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Q.8.14

Expert-verifiedFound in: Page 467

Book edition
OER 2018

Author(s)
Barbara Illowsky, Susan Dean

Pages
902 pages

ISBN
9781938168208

Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be % confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones?

The sample size is n = $271$

Given in the question that, an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones

We must locate How many consumers need the company survey in order to be 90 percent confident that the predicted proportion of customers who click on advertising on their cellphones is within five percentage points of the genuine population proportion?

According to the information, We know that confident level is $90$%,

Then,

$\frac{\alpha}{2}=\frac{1-0.90}{2}=0.05$

and

${z}_{\frac{\alpha}{2}}=1.645$

However, we need to know the estimated (sample) proportion p in order to find localid="1650608891485" $n$. Keep in mind that localid="1650608908717" $q=1-p$. However, we don't yet know what localid="1650608912935" $p$ is. Because localid="1650608916792" $p$ and localid="1650608921420" $q$ are multiplied simultaneously, we set them both to localid="1650608941278" $0.5$ because localid="1650608925298" $pq=0.5\times 0.5=0.25$ yields the biggest feasible product.

The error in estimating the true value of localid="1650608945447" $p$ is localid="1650608949079" $E=5\%=0.05$ than from equation (1)

localid="1650608953027" $n={\left(\frac{1.645}{0.05}\right)}^{2}0.25\phantom{\rule{0ex}{0ex}}=270.6025$

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