Question

The driving distances of college students who commute to a particular school are normally distributed. In a random sample of 8 college students, the sample mean driving distance was 22.2 miles and the sample standard deviation was 5.8 miles.

(a). Construct a 98% confidence interval for the mean driving distance of all the college students who drive to the school. Write a sentence interpreting your interval. (Round answer to 2 decimal places)

7(b). Suppose the population standard deviation was known to be 6.1 miles. Construct a 95% confidence interval for the mean driving distance of all the college students who drive to the school. (Do not write a sentence, Round answer to 2 decimal places)

7(c). For the confidence interval in part (b), if we increased the sample size, the width of the interval would become: (choose your answer)

                        Narrower                 Wider                 Stay the same

Answer 1

Solution:

Given: The driving distances of college students who commute to a particular school are normally distributed.

Sample Size = n = 8

Sample mean = miles

Sample standard deviation = s =  5.8 miles.

Part a) Construct a 98% confidence interval for the mean driving distance of all the college students who drive to the school.

where

t_c is t critical value for c = 98%  confidence level

Thus two tail area = 1 - c = 1 - 0.98 = 0.02

df = n - 1 =  8 - 1 = 7

Look in  t table for df = 7 and two tail area = 0.02 and find t critical value

t_c = 2.998

Thus

Thus

Interpretation:

We are 98% confident that the true population mean driving distance of all the college students who drive to the school is between 16.05 miles and 28.35 miles.

Part b)

Given:  the population standard deviation was known to be 6.1 miles. That is:

Since population standard deviation was known, we use z interval:

Construct a 95% confidence interval for the mean driving distance of all the college students who drive to the school.

where

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Thus

Thus

Part c) For the confidence interval in part (b), if we increased the sample size, the width of the interval would become:

Sample size n is inversely proportional to Margin of Error.

Thus if we increase sample size n, then margin of error decreases.

As margin of error decreases, width of confidence interval decreases.

Thus answer is: Narrower