Question

Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 199 with 121 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

99.9% C.I. =

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

Solution :

n = 199

x = 121

= x / n = 121 / 199 = 0.608

1 - = 1 - 0.608 = 0.392

At 99.9% confidence level the z is ,

= 1 - 99.9% = 1 - 0.999 = 0.001

/ 2 = 0.001 / 2 = 0.0005

Z_/2 = Z_0.0.0005 = 3.291

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

= 3.291 * (((0.608 * 0.392 ) / 199 )

= 0.114

A 99.9% confidence interval for population proportion p is ,

- E < P < + E

0.608 - 0.114 < p < 0.608 + 0.114

0.494 < p < 0.722  

( 0.494, 722 )