Question

Constructa 99% confidence interval for the population mean, µ, AND State what type of Interval you use (ZInterval, TInterval, 1-PropZInterval). A sample of 22 professors had a mean amount of experience of 10.8 years with a standard deviation of 3.3 years. Assume the population is normally distributed

Answer 1

Solution :

TInterval,

Given that,

Point estimate = sample mean = = 10.8

sample standard deviation = s = 3.3

sample size = n = 22

Degrees of freedom = df = n - 1 = 22 - 1 = 21

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,21 = 2.831

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

= 2.831 * (3.3 / 22)

= 2.0

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

- E < < + E

10.8 - 2.0 < < 10.8 + 2.0

8.8 < < 12.8

(8.8 , 12.8)