Question

In a city of 80,000 households, 800 were randomly sampled in a marketing survey by an auto manufacturer. To see how typical this city was, some statistics were collected: the average number of cars per household was 1.32, while the standard deviation for the number of cars per household was 0.8. Construct a 95% confidence interval for the average number of cars per household in the whole city.

Answer 1

Solution :

Given that,

= 1.32

s = 0.8

n = 800

Degrees of freedom = df = n - 1 = 800 - 1 = 799

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,799 = 1.963

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

= 1.963 * (0.8/ 800)

=0.05

Margin of error = 0.05

The 99% confidence interval estimate of the population mean is,

- E < < + E

1.32 - 0.05 < < 1.32 + 0.05

1.27 < < 1.37

( 1.27, 1.37 )