Answer:
The margin of error for a 99% confidence interval was of $1.84.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.325[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation(square root of the variance) and n is the size of the sample. In this question, [tex]\sigma = \sqrt{375}, n = 600[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}} = 2.325\frac{\sqrt{375}}{\sqrt{600}} = 1.84[/tex]
The margin of error for a 99% confidence interval was of $1.84.
