Question

1.) Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 261 with 40% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to 4 decimal places.

< p <

2.) You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 78%. You would like to be 99% confident that your estimate is within 2.5% of the true population proportion. How large of a sample size is required?

n=

Answer 1

1)

Solution :

Given that,

n = 261

= 40% = 0.40

1 - = 1 - 0.40 = 0.60

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_0.0005 = 3.29

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

= 3.29 * (((0.40 * 0.60) / 261)

= 0.0997

A 99.9% confidence interval for population proportion p is ,

- E < P < + E

0.40 - 0.0997< p < 0.40 + 0.0997

0.3003 < p < 0.4997

2)

Solution :

Given that,

= 78% = 0.78

1 - = 1 - 0.78 = 0.22

margin of error = E = 2.5% = 0.025

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.576 / 0.025)² * 0.78 * 0.22

= 1822

sample size = n = 1822