The US Department of Transportation provides the number of m

The U.S. Department of Transportation provides the number of miles that residents of the 75 largest metropolitan areas travel per day in a car. Suppose that for a random sample of 50 Buffalo residents the mean is 22.5 miles a day and the standard deviation is 8.4 miles a day, and for an independent random sample of 40 Boston residents the mean is 18.6 miles a day and the standard deviation is 7.4 miles a day.

a. What is the point estimate of the difference between the mean number of miles that Buffalo residents travel per day and the mean number of miles that Boston residents travel per day?
b. What is the 95% confidence interval for the difference between the two population means?

Solution

a)
Mean(x1)=22.5
Standard deviation( sd1 )=8.4
Sample Size(n1)=50
Mean(x2)=18.6
Standard deviation( sd2 )=7.4

Point of estimate = =22.5-18.6 = 3.9

b)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval

Sample Size(n1)=40
CI = [ ( 22.5-18.6) ±t a/2 * Sqrt( 70.56/50+54.76/40)]
= [ (3.9) ± t a/2 * Sqrt( 2.78) ]
= [ (3.9) ± 2.023 * Sqrt( 2.78) ]
= [0.53 , 7.27]