Question

Hi, can you help me answer this question please, thank you!

Answer 1

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:

`\bar{x}\pm Z_{(\alpha)/(2)}\frac{s}{\sqrt[]{n}}`

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):

`\begin{gathered} 33\pm(2.326)\frac{2}{\sqrt[]{31}}= \\ 33\pm0.83 \end{gathered}`

Then, the margin of error is:

(33.00 - 0.83 , 33.00 + 0.83) = (32.17 , 33.83)