Confidence Intervals: 1 Proportion
Introduction
One Proportion confidence intervals are used when you are dealing with a single proportion (\(\widehat{p}\)). The critical value used will be \(z^*\).
Remember that:
 The sample proportion is denoted as \(\widehat{p}\)
 \(1  \widehat{p}\) is denoted as \(\widehat{q}\)
 \(z^*\) is the critical value found on a table.
A local pet shop is trying to determine what percent of the U.S. population prefers dogs to cats. Last week they surveyed the 274 people who entered the store. Of those 274 people, 183 people prefer dogs to cats. Create a 95% confidence interval to estimate the true percentage of U.S. residences who prefer dogs to cats.
Step 1: Name the Confidence Interval: One Proportion
Step 2: Check the Conditions
1. The data is drawn using a random sample.
2. N ≥ 10n
3. \(n \widehat{p} ≥ 10\) and \(n \widehat{q} ≥ 10\)
Step 3: Construct the Interval (Apply the Formula)
\(\widehat {p} = {183 \over 274} = 0.67\)
→ .67 \(^+_\) 1.96 (\(\sqrt{{.67(1.67)} \over 274}\))
→ .67 \(^+_\) 0.056
.67  0.056 = 0.61
.67 + 0.056 = 0.73
Interval: (0.61, 0.73)
Note: The critical value was found using a ztable. A small portion of the table is listed below with the part needed for our problem highlighted:
Confidence Level  z* Value

95%  1.960 
99%  2.576 
Step 4: Interpret your results in context of the problem
Based on the data, we can be 95% confident that the proportion of U.S. residents who prefer dogs to cats is between 0.61 and 0.73.
Margin of Error Questions
Sometimes, you desire a specific margin of error (recall that the margin of error is the critical value \(\times\) the standard deviation of the statistic). One way to control the margin of error is by manipulating the sample size.
Standard problems are set up as follows (\(M\) stands for the margin of error):
Question: Given \(\widehat{p} = 0.7\), find the sample size needed to make the margin of error 0.9 for a 70% confidence interval.
Solution: With this information, you know \(\widehat{p}\), \(\widehat{q}\) (since \(\widehat {q} = 1  \widehat {p}\)), \(M\), and \(z^*\) (since you know the confidence level and have access to a ttable or ztable). Thus, it is simple to plug these numbers into \(M = (z^*)(\sqrt { {\widehat{p} \widehat{q} \over n} })\) and solve for n.
However, certain problems do not give you \(\widehat{p}\). When this happens, you can assume a proportion value of 0.5 and call this \(p^*\). Hence, your solution would look like this: \(M = (z^*)(\sqrt { {p^* q^* \over n} })\) → \(M = (z^*)(\sqrt { {(0.5)(0.5) \over n} })\) → Then solve for n.