Question

A random sample of 13 DVD movies had a mean length of 114.9 minutes, with a standard deviation of 72.8 minutes. Find the lower bound of the 90% confidence interval for the true mean length of all Hollywood movies. Assume movie lengths to be approximately normally distributed. Round to one decimal place.

Answer 1

Given that,

= 114.9

s =72.8

n = 13

Degrees of freedom = df = n - 1 =13 - 1 = 12

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

  = 0.1

t _,df = t_0.1,12 = 1.356    ( using student t table)

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

= 1.356 * (72.8 / 13) =27.4

The 90% confidence interval estimate of the population mean is,

- E

114.9 -27.4  

87.5

lower bound=87.5