Question 1204120
population mean proportion is .48.
98% confidence interval with two tails requires critical z-score of plus or minus 2.326347.
margin of error is plus or minus .05.
z-score formula is z = (x-m)/s
high side z-score = 2.326347.
(x-m) is equal to .05 which is the high side margin of error.
s = standard error = sqrt(.48 * .52 / n) = sqrt(.2496/n)
n is the sample size.
z-score formula becomes 2.326347 = .05 / sqrt(.2496/n)
mutiply both sides of the equation by sqrt(.2496/n) and divide both sides of the formula by 2.326347 to get:
sqrt(.2496/n) = .05 / 2.326347
square both sides of the equation to get:
.2496/n = (.05/2.326347)^2
solve for n to get:
n = .2496 / (.05/2.326347)^2 = 540.323134.
standard error becomes equal to sqrt(.2496/540.323134) = .021493.
z-score formula becomees:
2.326347 = (x-m) / .021493.
solve for (x-m) to get:
(x-m) = 2.326347 * .021493 = .050000.
that's your margin of error.


use the calculator at <a href = "https://davidmlane.com/hyperstat/z_table.html" target = "_blank">https://davidmlane.com/hyperstat/z_table.html</a> to test it out.


calculator says that 98% confidence interval is between .43 and .53.
.48 minus .43 = .05
.53 minus .48 = .05
margin of error is .05 as desired.


here are the results from that calculator.


<img src = "http://theo.x10hosting.com/2023/100831.jpg">


your sample size needs to be an integer, so round 540.323134 to 541 and that's your solution.
you will get a margin of error slightly less than .05 when you do that.
your revised standared error will be equal to sqrt(.2496/541) = .021479.
your margin of error becomes .021479 * 2.326347 = .049970 which is slightly less than .05 = within .05, as required.