Question

Gasoline Prices.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 $.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.a.The desired margin of error is $.10.b.The desired margin of error is $.07.c.The desired margin of error is $.05.

Answer 1

a) Sample size of 96 or higher

b) Sample size of 196 or higher

c) Sample size of 385 or higher

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 $.10.

Sample size of n or higher when
`M = 0.1`. So

`M = z*(\sigma)/(√(n))`

`0.1 = 1.96*(0.5)/(√(n))`

`0.1√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.1)`

`√(n) = 9.8`

`√(n)^(2) = (9.8)^(2)`

`n = 96`

b.The desired margin of error is $.07.

Sample size of n or higher when
`M = 0.07`. So

`M = z*(\sigma)/(√(n))`

`0.07 = 1.96*(0.5)/(√(n))`

`0.07√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.07)`

`√(n) = 14`

`√(n)^(2) = (14)^(2)`

`n = 196`

c.The desired margin of error is $.05.

Sample size of n or higher when
`M = 0.05`. So

`M = z*(\sigma)/(√(n))`

`0.05 = 1.96*(0.5)/(√(n))`

`0.05√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.05)`

`√(n) = 19.6`

`√(n)^(2) = (19.6)^(2)`

`n = 384.1`

Rounding up, sample size of 385 or higher