Question

The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $3.98 The US EIA updates its estimates of average gas prices on a weekly basis. Assume the standard deviation is $.24 for the price of a gallon of regular gasoline and recommend the appropriate sample size for the US EIA to use if they wish to report each of the following margins of error at 95% confidence. Round up to the next whole number.

a. The desired margin of error is $ .11 The appropriate sample size is. is disclaimer (.11, NOT 11)

b. The desired margin of error is $ .07 The appropriate sample size is  . Disclaimer (. 07 NOT 7)

According to statistics reported on CNBC, a surprising number of motor vehicles are not covered by insurance. Sample results, consistent with the CNBC report, showed 15 of 239 vehicles were not covered by insurance.

a. What is the point estimate of the proportion of vehicles not covered by insurance (to 4 decimals)?

the answer is .062 i just need help for part B

b. Develop a  confidence interval for the population proportion (to 4 decimals).

  ,   

Answer 1

SolutionA:

E=margin of Error=0.11

z crit for 95%=1.96

sigma=0.24

let required sample size=n

n=(Zcrit*sigma/E)^2

=(1.96*0.24/0.11)^2

=18.28729

=18

ANSWER:18

Solutionb:

E=0.07

n=(Zcrit*sigma/E)^2

=(1.96*0.24/0.07)^2

=45.1584

=45

ANSWER:45

a. What is the point estimate of the proportion of vehicles not covered by insurance (to 4 decimals)?

point estimate=p^=x/n=15/239=0.06276151

b. Develop a  confidence interval for the population proportion (to 4 decimals).

nothing specfied alpha=0.05

z crit for 99%=2.576

99%  confidence interval for the population proportion

=p^-z*sqrt(p^*(1-p^)/n,p^+z*sqrt(p^*(1-p^)/n

=0.06276151-2.576sqrt(0.06276151*(1-0.06276151)/239),0.06276151+2.576*sqrt(0.06276151*(1-0.06276151)/239

=0.0223,0.1032

99% lower limit=0.0223

99% upper limit=0.1032