Question

The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. The US EIA updates its estimates of average gas prices on a weekly basis. Assume the standard deviation is $0.20 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 your answers up to the nearest whole number.) (a) The desired margin of error is $0.10. 16 Changed: Your submitted answer was incorrect. Your current answer has not been submitted. (b) The desired margin of error is $0.06. 43 Correct: Your answer is correct. (c) The desired margin of error is $0.04.

Answer 1

Answer:

The answer is "16, 43, and 96".

Step-by-step explanation:

Given:

$\sigma = 0.20\\\\c = 95\% = 0.95\\\\\therefore \alpha = 1- c = 1- 0.95 = 0.05\\\\\therefore \frac{\alpha}{2} = 0.025$

Using Z table:

$\therefore Z_{\frac{\alpha}{2}} = 1.96$

For point a:

$E = 0.10\\\\n=(\frac{Z_{\frac{\alpha}{2} \times \sigma}}{E})^2$

    $= (\frac{(1.96\times 0.20)}{0.10})^2\\\\= 15.3664 \approx 16$

so, The Sample size (n) = 16

For point b:

$E = 0.06\\\\n=(\frac{Z_{\frac{\alpha}{2} \times \sigma}}{E})^2$

   $= (\frac{(1.96\times 0.20)}{0.06})^2\\\\= 42.6844444444\approx 43$

For point c:

$E = 0.04\\\\n=(\frac{Z_{\frac{\alpha}{2} \times \sigma}}{E})^2$

    $= (\frac{(1.96\times 0.20)}{0.04})^2\\\\= 96.04\approx 96$