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# 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?

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.

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