SOLUTION: A researcher collected sample data for 16 women ages 18 to 24. The sample had a mean serum cholesterol level (measured in mg/100 mL) of 191.1, with a standard deviation of 6.5. Ass

Question 454999: A researcher collected sample data for 16 women ages 18 to 24. The sample had a mean serum cholesterol level (measured in mg/100 mL) of 191.1, with a standard deviation of 6.5. Assuming that serum cholesterol levels for women ages 18 to 24 are normally distributed, find a 90% confidence interval for the mean serum cholesterol level of all women in this age group. Then complete the table below.
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

Answer by Mathematicians(84) (Show Source): You can put this solution on YOUR website!
We first need to find the sampling distribution. Unfortunately, since we only know the sample standard deviation (not the population standard deviation) and the sample size is under 30 (specifically 16), the sampling distribution will be a students T distribution with 15 (16 - 1) degrees of freedom.
I'm not sure where you are at in Statistics, but I'll present this the most elementary way possible and that is to computer using a direct formula.
We have the formula for a 90% two sided confidence interval.
Let M = mean, S = Stddev, T = calculated from tables, N = sample size
90%:
(, )

Now to find T, we want a 90% two sided confidence interval, this implies level of significance alpha = .1. Since it is two sided, we will need to allocate .05 of that alpha to the left tail and .05 to the right tail to keep things balanced. T_.05,15 = 1.573
Plug everything in and you get
(188.5, 193.7)
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