Mileage test are conducted for a particular model of automobile. If a 98% confid

Question

Mileage test are conducted for a particular model of automobile. If a 98% confidence interval with a margin of error of 1 mile per gallon is desired, how many automobiles should be used in the test? Assume that preliminary mileage tests indicate the standard deviation is 2.6 miles per gallon.

Need details of how you get margin of error and what you do with standard deviation.
I have the answer, but have no clue where numbers come from or even if it is correct. I got this from this website from someone else but this is all they gave me.

1 = 2.33*2.6/sqrt(n) so n = (2.33*2.6)2 = 37

I need more explanation of where the 2.33 came from and others.

Explanation / Answer

98% confidence means that that we want to exclude the most extreme 2%, which is the lowest 1% and highest 1%.

So we need to look up 0.01 in a z table and get z=-2.33, but z* has to be positive, so we use 2.33.

The margin or error formula is ME = z* /sqrt(n)

We want ME = 1 according to the problem, we found z* = 2.33 above, and the problem tells us that the SD is 2.6.

So we plug all of that in and solve for n to get n = 37.

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