Question

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 99 with 38 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

90% C.I. =

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 = 99

x = 38

Point estimate = sample proportion = = x / n = 38 /99 = 0.384

1 - = 1-0.384 = 0.616

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z0.05 = 1.645

Z/2 = 1.645

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

= 1.645 * ((0.384*(0.616) /99 )

= 0.080

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.384 - 0.080 < p < 0.384 + 0.080

0.304 < p < 0.464  

( 0.304 ,0.464 )