Question

The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is . The US EIA updates its estimates of average gas prices on a weekly basis. Assume the standard deviation is 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 confidence. Round up to the next whole number.

a. The desired margin of error is $.10. The appropriate sample size is ____.
b. The desired margin of error is $.06. The appropriate sample size is ____.
c. The desired margin of error is $.04. The appropriate sample size is _____.

Answer 1

The answer is "23, 62, and 139"

Explanation:

In point a:

`\to \sigma = \$0.24\\\\ \to E = \$0.10\\\\ 95\% \text{confidence level of the z}: \\\\\to \alpha = 1 - 95\% = 1 - 0.95 = 0.05\\\\\to (\alpha)/(2) = (0.05)/(2) = 0.025\\\\\to Z_{(\alpha)/(2)} = Z_(0.025) = 1.96\\\\`

Calculating the sample size:

`\to n = (\frac{(Z_{(\alpha)/(2)} * \sigma)}{E})^2`

`= (((1.96 * 0.24 ))/(0.10))^2\\\\= 22.13 \approx 23`

In point b:

`\to \sigma = \$0.24\\\\\to E = \$0.06\\\\\to Z_{(\alpha)/(2)} = Z_(0.025) = 1.96\\\\\to n = (\frac{(Z_{(\alpha)/(2)} * \sigma )}{E})^2\\\\`

`= (((1.96 * 0.24 ))/(0.06))^2\\\\= 61.47 \approx 62`

In point c:

`\to \sigma = \$0.24\\\\\to E =\$0.04\\\\\to Z_{(\alpha)/(2)} = Z_(0.025) = 1.96\\\\\to n = (\frac{(Z_{(\alpha)/(2)} * \sigma )}{E})^2\\\\`

`= (((1.96* 0.24 ))/(0.04))^2 \\\\=138.30 \approx 139`