Question

In the week before and the week after a holida 10,000 total deaths, and 4956 ofthem occured inth week b oe?he holidar Construct a 95% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday y, there were < p < L (Round to three decimal places as needed.)

Answer 1

Solution :

Given that,

n = 10000

x = 4956

$\hat p$ = x / n = 4956 / 10000 = 0.496

1 - $\hat p$ = 1 - 0.496 = 0.504

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$(($\hat p$ * (1 - $\hat p$)) / n)

= 1.96 * ($\sqrt$((0.496 * 0.504) / 10000)

= 0.010

A 95% confidence interval for population proportion p is ,

$\hat p$ - E < P < $\hat p$ + E

0.496 - 0.010 < p < 0.496 + 0.010

0.486 < p < 0.506