Question

A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480 were delivered on time. a) Construct a 95% confidence interval for the proportion of all packages that are delivered on time by the Postal Service

Answer 1

$\hat{p}= \frac{X }{ n} = \frac{480}{500} = 0.96 \\ \hat{p} ± Z(α/2) \sqrt{( \frac{(p \times q) }{ n})} \\ 0.96 ± Z(0.05/2) \sqrt{( \frac{(0.96 \times 0.04) }{ 500})}$

Z(α/2) = Z(0.05/2) = 1.96 ( From Z table)

Lower Limit $= 0.96 - 1.96 \times \sqrt{( \frac{(0.96 \times 0.04) }{ 500})} = 0.943$

Upper Limit $= 0.96 + 1.96 \sqrt{( \frac{(0.96 \times 0.04) }{ 500})} = 0.977$

95% Confidence interval is ( 0.943 , 0.977 )