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).

y=$3.29 s=$0.27

A. Find a 95​% confidence interval for the mean price of regular gasoline in that region. ($__,$__)

B. Find a 90​% confidence interval for the mean price of regular gasoline in that region. ($__,$__)

C. If we had the same statistics from a sample of 80 ​stations, what would the​ 95% confidence interval be​ now?

Answer 1

a)

sample mean, xbar = 3.29
sample standard deviation, s = 0.27
sample size, n = 30
degrees of freedom, df = n - 1 = 29

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.045

ME = tc * s/sqrt(n)
ME = 2.045 * 0.27/sqrt(30)
ME = 0.1008

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (3.29 - 2.045 * 0.27/sqrt(30) , 3.29 + 2.045 * 0.27/sqrt(30))
CI = (3.19 , 3.39)

b)

sample mean, xbar = 3.29
sample standard deviation, s = 0.27
sample size, n = 30
degrees of freedom, df = n - 1 = 29

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.699

ME = tc * s/sqrt(n)
ME = 1.699 * 0.27/sqrt(30)
ME = 0.0838

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (3.29 - 1.699 * 0.27/sqrt(30) , 3.29 + 1.699 * 0.27/sqrt(30))
CI = (3.21 , 3.37)

c)

sample mean, xbar = 3.29
sample standard deviation, s = 0.27
sample size, n = 80
degrees of freedom, df = n - 1 = 79

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 1.99

ME = tc * s/sqrt(n)
ME = 1.99 * 0.27/sqrt(80)
ME = 0.0601

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (3.29 - 1.99 * 0.27/sqrt(80) , 3.29 + 1.99 * 0.27/sqrt(80))
CI = (3.23 , 3.35)