Question

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 340 with 180 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. <p> Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 180 / 340 = 0.529

1 - $\hat p$ = 1 - 0.529 = 0.471

+ /2 = 1.645

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

= 1.645 * ($\sqrt$((0.529 * 0.471) / 340)

Margin of error = E = 0.045

A 90% confidence interval for population proportion p is ,

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

0.529 - 0.045 < p < 0.529 + 0.045

0.484 < p < 0.574