Topic 3.3 Notes – Constructing a Confidence Interval for a Population Proportion

What a Confidence Interval for a Population Proportion Does

This is for one population and one categorical variable with two outcomes. You choose one outcome to call success, even if it isn’t a good thing.

If $x$ people in a sample of size $n$ are successes, then the point estimate is

$$\hat p=\frac{x}{n}$$

That $\hat p$ is your sample proportion. The confidence interval takes that single number and builds a range of plausible values for the true population proportion $p$.

A complete parameter statement has to name all three parts:

• proportion
• success category
• population

Example: $p=$ the proportion of all seniors at Central High who have a driver’s license.

This procedure is called a one-sample $z$-interval for a population proportion. It is the right tool here, and only here.

It is not for:

• a population mean
• comparing two population proportions
• a categorical variable with 3 or more categories all at once

Conditions for a One-Sample Proportion Interval

All three conditions matter because each one justifies a different part of the method.

Randomization

Your data should come from a random sample.

On an FRQ, say how the sample was chosen in the prompt, like:

• “A simple random sample of 150 voters was selected.”

Just writing “random” is usually too vague. Also, a huge sample does not fix bias from a convenience sample or voluntary response sample.

Independence and the 10% condition

If sampling is without replacement, you need

$$N \ge 10n$$

This lets you treat observations as approximately independent.

If the sample is effectively with replacement, or the population is enormous compared with the sample, you may not need a separate 10% check.

Normality

For a confidence interval for a proportion, use the observed counts:

• $x \ge 10$
• $n-x \ge 10$

Equivalent form:

$$n\hat p \ge 10 \quad \text{and} \quad n(1-\hat p)\ge 10$$

These conditions are what let us use the normal model for the interval, with the familiar middle 95% between about $-1.96$ and $1.96$.

A common mistake is using $n \ge 30$. That rule is for means, not proportions.

Building the Interval

Here’s what the full process looks like with numbers. Suppose 84 of 120 randomly selected students say they have a school ID with them.

1. Parameter in context
   $p=$ the proportion of all students at the school who have a school ID with them.

2. Procedure
   One-sample $z$-interval for a population proportion.

3. Conditions

   • random sample given
   • if population is at least 1200, then $N \ge 10n$
   • successes $=84$, failures $=36$, both at least 10

4. Point estimate

   $$\hat p=\frac{84}{120}=0.70$$

5. Standard error

   $$SE=\sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.70(0.30)}{120}}\approx0.0418$$

6. Critical value
   Common values:

   • 90% → 1.645
   • 95% → 1.96
   • 99% → 2.576

7. Margin of error

   $$MOE=z^*\cdot SE=1.96(0.0418)\approx0.0819$$

8. Interval

   $$\hat p \pm MOE=0.70\pm0.0819$$

   So the interval is about $(0.618,\ 0.782)$.

Calculator shortcut: 1-PropZInt with $x$, $n$, and C-Level.

Standard Error, Margin of Error, and Sample Size

The standard error is the estimated standard deviation of $\hat p$. It tells you how much sample proportions typically vary from sample to sample.

For confidence intervals, use $\hat p$ in the SE formula because the true $p$ is unknown.

The margin of error is how far the interval goes on each side of $\hat p$. The full width is $2(MOE)$.

If an interval is $(L,U)$, then

$$\text{center}=\frac{L+U}{2} \qquad MOE=\frac{U-L}{2}$$

Planning a sample size uses

$$n=\frac{(z^*)^2\hat p(1-\hat p)}{MOE^2}$$

If no estimate for $\hat p$ is available, use $0.5$. That gives the largest required sample size. Always round up.

What Students Mix Up

• $p$ vs. $\hat p$
  $p$ is the unknown population proportion. $\hat p$ is the known sample proportion.
• Confidence interval vs. test
  A confidence interval uses $\hat p$ in the SE. A one-proportion significance test uses the hypothesized $p_0$.
• $x$ vs. $\hat p$
  TI calculators usually want the count of successes $x$, not the decimal proportion.
• Normality check
  Use observed successes and failures, not $n \ge 30$.
• Conditions vs. bias
  Passing the formulas does not rescue bad sampling.
• Parameter wording
  If you forget the population or the success category, you can lose credit.