Respuesta :
Answer:
The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547
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 question:
[tex]\sigma = 1.27, n = 85[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]M = 2.575*\frac{1.27}{\sqrt{85}}[/tex]
[tex]M = 0.3547[/tex]
The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547
