In: Statistics and Probability
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.
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)