Question

If n=320 and p' (p-prime) = 0.23, construct a 95% confidence interval.

Give your answers to three decimals.

_______ < p < ________

Answer 1

Solution :

Given that,

n = 320

= 0.23

1 - = 1 - 0.23 = 0.77

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.23 * 0.77) / 320)

= 0.046

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.23 - 0.046 < p < 0.23 + 0.046

0.184 < p < 0.276