Question 1203042
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 .