If npq > 5, then the normal approximation can be used to develop a confidence interval for a binomial variable. The formula is:
Suppose a TV station conducts a survey on whether the public supports or opposes a new city ordinance and finds that of the 900 city residents surveyed, 340 support the ordinance and 560 oppose it. What is a 95% confidence interval for the proportion of the city residents that support the proposed ordinance.
First we need to calculate npq to see if we can use the normal approximation. For this problem we have:
Since the value of npq is much greater than 5, we can use the normal approximation. For a 95% confidence interval the value of Zc is 1.96 and the confidence interval is:
We are 95% sure that the proportion of all city residents that support the ordinance is between 36% and 42%. If this survey result was reported in the media, it is likely that it would be reported that 39% of survey respondents support the ordinance and the margin of error for the survey is 3%. This is not a complete description of the survey results as it does not report the level of confidence, which is 95%.
For a 60% the confidence interval the value of Zc is 0.842 and the confidence interval is:
The "margin of error" is only 1.4%, but we are only 60% sure that the population proportion is in this range.
Note that if we were to calculate the confidence interval for people who oppose the ordinance we would get the same magin of error.