Question

A researcher collected sample data for 12 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred
milliliters) of 189.6, with a standard deviation of 10. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 95%
confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 95% confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)
Lower limit:
Upper limit:
X
5

Answer 1

The lower and upper limit of the 95% confidence interval is 183.2 and 196.0 unit.

What is a confidence interval ?

When you run your experiment again or resample the population in the same way, you can expect your estimate to fall within a certain range of values a certain percentage of the time. This is known as the confidence interval.

Main Body:

Given in question :

x- bar = 189.6

standard deviation = 10

C.I. = 95% =0.95

N=12

so, degree of freedom (D.F) = N-1 = 11

confidence  level (α) = (1- C.I.)/2

                                  = 0.025

according to t- distribution table for d.f = 11 and α= 0.025  , my result is 2.201

Now dividing standard deviation by square root of samples, we get

= 10/√12= 2.887

Lower limit = Mean - 2.201*2.887

                  = 189.6- 6.354

                  = 183.246

Upper limit = Mean + 2.201*2.887

                  = 189.6 + 6.354

                  = 195.954

Hence the result is 183.2 and 196.

to learn more about confidence interval , visit;

brainly.com/question/15712887

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