Answer :
Answer:
a)
The margin of error is of 0.33 years.
b)
The margin of error is of 0.67 years.
c)
INCREASES
Step-by-step explanation:
In this question, we have that:
Sample of 109 means that [tex]n = 109[/tex]
The standard deviation of time with the company for all general managers is 2.7 years, which means that [tex]\sigma = 2.7[/tex]
Question a:
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 - 0.8}{2} = 0.1[/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 - 0.1= 0.9[/tex], so Z = 1.28.
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.
So
[tex]M = 1.28\frac{2.7}{\sqrt{109}} = 0.33[/tex]
The margin of error is of 0.33 years.
Question b:
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].
That is z with a pvalue of [tex]1 - 0.005= 0.995[/tex], so Z = 2.575. So
[tex]M = 2.575\frac{2.7}{\sqrt{109}} = 0.67[/tex]
The margin of error is of 0.67 years.
(c) In general, increasing the confidence level the margin of error (width) of the confidence interval.
We have that:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
That is, M and z are direct proportional, and since z increases when the confidence level increases, the marign of error will increase. So the answer is INCREASES.