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

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).

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:

`(1-0.99)/(2) = (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 = (30)/(√(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).