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Question

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. An investor is considering funding of a new video game. She wants to know the worldwide percentage of people who play video games, so a survey is being planned. How many people must be surveyed in order to be 90% confident that the estimated percentage is within three percentage points of the true population percentage? Assume that about 16% of people play video games (based on a report by Spil Games).

Solution

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Given:

c=90%=0.90c=90\%=0.90

E=3%=0.03E=3\%=0.03

p^=16%=0.16\hat{p}=16\%=0.16

Formula sample size:

p^ known: n=[zα/2]2p^q^E2=[zα/2]2p^(1p^)E2\hat{p}\text{ known: }n=\dfrac{[z_{\alpha/2}]^2\hat{p}\hat{q}}{E^2}=\dfrac{[z_{\alpha/2}]^2\hat{p}(1-\hat{p})}{E^2}

p^ unknown: n=[zα/2]20.25E2\hat{p}\text{ unknown: }n=\dfrac{[z_{\alpha/2}]^2 0.25}{E^2}

For confidence level 1α=0.901-\alpha=0.90, determine zα/2=z0.05z_{\alpha/2}=z_{0.05} using using the normal probability table in the appendix (look up 0.05 in the table, the z-score is then the found z-score with opposite sign):

zα/2=1.645z_{\alpha/2}=1.645

p^\hat{p} is known, then the sample size is (round up to the nearest integer!):

n=[zα/2]2p^(1p^)E2=1.6452×0.16(10.16)0.032405n=\dfrac{[z_{\alpha/2}]^2 \hat{p}(1-\hat{p})}{E^2}=\dfrac{1.645^2\times 0.16(1-0.16)}{0.03^2}\approx 405

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