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?

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 level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which 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

We have that .

The smallest sample size is n. So

Rounding up

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

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

b.-4 units

all you   do is count how many times you move