Question

A random sample of 13 DVD movies had a mean length of 111.6 minutes, with a standard deviation of 66.9 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 (for example : answer . Do not write any units . 3.1) . Write only a number as your

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 111.6

Population standard deviation =    = 66.9

Sample size = n = 13

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Margin of error = E = Z/2 * ( /n)

= 1.645 * ( 66.9 /  13 )

= 30.5

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

- E < < + E

111.6 - 30.5 <   < 111.6 + 30.5

81.1 <   < 142.1

( 81.1 , 142.1 )

The 90% confidence interval estimate of the population mean is : ( 81.1 , 142.1 )