Question

Construct the Confidence Interval for p if q = .364 and n = 2,200. (Use 1.96)

Answer 1

Solution :

Given that,

n = 2200

q=0.364

Point estimate = sample proportion = $\hat p$ = 1-q=1-0.364=0.636

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 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 ($\sqrt$((0.636*0.364) /2200 )

= 0.0201

A 95% confidence interval for population proportion p is ,

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

0.636-0.0201 < p <0.636+ 0.0201

0.6159< p < 0.6561

The 95% confidence interval for the population proportion p is : 0.6159,0.6561