Answer:
The minimum sample size is n = 75 so that the desired margin of error is 5 or less.
Step-by-step explanation:
We are given the following in the question:
Population variance = 484
Standard deviation =
[tex]\sigma^2 = 484\\\sigma =\sqrt{484} = 22[/tex]
Confidence level = 0.95
Significance level = 0.05
Margin of error = 5
Formula:
Margin of error =
[tex]E = z\times \dfrac{\sigma}{\sqrt{n}}\\\\n = (\dfrac{z\times \sigma}{E})^2[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting values, we get
[tex]E\leq 5\\\\1.96\times \dfrac{22}{\sqrt{n}} \leq 5\\\\\sqrt{n} \geq 1.96\times \dfrac{22}{5}\\\\n \geq (1.96\times \dfrac{22}{5})^2\\\\n \geq 74.373[/tex]
Thus, the minimum sample size is n = 75 so that the desired margin of error is 5 or less.
