Question

The average cholesterol level of 32 adults was 203 mg/dL, with a sample standard deviation of 24 mg/dL. Find the 99% confidence interval for the mean cholesterol level of adults. State the formulas used to show your work.

Answer 1

Solution :

Given that,

= 203

s = 24

n = 32

Degrees of freedom = df = n - 1 = 32 - 1 = 31

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,31 =2.244

Margin of error = E = t/2,df * (s /n)

= 2.797 * (24 / 32)

= 11.64

Margin of error = 11.64

The 99% confidence interval estimate of the population mean is,

- E < < + E

203 - 11.64 < < 203 + 11.64

191.36 < < 214.64

(191.36, 214.64 )