Question

The average stopping distances for a population of school buses traveling 50 miles per hour, measured in feet, are normally distributed, with a standard deviation of 6 feet. Using a random sample of 23 buses having a sample mean of 262 feet, construct a 95% confidence interval for the mean stopping distance of the population.

Answer 1

Solution :

Given that,

= 50

s =6

n = Degrees of freedom = df = n - 1 = 23 - 1 = 22

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,22 =   2.074 ( using student t table)

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

= 2.074 * ( 6/ 23)

= 2.59

The 95% confidence interval mean is,

- E < < + E

50 - 2.59 < < 50+ 2.59

47.41 < < 52.59

(47.41 , 52.59)