In the margin of error formula, the sample proportions \( <\hat p>\) and 1- <\hat p>are estimates of the unknown population proportions p and 1-p. The estimated proportions \(<\hat p>\) and \(1-<\hat p>\) are used because \(p\) and \(1-<\hat p>\) are not known. The sample proportions \(<\hat p>\) and \(1-<\hat p>\) are calculated from the data: \(<\hat p>\) is the estimated proportion of successes, and \(1-<\hat p>\) is the estimated proportion of failures.
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.
Answer
Solution A
\[1 – <\hat p>= 1 – 0.842 = 0.158\nonumber \]
Since the confidence level \(CL = 0.95\), then
\(\alpha = 1 – CL = 1 – 0.95 = 0.05 \)
Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find \(z_\). Remember that the area to the right of \(z_\) is \(0.025\) and the area to the left of \(z_\) is \(0.975\). This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.
The confidence interval for the true binomial population proportion is \(( <\hat p>– \text, <\hat p>+\text) = (0.810, 0.874)\).
Interpretation
We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.
Explanation of 95% Confidence Level
Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.
Solution B
Press STAT and arrow over to TESTS .
Arrow down to A:1-PropZint . Press ENTER .
Arrow down to xx and enter 421.
Arrow down to nn and enter 500.
Arrow down to C-Level and enter .95.
Arrow down to Calculate and press ENTER .
The confidence interval is (0.81003, 0.87397).
For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval.
Answer
Solution A
\[1- <\hat p>= 1 − 0.600 = 0.400\nonumber \]
Since \(CL = 0.90\), then
\(\alpha = 1 – CL = 1 – 0.90 = 0.10 \)
Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find \(z_\). Remember that the area to the right of \(z_\) is 0.05 and the area to the left of \(z_\) is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.
The confidence interval for the true binomial population proportion is \(( <\hat p>– \text, <\hat p>+\text) = (0.564,0.636)\).
Interpretation
Explanation of 90% Confidence Level
Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.
Solution B
Press STAT and arrow over to TESTS .
Arrow down to A:1-PropZint . Press ENTER .
Arrow down to xx and enter 300.
Arrow down to nn and enter 500.
Arrow down to C-Level and enter 0.90.
Arrow down to Calculate and press ENTER .
The confidence interval is (0.564, 0.636).
To estimate the proportion of students at a large college who are female, a random sample of \(120\) students is selected. There are \(69\) female students in the sample. Construct a \(90\%\) confidence interval for the proportion of all students at the college who are female.
Solution A
The proportion of students in the sample who are female is
\[ \hat
=69/120=0.575 \nonumber\]
Confidence level \(90\%\) means that \(\alpha =1-0.90=0.10\) so \(\alpha /2=0.05\). From the last line of Figure 7.1.6 we obtain \(z_=1.645\).
One may be \(90\%\) confident that the true proportion of all students at the college who are female is contained in the interval \((0.575-0.074,0.575+0.074)=(0.501,0.649)\).
Solution B
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to \(x\) and enter 69.
Arrow down to \(n\) and enter 120.
Arrow down to C-Level and enter 0.90.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.501,0.649).
This page titled 7.2: Confidence Interval for a Proportion is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
