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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.

2 Answers

5 votes

Answer:

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] .

User Ameen Maheen
by
6.4k points
3 votes

Answer:

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).

User Jesper Madsen
by
6.6k points
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