A local hotel wants to estimate the average age of its guests that are from out-of-state. Preliminary estimates are that standard deviation

Question

A local hotel wants to estimate the average age of its guests that are from out-of-state. Preliminary estimates are that standard deviation of population of guests from out-of-state is 30. How large a sample should be taken to estimate the average age of out-of-state guests with a margin of error no larger than 5 and with a 95% level of confidence? a. 12 b. 11 c. 139 d. 138

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Amelia 2 weeks 2021-10-08T07:50:54+00:00 2 Answers 0

Answers ( )

    0
    2021-10-08T07:52:19+00:00

    Answer:

    n=(\frac{1.960(30)}{5})^2 =138.30 \approx 139

    And if we round up to the nearest integer we got n =139, and the best answer for this case is:

    c. 139

    Step-by-step explanation:

    For this case we have this previous info:

    \sigma = 30 represent the previous estimation for the population deviation

    Confidence =0.95 represent the confidence level

    The margin of error for the true mean is given by this formula:

     ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

    The desired margin of error is ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

    n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

    The critical value for a 95% of confidence interval given now can be founded using the normal distribution. For this case the critical value would be given by z_{\alpha/2}=1.960, replacing into formula (b) we got:

    n=(\frac{1.960(30)}{5})^2 =138.30 \approx 139

    And if we round up to the nearest integer we got n =139, and the best answer for this case is:

    c. 139

    0
    2021-10-08T07:52:23+00:00

    Answer:

    c. 139

    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{1-0.95}{2} = 0.025

    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 1-0.025 = 0.975, so z = 1.96

    Now, find the margin of error 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.

    How large a sample should be taken to estimate the average age of out-of-state guests with a margin of error no larger than 5 and with a 95% level of confidence?

    We need a sample size of n.

    n is found when M = 5, \sigma = 30

    So

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

    5 = 1.96*\frac{30}{\sqrt{n}}

    5\sqrt{n} = 1.96*30

    Simplifying by 5

    \sqrt{n} = 1.96*6

    (\sqrt{n})^{2} = (1.96*6)^{2}

    n = 138.30

    We round up,

    So the correct answer is:

    c. 139

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