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

Answer:

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

Step-by-step explanation:

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

$\alpha = \frac{1-0.9}{2} = 0.05$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

$20 = 1.645*\frac{50}{\sqrt{n}}$

$20\sqrt{n} = 50*1.645$

$\sqrt{n} = \frac{50*1.645}{20}$

$(\sqrt{n})^2 = (\frac{50*1.645}{20})^2$

$n = 16.9$

Rounding up

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