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A study of the career paths of hotel general managers sent questionnaires to an SRS of 230 hotels belonging to major U.S. hotel chains. There were 109 responses. The average time these 109 general managers had spent with their current company was 12.08 years. (Take it as known that the standard deviation of time with the company for all general managers is 2.7 years.) (a) Find the margin of error for an 80% confidence interval to estimate the mean time a general manager had spent with their current company: 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: years (c) In general, increasing the confidence level the margin of error (width) of the confidence interval. (Enter: ''DECREASES'', ''DOES NOT CHANGE'' or ''INCREASES'', without the quotes.)

Sagot :

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.

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