Respuesta :
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 [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].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find 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.
[tex]M = 2.575*\frac{18.52}{\sqrt{22}} = 10.167[/tex]
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:
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;
[tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
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, [tex]\mu[/tex] is given by;
P(-2.831 < [tex]t_2_1[/tex] < 2.831) = 0.99
P(-2.831 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.831) = 0.99
P(-2.831 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.831 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.99
P(X bar - 2.831 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.831 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ X bar - 2.831 * [tex]{\frac{s}{\sqrt{n} }[/tex] , X bar + 2.831 * [tex]{\frac{s}{\sqrt{n} }[/tex] ]
= [ 22.455 - 2.831 * [tex]{\frac{18.52}{\sqrt{22} }[/tex] , 22.455 + 2.831 * [tex]{\frac{18.52}{\sqrt{22} }[/tex] ]
= [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] .
