Question

A recent poll of 1500 new home buyers found that 60% hired a moving company to help them move to their new home. Find the margin of error for this poll if we want 95% confidence in our estimate of the percentage of new home buyers who hired movers.

Answer 1

Answer:

The margin of error M.O.E = 2.5%

Step-by-step explanation:

Given that;

The sample size = 1500

The sample proportion $\hat p$ = 0.60

Confidencce interval = 0.95

The level of significance ∝ = 1 - C.I

= 1 - 0.95

= 0.05

The critical value:

$Z_{\alpha/2} = Z_{0.05/2} \\ \\ Z_{0.025} = 1.96$ (From the z tables)

The margin of error is calculated by using the formula:

$M.O.E = Z_{\alpha/2} \times \sqrt{\dfrac{\hat p(1 -\hat p)}{n}}$

$M.O.E = 1.96 \times \sqrt{\dfrac{\hat 0.60(1 -0.60)}{1500}}$

$M.O.E = 1.96 \times \sqrt{\dfrac{0.24}{1500}}$

$M.O.E = 1.96 \times \sqrt{1.6 \times 10^{-4}}$

M.O.E = 0.02479

M.O.E ≅ 0.025

The margin of error M.O.E = 2.5%