## 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. Con

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.

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4 days 2021-11-23T20:04:43+00:00 2 Answers 0 views 0

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; ~ 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, is given by;

P(-2.831 < < 2.831) = 0.99

P(-2.831 < < 2.831) = 0.99

P(-2.831 * < < 2.831 * ) = 0.99

P(X bar – 2.831 * < < X bar + 2.831 * ) = 0.99

99% confidence interval for = [ X bar – 2.831 * , X bar + 2.831 * ]

= [ 22.455 – 2.831 * , 22.455 + 2.831 * ]

= [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] .

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 level, that is the subtraction of 1 by the confidence interval divided by 2. So: Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so Now, find M as such In which is the standard deviation of the population and n is the size of the sample.

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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).