SOLUTION: A study is conducted to estimate the mean completion time of a task to within 3 minutes 6 seconds of the population mean. Suppose completion time is normally distributed with a sta

Question 1187716: A study is conducted to estimate the mean completion time of a task to within 3 minutes 6 seconds of the population mean. Suppose completion time is normally distributed with a standard deviation of 9 minutes 3 seconds. What minimum sample size is required at the 90.9% confidence level?
You must use a critical z value rounded to two (2) decimal places.
Do not round other intermediate results. Report final answer as an integer.
Minimum sample size required =
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the standard deviation is 9 minutes 3 seconds.
you want the margin of error to be plus or minus 3 minutes 6 seconds.

it probably makes sense to convert everything to seconds.

standard deviation of 9 minutes 3 seconds is equal to 543 seconds.
margin of error of 3 minutes 6 seconds is equal to 186 seconds.

the formula for the standard error is:

s = standard deviation / square root of sample size.

90.9% confidence level gives you a two tail alpha of 100 - 90.9 = 9.1% / 2 = 4.55% on each end of the confidence interval.

that's equivalent to an area of .0455 on each end of the confidence interval.

that give you a critical z-score of plus or minus 1.690146137.

you could probably round that off to 1.69 but we'll keep it at full value from the calculator for now.

the z-score formula is z = (x -m) / s

z is the z-score
x is the raw score
m is the raw mean
s is the standard error.

the standard error is equal to the standard deviation divided by the square root of the sample size.

since the standard deviation is equal to 543 seconds, then the standard error is equal to 543 / sqrt(sample size).

if we want the margin of error to be equal to 186 seconds, then (x - m) must be equal to 186.

the z-score formula becomes:

1.690146137 = 186 / (543 / sqrt(sample size).

this is equivalent to:

1.690146137 = 186 * sqrt(sample size) / 543,

solve for sqrt(sample size) to get:

sqrt(sample size) = 1.690146137 * 543 / 186.

that becomes:

sqrt(sample size) = 4.934136303.

square both sides of that equation to get:

sample size = 24.34570106.

calculate the standard error using the square root of the sample size to get:

standard error = 543 / 4.934136303 = 110.0496554.

with that standard error, you should be able to calculate the margin of error to get what you wanted it to be.

on the high side, z = (x - m) / s becomes:

1.690146137 = (x - m) / 110.0496554.

solve for (x - m) to get:

(x - m) = 186.

on the low side, z = (x - m) / s becomes:

1.690146137 = (x - m) / 110.0496554.

solve for (x - m) to get:

(x - m) = -186.

looks like the margin of error is exactly what you wanted.

this margin or error should be the same regardless of what the mean is, as ong as the standard error is equal to 110.0496554.

to see if that's true, pick any mean that looks reasonable and solve for the confidence interval.

assume the mean is 5000 seconds.

the critical z-score formula becomes:

1.690146137 = (x - 5000) / 110.0496554.

solve for x to get:

x = 1.690146137 * 110.0496554 + 5000 = 5186.

that's 186 greater than 5000.

on the low side, the formula becomes:

-1.690146137 = (x - 5000) / 110.0496554.

solve for x to get:

x = -1.690146137 * 110.0496554 + 5000 = 4814

5000 - 4814 = 186, which is what you wanted.

your solution is that the minimum sample size required is 24.34570106.

round that the next highest integer to get:

the minimum sample size required is 25.

that's what i get.

let me know if you have any questions.

theo

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