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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.

User Zaw Lin
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1 Answer

6 votes

Answer:

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

User RyanfaeScotland
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