Question

Please show step by step solution

Answer 1

Answer:

b) 74

The sample size that needs to be taken if the desired margin of error is 5 or less is 74

Step-by-step explanation:

Explanation:-

Given population variance σ² = 484

                                            σ = √484

                                            σ = 22

The level of significance ∝=0.95

The z-score of 0.95 level of significance = 1.96

Given Margin of error = 5

we know that the margin of error is determined by

$M.E = \frac{Z_{\alpha }S.D }{\sqrt{n} }$

cross multiplication, we get

$\sqrt{n} = \frac{Z_{\alpha }S.D }{M.E}$

$\sqrt{n} = \frac{1.96 X22 }{5} = 8.624$

Squaring on both sides, we get

n = (8.624) ²

n = 74.37 ≅74

Conclusion:-

The sample size that needs to be taken if the desired margin of error is 5 or less is 74