Question 1188882: A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 90% confident that her estimate is within 4 ounces of the true mean? Assume that s = 7 ounces based on earlier studies.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i get the following:
standard error = standard deviation divided by square root of sample size.
when the standard deviation is 7, then the standard error is 7/sqrt(n).
n is the sample size.
critical z-score at 90% confidence interval is plus or minus 1.645.
z-score formula is z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard error.
you are looking for (x - m) to be equal to plus or minus 4.
on the high side, your formula becomes:
1.645 = 4 / (7 / sqrt(n)).
multiply both sides of this equation by 7 / sqrt(n) to get:
1.645 * 7 / sqrt(n) = 4.
multiply both sides of this equation by sqrt(n) and divide both sides of this equation by 4 to get:
1.645 * 7 / 4 = sqrt(n)
solve for sqrt(n) t9 get:
sqrt(n) = 2.87875.
solve for standard error to get:
standard error = 7 / sqrt(n) = 7 / 2.87875 = 2.431610942.
that's the standard error that should get you a margin of error of plus or minus 4.
on the high side, (x - m) would be equal to 4.
on the low side, (x - m) would be equal to -4.
1.645 * 2.431618942 = 4
-1.645 * 2.431618942 = -4
this result was achieved with a sample size of 2.87875 ^ 2 = 8.287201563.
since a sample size can't be a fraction of a whole number, then you need to go up to the next highest integer to get a minimum sample size of at least 9.
with a sample size of at least 9, your margin of error will be within plus or minus 4.
in fact, it will be within less than plus or minus 4.
with a sample size of 9, standard error = 7 / sqrt(sample size) = 7/sqrt(9) = 7/3 = 2.333333333.
1.645 = (x - m) / 2.333333333.....
(x - m) = 1.645 * 2.3333333..... = 3.838333333.....
-1.645 = (x - m) / 2.3333333.....
(x - m) = -1.645 * 2.333333..... = -3.83833333.....
the mean can be anything and you will always get a margin of error within plus or minus 3.83833333..... as long as the sample size is greater than or equal to 9.
as an example, assume the mean is 100.
on the high side, 1.645 = (x - 100) / 2.33333.....
solve for x to get:
X = 1.645 * 2.3333..... + 100 = 103.8383333.....
on the low side, x = -1.645 * 2.333..... + 100 = 96.16166666.....
103 - 100 = 3.83833333......
100 - 96.16166666..... = 3.83833333.....
note that, every time you increase the sample size, the standard error gets smaller as long as the standard deviation remains the same.
it has to be recalculated every time the sample size is changed.
as another example, assume the sample size is 16.
the standard error becomes 7 / sqrt(16) = 7/4 = 1.75.
1.645 = (x - m) / 1.75 gets you (x - m) = 1.75 * 1.645 = 2.87875.
the standard error gets smaller and the margin of error becomes smaller as well.
the solution to this problem is that the sample size needs to be greater than or equal to 9.
this allows the margin of error to be less than 4.
let me know if you have any questions.
theo
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