Question 1139622: Can someone help with explain how to properly solve this equation?
People were polled on how many books they read the previous year. Initial survey results indicate that s=11.8
How many subjects are needed to estimate the mean number of books read the previous year within four books with 95% confidence?
How many subjects are needed to estimate the mean number of books read the previous year within two books with 95% confidence?
How many subjects are needed to estimate the mean number of books read the previous year within four books with 99% confidence?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i think it works like this.
i believe what you called s is the standard deviation.
i will assume that's true, since i don't think i can solve the problem otherwise.
you would use the z-score for 95% confidence level and 99% confidence level.
z-score for 95% confidence level is plus or minus 1.96.
z-score for 99% confidence level is plus or minus 2.576.
these are both rounded to 3 decimal digits, which is usually the detail you want to get to.
the z-score formula is z = (x - m) / s.
z is the z-score
x is the actual score
m is the actual mean
s is the standard error, not to be confused with the standard deviation.
the standard error formula is s = standard deviation / square root of sample size.
in algebraic notation that i use, this would be shown as s = sd / sqrt(n).
with a standard deviation of 11.8, this becomes s = 11.8 / sqrt(n).
when you want to estimate the number of books within plus or minus 4 at 95% confidence level, the formula for z-score becomes as follows:
z = (x - m) / s becomes 1.96 = 4 / s
we are working with the high side of the z-score.
since the normal distribution is symmetric, we don't really need to do the low side also, because finding n for the high side will also get you the same number for the low side as i will show you below.
the margin of error is equal to (x - m) which is 4 in this case.
our formula has become 1.96 = 4 / s.
since s = 11.8/sqrt(n), then the formula becomes 1.96 = 4 / (11.8/sqrt(n)).
this becomes 1.96 = 4 * sqrt(n) / 11.8
solve for sqrt(n) to get sqrt(n) = 1.96 * 11.8 / 4 = 5.782.
square both sides of sqrt(n) = 5.782 and you get n = 33.431524.
normally you would then round up to the next highest integer, but i'll leave it as is because that gets your answer to be a margin of error of exactly 4.
if you were to also work with the low side z-score, then the formula for z-score becomes as follows:
-1.96 = -4 * sqrt(n) / 11.8
solve for sqrt(n) to get sqrt(n) = -1.96 * 11.8 / -4 which becomes sqrt(n) = 5.782 which is the same as we got when working with the high side z-score.
bottom line is, if we want the margin of error to be plus or minus 4 books at 95% confidence interval, then we need a sample size of 33.431524.
from what i can see, the mean can be anything as long as the difference between the mean and the actual number of books is 4 books and the standard deviation is 11.81.
for example:
assume the mean is 10 books and the standard deviation is 11.8 and the sample size is 33.431524 which makes the standard error equal to 2.040816327.
at 95% confidence level, z = (x - m) / s becomes 1.96 = (x - 10) / 2.040816327.
solve for x to get x = 1.96 * 2.040816327 + 10 = 14 which is a difference of 4 from the mean.
on the low side, z = (x - m) / s becomes -1.96 = (x - 10) / 2.040816327.
solve for x to get x = -1.96 * 2.040816327 + 10 = 6 which is a difference of 4 from the mean.
assume the mean is 5000 books and the standard deviation is 11.8 and the sample size is 33.431524 which makes standard error equal to 2.04081637.
at 95% confidence level, z = (x - m) / s becomes 1.96 = (x - 5000) / 2.040816327.
solve for x to get x = 1.96 * 2.040816327 + 5000 = 5004 which is a difference of 4 from the mean.
as long as the standard deviation is 11.8 and the sample size is 33.431524, the standard error will be 2.040816327 and you will get a margin of error of plus or minus 4 at 95% confidence level.
if you want the margin of error to be plus or minus 2 at 95% confidence level, then the z-score formula of 1.96 = 4 / (11.8/sqrt(n)) becomes:
1.96 = 2 / (11.8/sqrt(n)).
it's the same formula except you are replacing the difference of 4 with a difference of 2.
this becomes 1.96 = 2 * sqrt(n) / 11.8.
solve for sqrt(n) to get sqrt(n) = 1.96 * 11.8 / 2 = 11.564.
square both sides to get n = 133.726096.
the standard error becomes 11.8 / sqrt(133.726096) = 1.020408163.
the mean can be anything as long as the standard error is 1.020408163, while the standard error will always be 1.020408163 with a standard deviation of 11.8 and a sample size of 133.726096.
for example, suppose the mean is 500.
then the z-score formula for the high side is 1.96 = (x - 500) / 1.020408163.
solve for x to get x = 1.96 * 1.020408163 + 500 = 502
on the low side, the z-score formula becomes -1.96 = (x - 500) / 1.020408163.
solve for x to get x = -1.96 * 1.020408163 + 500 = 498.
to summarize the results:
at 95% confidence interval, .....
if you want the margin of error to be plus or minus 4, the sample size needs to be equal to 33.431524.
if you want the margin of error to be plus or minus 2, the sample size needs to be equal to 133.726096
if you want to find the margin of error to be plus or minus 4 at 99% confidence level, then you do the same procedure using the z-score for 99% confidence level, which is plus or minus 2.576.
you only need to work with the high side z-score because the sample size will be the same whether or not you work with the low side z-score because the normal distribution curve is symmetric about the mean.
once again, your formula for standard error is s = standard deviation / square root of sample size which is shown as s = sd / sqrt(n).
your standard deviation is equal to 11.8 as before.
your z-score now becomes plus or minus 2.576
your margin of error is plus or minus 4.
working with the high side z-score, the z-score formula of z = (x - m) / s becomes 2.576 = 4 / s.
since s = 11.8 / sqrt(n), this formula becomes 2.576 = 4 / (11.8 / sqrt(n)) which then becomes 2.576 = 4 * sqrt(n) / 11.8
solve for sqrt(n) to get sqrt(n) = 2.576 * 11.8 / 4 = 7.5992.
square both sides of that equation to get n = 57.74784064.
when n = 57.747840964, s = 11.8 / sqrt(57.74784064) = 1.552795031.
once again, the mean can be anything as long as the z-score is 2.576 and the standard error is 1.552795031 which was derived from a standard deviation of 11.8 and a sample size of 57.74784064.
for example, assume the mean is 349 books.
z-score formula becomes 2.576 = (x - 349) / 1.552795031.
solve for x to get x = 2.576 * 1.552795031 + 349 = 353, a difference of 4 from the mean.
low side z-score formula becomes -2.576 = (x - 349) / 1.552795031.
solve for x to get x = -2.576 * 1.552795031 + 349 = 345, a difference of 4 from the mean.
to summarize the results again, adding the final result using 99% confidence level, your solution should be, if i did this correctly:
at 95% confidence interval, .....
if you want the margin of error to be plus or minus 4, the sample size needs to be equal to 33.431524.
if you want the margin of error to be plus or minus 2, the sample size needs to be equal to 133.726096
at 99% confidence interval, .....
if you ant the margin of error to be plus or minus 4, the sample size needs to be equal to 57.74784064.
once again, you would probably round to the next higher or lower integer since the sample size has to be a whole number.
i believe that rounding up to the next higher integer is what is normally recommended.
this insures your margin of error is less than or equal to the desired margin of error.
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