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. ( give responses to 3 decimal places)

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

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

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 .

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

Now, we multiply T and s

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).

the answer is a

step-by-step explanation:

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step-by-step explanation: