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 $0.25 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. (b) The desired margin of error is $0.06. (c) The desired margin of error is $0.05.

Answer 1

a) 25

b) 67

c) 97

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample. In this problem,
`\sigma = 0.25`

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

`M = z*(\sigma)/(√(n))`

`0.1 = 1.96*(0.25)/(√(n))`

`0.1√(n) = 1.96*0.25`

`√(n) = (19.6*0.25)/(0.1)`

`(√(n))^(2) = ((19.6*0.25)/(0.1))^(2)`

`n = 24.01`

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

`M = z*(\sigma)/(√(n))`

`0.06 = 1.96*(0.25)/(√(n))`

`0.06√(n) = 1.96*0.25`

`√(n) = (19.6*0.25)/(0.06)`

`(√(n))^(2) = ((19.6*0.25)/(0.06))^(2)`

`n = 66.7`

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

`M = z*(\sigma)/(√(n))`

`0.05 = 1.96*(0.25)/(√(n))`

`0.05√(n) = 1.96*0.25`

`√(n) = (19.6*0.25)/(0.05)`

`(√(n))^(2) = ((19.6*0.25)/(0.05))^(2)`

`n = 96.04`

Rounding up, 97