Respuesta :
Answer:
The margin of error for a 99% confidence interval for the population mean is 1.8025.
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 = 2.575[/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 of the population and n is the size of the sample.
In this problem:
[tex]\sigma = 5.6[/tex]
So
[tex]M = 2.575*\frac{5.6}{\sqrt{64}} = 1.8025[/tex]
The margin of error for a 99% confidence interval for the population mean is 1.8025.
The margin of error for a 99% confidence interval for the population mean is gotten as; M = 1.8032
What is the margin of error?
We are given;
Sample size; n = 64
Standard deviation; σ = 5.6
Confidence level = 99%
Formula for margin of error is;
M = z(σ/√n)
where;
z value at CI of 99% is 2.576
Thus;
M = 2.576(5.6/√64)
M = 1.8032
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