Question

There is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area, with a standard deviation of 18.52. Construct a 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers.

Answer 1

99 percent confidence interval for the true mean is [11.28 , 33.63] .

Explanation:

We are given that there is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area, with a standard deviation of 18.52.

The Pivotal quantity for 99% confidence interval is given by;

`(Xbar - \mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where, X bar = sample mean = 22.455

s = sample standard deviation = 18.52

n = sample size = 22

So, 99% confidence interval for the true mean number of full-time employees at office furniture dealers,
`\mu` is given by;

P(-2.831 <
`t_2_1` < 2.831) = 0.99

P(-2.831 <
`(Xbar - \mu)/((s)/(√(n) ) )` < 2.831) = 0.99

P(-2.831 *
`{(s)/(√(n) )` <
`{Xbar - \mu}` < 2.831 *
`{(s)/(√(n) )` ) = 0.99

P(X bar - 2.831 *
`{(s)/(√(n) )` <
`\mu` < X bar + 2.831 *
`{(s)/(√(n) )` ) = 0.99

99% confidence interval for
`\mu` = [ X bar - 2.831 *
`{(s)/(√(n) )` , X bar + 2.831 *
`{(s)/(√(n) )` ]

= [ 22.455 - 2.831 *
`{(18.52)/(√(22) )` , 22.455 + 2.831 *
`{(18.52)/(√(22) )` ]

= [11.28 , 33.63]

Therefore, 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers is [11.28 , 33.63] .

Answer 2

The 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers is (12.288, 32.622).

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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

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

So it is z with a pvalue of
`1-0.005 = 0.995`, so
`z = 2.575`

Now, find 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 = 2.575*(18.52)/(√(22)) = 10.167`

The lower end of the interval is the mean subtracted by M. So it is 22.455 - 10.167 = 12.288

The upper end of the interval is the mean added to M. So it is 22.455 + 10.167 = 32.622

The 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers is (12.288, 32.622).