Question 1203042: An IQ test is designed so that the mean is 100 and the standard deviation is 10 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95% confidence that the sample mean is within 2 IQ points of the true mean. Assume sigma equals=10 that and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean is 100.
standard deviation is 10.
standard error is 10 / sqrt(n).
critical z-score is plus or minus 1.96 for 95% confidence interval.
when z = 1.96, z-score formula becomes 1.96 = (x-m)/(10/sqrt(n)) which becomes 1.96 = 2 / (10/sqrt(n)).
2 is the margin of error = (x-m)
solve for sqrt(n) to get sqrt(n) = 1.96*10/2 = 9.8
when sqrt(n) = 9.8, standard error = 10/9.8 = 1.020408163.
when margin of error = 2, z-score formula becomes z = 2/1.020408163 = 1.96 which is the critical z-score.
the sample size needs to be at least 9.8^2 = 96.04 = 97 when rounded to the next highest intger.
this will ensure a margin of error to be less than 2.
as a test, let the mean be 100 and the standard deviation be 20 and the sample size be 97.
when the critical z-score is 1.96, the formula becomes 1.96 = (x - 100) / (10 / sqrt(97)).
solve for x to get x = 1.96 * 10 / sqrt(96) + 100 = 101.9900785.
that minus 100 gets you a margin lf error equal to 1.9900785 which is less than 2.
when the critical z-score is -1.96, solving for x gets you x = -1.96 * 10 / sqrt(96) + 100 = 98.00992152.
100 minus that gets you a margin of error equal to 1.9900785 which is less than 2.
a sample size of 97 or greater will get you a margin of error less than 2.
i don't really know if the sample size is resonable or not.
it is greater than 30 which, i believe, is one of the criteria.
based on that, i would say that it is .
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