Question

Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 380 with 125 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,

n = 380

x = 125

Point estimate = sample proportion = = x / n = 125 / 380 = 0.329

1 - = 1 - 0.329 = 0.671

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 * (((0.329*0.671) /380 )

= 0.047

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.329-0.047 < p <0.329+0.047

0.282< p <0.376

The 95% confidence interval for the population proportion p is : 0.282 , 0.376