Topic 3.7 Notes – Carrying Out a Test for a Population Proportion

What a One-Sample z-Test for a Population Proportion Does

This test is for a categorical variable with two outcomes, usually called success and failure. “Success” just means the response you care about.

If $x$ successes are observed in a sample of size $n$, then the sample proportion is

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

The parameter is $p$, the true population proportion with the response of interest. On an FRQ, name it in context, like “the true proportion of all voters in the county who support the bond measure.”

The null model says assume

$$H_0:p=p_0$$

is true. Then ask whether your observed $\hat p$ would be unusual under that assumption.

You should know the three possible alternatives together:

• $H_a:p>p_0$ right-tailed
• $H_a:p<p_0$ left-tailed
• $H_a:p\ne p_0$ two-tailed

If a problem writes the null with an inequality, like $H_0:p\le 0.40$, the actual test is still done at the boundary, so use $p=0.40$.

Conditions and the Null Distribution

Before doing the math, check whether the test is justified.

• Randomization
  The data should come from a random sample or some randomized collection method. Quote the prompt, like “a simple random sample of 150 students was selected.”
• Independence
  If sampling without replacement, use the 10% condition. You need $N \ge 10n$.
• Normality under the null
  Use the null value $p_0$, not $\hat p$, for expected counts:
  • $np_0 \ge 10$
  • $n(1-p_0) \ge 10$

That use of $p_0$ matters because the whole test is built assuming $H_0$ is true. So the null distribution of $\hat p$ has

$$\text{mean}=p_0 \qquad \text{SD}=\sqrt{\frac{p_0(1-p_0)}{n}}$$

Think of it as a normal curve centered at the null value, with spread based on $p_0$ and $n$.

A very common mistake is using $\hat p$ in the standard deviation. That belongs to a confidence interval, not this test.

Carrying Out the Test

Here’s the full calculation flow.

1. Identify the procedure.
   This is a one-sample $z$-test for a population proportion.

2. Compute the sample proportion.
   Suppose 118 of 250 households compost. Then

   $$\hat p=\frac{118}{250}=0.472$$

3. Calculate the test statistic.
   If testing $H_0:p=0.40$,

   $$z=\frac{\hat p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} =\frac{0.472-0.40}{\sqrt{\frac{0.40(0.60)}{250}}} \approx 2.32$$

   A positive $z$ means $\hat p>p_0$. A negative $z$ means $\hat p<p_0$. Bigger distance from 0 means stronger evidence against $H_0$.

4. Find the p-value from the standard normal curve, using the alternative:

   • right-tailed: $P(Z\ge z_{\text{obs}})$
   • left-tailed: $P(Z\le z_{\text{obs}})$
   • two-tailed: $2P(Z\ge |z_{\text{obs}}|)$

   The sketch below is just a reminder that the p-value comes from the tail area that matches $H_a$. For this example, focus on the left panel since this is a right-tailed test.

   

   For $z=2.32$ in a right-tailed test, the p-value is about $0.0102$.

Calculator output is fine, but still show the setup and correct tail. Students lose points by choosing the wrong tail, doubling when they should not, or letting the sign of $z$ change the stated $H_a$.

Making the Decision and Writing the Conclusion

The significance level $\alpha$ is the cutoff for statistical significance.

• If p-value $\le \alpha$, reject $H_0$
• If p-value $> \alpha$, fail to reject $H_0$

With p-value $=0.0102$ and $\alpha=0.05$, reject $H_0$.

That decision turns into a conclusion in context. For the compost example, say:

• “There is convincing statistical evidence that the true proportion of all city households that participate in the composting program is greater than $0.40$.”

Good conclusion habits:

• mention the population
• mention the parameter
• match the direction of $H_a$
• use phrases like “convincing statistical evidence” or “not convincing statistical evidence”

Never say:

• “accept $H_0$”
• “prove $H_0$ is true”
• “the p-value is the probability that $H_0$ is true”

Random sampling supports generalizing to the population. It does not give cause-and-effect by itself.

How to Organize a Full FRQ Response

A clean FRQ response usually follows State, Plan, Do, Conclude.

• State
  Define $p$. Write $H_0$ and $H_a$ in context.
• Plan
  Name the one-sample $z$-test for a population proportion. Check Random, 10%, and Normality.
• Do
  Calculate $\hat p$, $z$, and the p-value with the correct tail.
• Conclude
  Compare p-value to $\alpha$, make the decision, and write the conclusion in context.

A small p-value means the result is statistically significant. It does not automatically mean the difference is important in real life.