Question

A random sample of 100 automobile owners in a region shows that an automobile is driven on average 21,500 kilometers per year with a standard deviation of 4500 kilometers. Assume the distribution of measurements to be approximately normal. Construct a 95​% prediction interval for the kilometers traveled annually by an automobile owner in the region.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 21500

Population standard deviation =    = 4500

Sample size = n = 100

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

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

= 1.96 * ( 4500 /  100 )

= 882

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

- E < < + E

21500 - 882 <   < 21500 + 882

20618 <   < 22382

( 20618 , 22382 )

The 95% confidence interval of the population mean is : ( 20618 , 22382 )