Math and Arithmetic

If 135 out of 500 students sampled have computers what is the 95-percent confidence interval for the true proportion of all students who own computers?

User Avatar
Wiki User
2007-05-16 02:20:23

First identify the proper confidence interval. Since we are

dealing with one proportion, a 1-proportion Z-interval is

appropriate. Now check to see if the necessary conditions are

fulfilled. First is whether the data was collected from a simple

random sample representative of the population. Second is whether n

* p-hat and n * (1 - p-hat) are both sufficiently large, where n is

the sample size and p-hat is the sample proportion. This is the

case, since both 135 and 500 - 135 = 365 are both greater than 10,

a generally accepted value. Third is whether n is a sufficiently

small fraction of the population (about 1/10 of the population is

the largest acceptable fraction). If you have at least 5000

students in your population, the test can be used. Finally, the

calculation (assuming all conditions have been fulfilled). The

confidence interval for a proportion is p-hat +/- z-star *

sqrt(p-hat * (1 - p-hat) / n). Here z-star is the critical value

for which P(abs(Z) > z-star) on the standard normal curve is

equal to 1 minus your confidence level. In this case, we're looking

for the value where the probability of a standard normal random

variable producing a value either greater than z-star or less than

negative z-star is 1 - .95 = 0.05. This value is approximately

1.96. Putting it all together: p-hat = 135/500 = 0.27

z-star = 1.96

n = 500

0.27 +/- 1.96 * sqrt(0.27 * 0.73 / 500)

0.27 +/- 1.96 * sqrt(0.0003942)

0.27 +/- 1.96 * .0199

0.27 +/- .03891

We are 95% confident that the true proportion of students who own

computers is between .23109 and .30891.

Copyright © 2020 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply.