mg/dL. Recently, doctors have started worrying about elevated cholesterol levels in some children. A study was done to see if there is a familial aggregation of cholesterol levels. Specifically, in families where the father had a heart attack and has elevated cholesterol levels (> 250 mg.dL), is the cholesterol level of the children in such families higher than 175? A group of 100 such children had an average cholesterol level of 207.3 mg/dL. Assume there were no outliers in the data set. Use a standard deviation = 30 mg/dL and find a 99% confidence interval for the average cholesterol level of all children whose father has had a heart attack with cholesterol level above 250. (Please give responses to 3 decimal places)

Answer 1

Answer:

The 99% confidence interval for the average cholesterol level of all children whose father has had a heart attack with cholesterol level above 250 is (199.422mg/dL, 215.178 mg/dL).

Step-by-step explanation:

The sample size is 100.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 100-1 = 99$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 99 and 0.005 in the t-distribution table, we have $T = 2.626$.

Now, we need to find the standard deviation of the sample. That is:

$s = \frac{30}{\sqrt{100}} = 3$

Now, we multiply T and s

$M = T*s = 3*2.626 = 7.878$

For the lower end of the interval, we subtract the mean by M. So 207.3 - 7.878 = 199.422 mg/dL.

For the upper end of the interval, we add the mean to M. So 207.3 + 7.878 = 215.178 mg/dL.

The 99% confidence interval for the average cholesterol level of all children whose father has had a heart attack with cholesterol level above 250 is (199.422mg/dL, 215.178 mg/dL).