Respuesta :
Answer:
a) The margin of error is 0.3643 years.
b) The margin of error is 0.6514 years.
c) INCREASES
Step-by-step explanation:
(a) Find the margin of error for an 85% confidence interval to estimate the mean time a general manager had spent with their current company:
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 - 0.85}{2} = 0.075[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.075 = 0.925[/tex], so Z = 1.44.
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.
[tex]M = 1.44\frac{4}{\sqrt{250}} = 0.3643[/tex]
The margin of error is 0.3643 years.
(b) Find the margin of error for a 99% confidence interval to estimate the mean time a general manager had spent with their current company:
[tex]\alpha = \frac{1 - 0.99}{2} = 0.005[/tex]
That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575
[tex]M = 2.575\frac{4}{\sqrt{250}} = 0.6514[/tex]
The margin of error is 0.6514 years.
(c) In general, increasing the confidence level the margin of error (width) of the confidence interval.
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
Increase of confidence level -> Increases z -> Increases margin of error.
So Increases is the answer.
