Question

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 337 with 165 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

______ <p< ________

Answer 1

Solution :

Given that,

n = 337

x = 165

Point estimate = sample proportion = = x / n = 165/337=0.490

1 -   = 1- 0.490 =0.510

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.490*0.510) / 337)

= 0.063

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.490-0.063 < p <0.490+ 0.063

0.427< p < 0.553