Given:
In a survey, the sample size = 32
The results were roughly bell-shaped.
The mean = μ = 36
The standard deviation = σ = 6
We will find the margin of error at a 99% confidence level.
We will use the following formula:
[tex]e=z*\frac{\sigma}{\sqrt{n}}[/tex]The value of the z-score that corresponds to a 99% confidence level = 2.575829
Substitute with data:
[tex]e=2.575829*\frac{6}{\sqrt{32}}=2.73208[/tex]Rounding to two decimal places.
So, the answer will be:
The margin of error = 2.73
