Answer:
The sample size needed to be taken is approximately 74.
Step-by-step explanation:
To get alpha; we have 1 - 0.95 = 0.05
Therefore, [tex]z_{\frac{a}{2} } = \frac{0.05}{2} = 0.025[/tex]
The we look up z* in a Standard Normal table where α = 0.025 area in each tail.
From the table, z* = 1.96
From the question variance is 484, then the standard deviation is 22
Standard deviation = [tex]\sqrt{Var(x)} = \sqrt{484} = 22[/tex]
And margin of error is 5 or less
The formula for margin of error is given as:
margin of error = [tex]\frac{(z^{*}) (SD)}{\sqrt{n} }[/tex]
[tex]5 = \frac{1.96 * 22}{\sqrt{n} } \\5 = \frac{43.12}{\sqrt{n} } \\5 * \sqrt{n} = 43.12\\(5)^{2} * (\sqrt{n} )^{2} = (43.12)^2\\25 * n = 1859.3344\\25n = 1859.3344\\\frac{25n}{25} = \frac{1859.3344}{25} \\n = 74.373376\\n = 74[/tex]
The approximate value of n is 74.
