Question

A sample of 45 movie tickets had a mean price of $9.50 with a standard deviation of $1.50. Find the 95% confidence interval for the population mean of the price of movie tickets AND State what type of confidence interval you used (ZInterval, TInterval, 1-PropZInterval)

Answer 1

Solution :

TInterval

Given that,

Point estimate = sample mean = = 9.50

sample standard deviation = s = 1.50

sample size = n = 42

Degrees of freedom = df = n - 1 = 45 - 1 = 44

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,44 = 2.015

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

= 2.015 * ( 1.50 / 45)

= 0.45

The 95% confidence interval estimate of the population mean is,

- E < < + E

9.50 - 0.45 < < 9.50 + 0.45

9.05 < < 9.95

(9.05 , 9.95)