The U. S. Department of Transportation provides the number of miles that residents of the75 largest metropolitan areas travel per day in a car. Suppose that for a simple randomsample 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 simple random sample of 40 Boston residents themean 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 thatBuffalo residents travel per day and the mean number of miles that Boston residentstravel per day?b. What is the 95% confidence interval for the difference between the two populationmeans?

a. 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 residentstravel per day is 88

b. The 95% confidence interval for the difference between the two population means is (0.58171,7.21829)

Step-by-step explanation:

a. According to the given data we have the following:

Buffalo residents mean=22.5

Standard deviation=8.4

n=50

Boston residents mean=18.6

Standard deviation=7.4

n=40

Hence, the point estimate is the difference of the means=22.5-18.6=3.9

The standard error=√((s1∧2/n1)+((s2∧2/n2))=√((8.4∧2/50)+(7.4∧2/40))=1.67

Therefore, the difference of miles=n1+n2-2=50+40-2=88

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 residentstravel per day is 88.

b. The critical value of t at 95% confidence level and DoF=88 is 1.987

The interval:(x1-x2)+ (critical value* std error)

=3.9-(1.987*1.67),3.9+(1.987*1.67)

Difference between mean=(0.58171,7.21829)

The 95% confidence interval for the difference between the two population means is (0.58171,7.21829)

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