Question

A random sample of the price of gasoline from 30 gas stations in a region gives the statistics below. Complete parts a through c below.

y=​ $3.89​, SE (y) =​$0.06

​a) Find a 95​% confidence interval for the mean price of regular gasoline in that region.

​(Round to three decimal places as​ needed.)

​b) Find the​ 90% confidence interval for the mean.

​(Round to three decimal places as​ needed.)

​c) If we had the same statistics from a sample of 60​stations, what would the​ 95% confidence interval be​ now?

​(Round to three decimal places as​ needed.)

Answer 1

Here we have

s=0.33

(a)

Since there 30  data values in the sample so degree of freedom is df=30-1=29 and critical value of t for 95% confidence interval is 2.0452. Therefore required confidence interval is

Therefore, a 95% confidence interval for the mean is (3.767, 4.013).

(b)

Since there 30 data values in the sample so degree of freedom is df=30-1=29 and critical value of t for 90% confidence interval is 1.6991. Therefore required confidence interval is

Therefore, a 90% confidence interval for the mean is (3.789,3.992).

(c)

Since there 60 data values in the sample so degree of freedom is df=60-1=59 and critical value of t for 95% confidence interval is2.001  Therefore required confidence interval is

Therefore, a 95% confidence interval for the mean is (1.307,6.473)