Ch.9-Estimating Population Proportions: Point Estimates and Confidence Intervals

Estimating a Population Proportion

Learning Objectives

• Obtain a point estimate for the population proportion

• Construct and interpret a confidence interval for a population proportion

• Determine the sample size necessary for estimating a population proportion within a specified margin of error

Point Estimate for the Population Proportion

A point estimate is the value of a statistic that estimates the value of a parameter. In the context of proportions, the point estimate for the population proportion is the sample proportion, denoted as $\hat{p}$.

• Sample Proportion ($\hat{p}$): $\hat{p} = \frac{x}{n}$, where x is the number of individuals in the sample with a specified characteristic, and n is the sample size.

• Example: In a poll of 1015 Americans, 458 said the amount of federal income tax they pay is too high. The point estimate is $\hat{p} = \frac{458}{1015} = 0.451$.

Constructing and Interpreting a Confidence Interval for the Population Proportion

In statistics, a confidence interval provides a range of plausible values for a population parameter, based on a point estimate and a specified level of confidence.

• The level of confidence (e.g., 95%) represents the expected proportion of intervals that will contain the parameter if many samples are taken.

• A higher level of confidence results in a wider interval; a lower level of confidence results in a narrower interval.

• The general form of a confidence interval for a population proportion is:

$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

• $z_{\alpha/2}$ is the critical value from the standard normal distribution for the desired confidence level.

• The margin of error is $E = z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$.

• Interpretation: For a 95% confidence interval, we expect that 95 out of 100 such intervals constructed from different samples would contain the true population proportion.

• Example: For $\hat{p} = 0.451$, $n = 1015$, and $z_{0.025} = 1.96$ (for 95% confidence):

$E = 1.96 \sqrt{\frac{0.451(1-0.451)}{1015}} \approx 0.04$ Confidence interval: $0.451 \pm 0.04 = (0.411, 0.491)$

• We are 95% confident that the proportion of Americans who believe their federal income tax is too high is between 0.411 and 0.491.

Key Concepts and Properties

• The critical value ($z_{\alpha/2}$) increases as the confidence level increases, resulting in a wider interval.

• The standard error of the sample proportion is $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$.

• For the normal approximation to be valid, the sample size should be large enough so that $n\hat{p} \geq 10$ and $n(1-\hat{p}) \geq 10$.

• The confidence interval does not give the probability that the parameter is in the interval; rather, it reflects the reliability of the estimation procedure.

Visual Representation

The width of the confidence interval depends on the level of confidence and the variability in the sample. A diagram (not shown here) typically illustrates that higher confidence levels correspond to wider intervals.

Summary Table: Effect of Confidence Level on Critical Value

Level of Confidence	Area in Each Tail	Critical Value ($z_{\alpha/2}$)
90%	0.05	1.645
95%	0.025	1.96
99%	0.005	2.575

Example Application

• Scenario: A poll of 1015 adults, 458 say taxes are too high.

• Point estimate: $\hat{p} = 0.451$

• Margin of error (95%): $E = 0.04$

• Confidence interval: (0.411, 0.491)

• Interpretation: We are 95% confident that the true proportion is between 0.411 and 0.491.

Additional info: These notes are based on textbook slides and are suitable for introductory college statistics courses. The content covers the foundational concepts of point estimation and confidence intervals for proportions, including practical computation and interpretation.