Question

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 131 with 56 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 = = x / n = 26 / 131 = 0.427

1 - = 1 - 0.427 = 0.573

Z/2 = 1.282

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

= 1.282 * (((0.427 * 0.573) / 131)

Margin of error = E = 0.055

A 80% confidence interval for population proportion p is ,

- E < p < + E

0.427 - 0.055 < p < 0.427 + 0.055

0.372 < p < 0.482