Question 1136947: "According to estimates made by the General Accounting Office, the Internal Revenue Service (IRS) answered 18.3 million telephone inquiries during a recent tax season, and 17% of the IRS offices provided wrong answers. These estimates were based on data collected from sample calls to numerous IRS offices. How many IRS offices should be randomly selected and contacted in order to estimate the proportion of IRS offices that fail to correctly answer questions about gift taxes with a 90% confidence interval of width 0.06?"
I have no idea how to go about solving this. It is a bonus question on my homework and I think a similar question may appear on my test. Thank you in advance!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! this is a proportion type study.
in a proportion type study, the mean is the proportion and the standard deviation of the distribution of sample means is sqrt (p * q / n).
p is the proportion mean = .17
q is equal to 1 - p = .83.
the standard deviation of the distribution of sample means is sqrt(.17 * .83 / n).
n is the sample size, which in this case is the number of offices used for the sample.
a 90% confidence interval gives you a z-score of plus or minus 1.645.
a confidence interval width of .06 gives you an interval of plus or minus .03 from the mean.
you can use the normal distribution calculator found online at http://davidmlane.com/hyperstat/z_table.html
to do this, you used the value from an area portion of the calculator.
the following display shows the use of this calculator to find the critical z-scores for a confidence interval of 90% = .9
once again,.....
in a proportion type study, the mean is p and the standard deviation of the distribution of sample means is sqrt(p * q / n).
we'll call the standard deviation of the distribution of sample means s.
since p = .17 and q = 1 - p = .83, then s = sqrt(.17 * .83 / n)
n is the sample size which is equal to the number of offices sampled.
the 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 deviation of the distribution of sample means.
we'll use the high z-score to solve for the sample size.
z = (x - m) / s becomes 1.645 = (.20 - .17) / s
the raw score of .20 is .03 above the mean of .17.
that's the width we're looking for because plus or minus .03 give us a confidence interval width of .06.
the z-score formula simplifies to 1.645 = .03 / s
we solve for s to get s = .03 / 1.645.
that becomes s = .0182370821.
since s = sqrt(.17 * .83 /n), we get sqrt(.17 * .83 / n) = .0182370821.
square both sides of this equation to get .17 * .83 / n = 3.325911623 * 10 ^ -4.
multiply both sides of this equation by n and divide both sides of this equation by 3.325911623 * 10 ^ -4 to get n = (.17 * .83) / (3.325911623 * 10 ^ -4) which results in n = 424.2445861.
when n = 424.2445861, sqrt(.17 * .83 / n) becomes equal to .0182370821.
that's the standard deviation of the distribution of sample means when the mean is .17 and the number of offices is 424.2445861.
at the 90% confidence level, the sample proportion will be plus or minus .03 from .17 which means it will be from .14 to .20.
using the value from an area portion of the online normal distribution calculator, enter mean of.17 and standard deviation of .0182371 and enter .9 as the confidence interval and then solve for the raw scores associated with that.
you will get the raw score is between .14 and .20 which is exactly what you wanted, as shown below.
if you use the area from a value portion of the calculator, you would enter the mean of .17 and the standard deviation of .0182371 and select between and enter .14 to .17.
the calculator will tell you that the area is .9 as shown below.
we calculated the number of offices to be 424.245861.
since that is impossible, you would round to the next highest integer to get the number of offices in the sample to be 425.
that would ensure that the interval width was less than or equal to .06
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