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

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