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

Answer:

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

Step-by-step explanation:

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

$\alpha = \frac{1-0.99}{2} = 0.005$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 2.575*\frac{18.52}{\sqrt{22}} = 10.167$

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).

Answer 2

Answer:

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

Step-by-step 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;

             $\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }$  ~  $t_n_-_1$

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 < $\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }$ < 2.831) = 0.99

P(-2.831 * ${\frac{s}{\sqrt{n} }$ < ${Xbar - \mu}$ < 2.831 * ${\frac{s}{\sqrt{n} }$ ) = 0.99

P(X bar - 2.831 * ${\frac{s}{\sqrt{n} }$ < $\mu$ < X bar + 2.831 * ${\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ X bar - 2.831 * ${\frac{s}{\sqrt{n} }$ , X bar + 2.831 * ${\frac{s}{\sqrt{n} }$ ]

                                           = [ 22.455 - 2.831 * ${\frac{18.52}{\sqrt{22} }$ , 22.455 + 2.831 * ${\frac{18.52}{\sqrt{22} }$ ]

                                            = [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] .