Consider that you have a population greater than 30, then, you can use the normal distribution to determine the margin of error.
Use the following formula:
[tex]\bar{x}\pm Z_{\frac{\alpha}{2}}\frac{s}{\sqrt[]{n}}[/tex]where:
x: mean = 33
s: standard deviation = 2
n = 31
Z: z-value for 98%
The value of Z can be found on a table for the normal distribution. For a margin of error at 98%, you get for Z:
Z = 2.326
Replace the previous values of the parameters into the formula for the margin of error (confidence interval):
[tex]\begin{gathered} 33\pm(2.326)\frac{2}{\sqrt[]{31}}= \\ 33\pm0.83 \end{gathered}[/tex]Then, the margin of error is:
(33.00 - 0.83 , 33.00 + 0.83) = (32.17 , 33.83)
