Answer:
The sample needs to be at least n = 897 in size.
Step-by-step explanation:
Given
Confidence Level = 95%
Let p = those who had laptop
p = 70% = 0.7
Margin of Error = 3% = 0.03
Let n = required sample.
To estimate the proportion of laptop owners with 95% confidence, first the z-value is calculated. [tex]z_{(a/2)}[/tex]
Given that confidence level = 95%
α = 1 − 95 %
α = 1 − 0.95
α = 0.05
So;
[tex]z_{(a/2)}[/tex] = [tex]z_{(0.05/2)}[/tex]
[tex]z_{(a/2)}[/tex] = [tex]z_{(0.025)}[/tex]
From the z table
[tex]z_{(0.025)} = 1.96[/tex]
Using formula for margin of error
[tex]M.E = z_{(0.025)} * \sqrt{\frac{pq}{n} }[/tex]
where p + q =1
q = 1 - p
q = 1 - 0.7
q = 0.3
So,
[tex]M.E = z_{(0.025)} * \sqrt{\frac{pq}{n} }[/tex]
[tex]0.03 = 1.96 * \sqrt{\frac{0.7 * 0.3}{n} }[/tex] --- Make n the subject of formula
First, square both sides
[tex]0.03^{2} = 1.96^{2} * {\frac{0.7 * 0.3}{n} }[/tex] ------ Multiply both sides by [tex]{\frac{n}{0.03^{2}}}[/tex]
[tex]0.03^{2} * {\frac{n}{0.03^{2}}} = 1.96^{2} * {\frac{0.7 * 0.3}{n} } * {\frac{n}{0.03^{2}}}[/tex]
[tex]n = 1.96^{2} * {\frac{0.7 * 0.3}{0.03^{2}} }[/tex]
[tex]n = {\frac{0.806736}{0.0009} }[/tex]
[tex]n = 896.473[/tex]
Hence, the sample needs to be at least n = 897 in size.
