Question

t is known that the population variance equals 484. With a 0.95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is 25 74 189 75 None of the above answers is correct

Answer 1

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, $z_{\frac{a}{2} } = \frac{0.05}{2} = 0.025$

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 = $\sqrt{Var(x)} = \sqrt{484} = 22$

And margin of error is 5 or less

The formula for margin of error is given as:

margin of error = $\frac{(z^{*}) (SD)}{\sqrt{n} }$

$5 = \frac{1.96 * 22}{\sqrt{n} } \\5 = \frac{43.12}{\sqrt{n} } \\5 * \sqrt{n} = 43.12\\(5)^{2} * (\sqrt{n} )^{2} = (43.12)^2\\25 * n = 1859.3344\\25n = 1859.3344\\\frac{25n}{25} = \frac{1859.3344}{25} \\n = 74.373376\\n = 74$

The approximate value of n is 74.