Question

A study of the career paths of hotel general managers sent questionnaires to an SRS of 250 hotels belonging to major U.S. hotel chains. There were 149 responses. The average time these 149 general managers had spent with their current company was 13.26 years. (Take it as known that the standard deviation of time with the company for all general managers is 4 years.) (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: 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.)

Answer 1

a) The margin of error is 0.3643 years.

b) The margin of error is 0.6514 years.

c) INCREASES

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
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.85)/(2) = 0.075`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.075 = 0.925`, so Z = 1.44.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.44(4)/(√(250)) = 0.3643`

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:

`\alpha = (1 - 0.99)/(2) = 0.005`

That is z with a pvalue of
`1 - 0.005 = 0.995`, so Z = 2.575

`M = 2.575(4)/(√(250)) = 0.6514`

The margin of error is 0.6514 years.

(c) In general, increasing the confidence level the margin of error (width) of the confidence interval.

`M = z(\sigma)/(√(n))`

Increase of confidence level -> Increases z -> Increases margin of error.

So Increases is the answer.