“A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percent. They want

Question

“A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percent. They want the estimate to be at the 90 percent confidence level and within 2 percent of the actual proportion. What sample size is needed

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Arianna 1 week 2021-11-25T17:12:03+00:00 1 Answer 0 views 0

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    2021-11-25T17:14:01+00:00

    Answer:

    The company should take a sample of 148 boxes.

    Step-by-step explanation:

    Hello!

    The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

    They estimated a “pilot” proportion of p’=0.20

    And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

    The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is “point estimation” ± “margin of error”

    [p’ ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

    Where

    p’ is the sample proportion/point estimator of the population proportion

    Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

    Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

    So

    d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

    d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

    (d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

    n*(d*Z_{1-\alpha /2})^2= p'(1-p')

    n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

    n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

    n= 147.28 ≅ 148 boxes.

    I hope it helps!

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