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

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    0
    2021-11-23T20:06:13+00:00

    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] .

    0
    2021-11-23T20:06:22+00:00

    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.

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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).

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