Question

A random sample of 1400 Internet users was selected from the records of a large Internet provider and asked whether they would use the Internet or the library to obtain information about health issues. Of these, 872 said they would use the Internet. If the Internet provider wanted an estimate of the proportion (p hat) that would use the Internet rather than the library, with a margin of error of at most 0.01 in a 99% confidence interval, how large a sample size would be required

Answer 1

Explanation:

Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 1400

x = 872

p = 872/1400 = 0.62

q = 1 - 0.62 = 0.38

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.005 = 0.995

The z score corresponding to the area on the z table is 2.58. Thus, the z score for a confidence level of 99% is 2.58

Given margin of error = ±0.01,

Therefore,

0.01 = 2.58√(0.62)(0.38)/n

0.01/2.58 = √0.2356/n

0.0039 = √0.2356/n

Square both sides

0.00001521 = 0.2356/n

n = 0.2356/0.00001521

n = 15490