Question

Christopher wants to estimate the mean amount of carbon dioxide that is emitted by burning 1\text{ L}1 L1, start text, space, L, end text of a new type of gasoline. He plans on burning 1\text{ L}1 L1, start text, space, L, end text at a time and measuring the resulting emissions. He'll repeat this process for a sample of nnn attempts and construct a confidence interval for the mean. He wants the margin of error to be no more than 202020 grams at a 90\%90%90, percent level of confidence. Preliminary data suggests that \sigma=50σ=50sigma, equals, 50 grams is a reasonable estimate for the standard deviation of the emissions from burning 1\text{ L}1 L1, start text, space, L, end text of this type of gasoline. Which of these is the smallest approximate sample size required to obtain the desired margin of error?

Answer 1

The smallest sample size required to obtain the desired margin of error is of 17.

Explanation:

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

`\alpha = (1-0.9)/(2) = 0.05`

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
`1-0.05 = 0.95`, so
`z = 1.645`

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.

Which of these is the smallest approximate sample size required to obtain the desired margin of error?

The desired margin of error is 20, so
`M = 20`

We have that
`\sigma = 50`.

The smallest sample size is n. So

`M = z*(\sigma)/(√(n))`

`20 = 1.645*(50)/(√(n))`

`20√(n) = 50*1.645`

`√(n) = (50*1.645)/(20)`

`(√(n))^2 = ((50*1.645)/(20))^2`

`n = 16.9`

Rounding up

The smallest sample size required to obtain the desired margin of error is of 17.