Unit 3 · Topic 3.7 · about 25 minutes

Carrying Out a Test for a Population Proportion

Carry out a one-sample z-test for a population proportion and justify a conclusion about the population by comparing the p-value to the significance level.

Predict first

A researcher tests at the significance level $α = 0.05$ . Suppose the null hypothesis is actually true. If the same study were repeated many times with new random samples, how often would the test reject the null hypothesis anyway?

The test statistic

Once the hypotheses are set and the conditions check out (3.5 Setting Up a Test for a Population Proportion), the calculation measures how far $\hat{p}$ landed from $p_{0}$ , in standard deviations of the null distribution:

z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p_{0} (1 - p_{0})}{n}}}

The standard deviation in the denominator uses $p_{0}$ , not $\hat{p}$ , because every step of a test is worked out in the world where $H_{0}$ is true. In that world, and with the conditions met, $z$ has a standard normal distribution, so a probability model you already know serves as the null distribution. The p-value is then an area under the standard normal curve, found from a table or with technology, in the direction of the alternative (3.6 p-Values).

The significance level and the decision

Before the data are collected, the researcher picks a significance level, $α$ : the probability of rejecting $H_{0}$ when $H_{0}$ is true. It is the cutoff for how small a p-value has to be. Common choices are 0.10, 0.05 and 0.01. A problem may give you $α$ . If it does not, state the level you will use before you look at the p-value.

The formal decision is a direct comparison.

If the p-value $\leq α$ , reject $H_{0}$ . The result is statistically significant, and there is convincing statistical evidence for $H_{a}$ .
If the p-value $> α$ , fail to reject $H_{0}$ . There is not convincing statistical evidence for $H_{a}$ .

Those are the only two outcomes. A test can never conclude that $H_{0}$ is true.

Worked exampleAre this state's teens online more than the national figure?

A national survey reported that 45% of U.S. teens say they are online almost constantly. A researcher believes the proportion is higher among teens in her state. She surveys a random sample of 500 of the state's teens, and 245 say they are online almost constantly. Do the data give convincing statistical evidence, at $α = 0.05$ , that more than 45% of the state's teens are online almost constantly?

Hypotheses. $H_{0} : p = 0.45$ and $H_{a} : p > 0.45$ , where $p$ is the proportion of all teens in the state who say they are online almost constantly. Use $α = 0.05$ .
Procedure and conditions. One-sample z-test for a population proportion. Randomization: the 500 teens were randomly selected. 10%: 500 is far less than 10% of the teens in the state. Normality: $n p_{0} = 500 (0.45) = 225$ and $n (1 - p_{0}) = 500 (0.55) = 275$ , both at least 10.
Calculate. $\hat{p} = \frac{245}{500} = 0.49$ , so $z = \frac{0.49 - 0.45}{\sqrt{\frac{0.45 (0.55)}{500}}} = \frac{0.04}{0.02225} \approx 1.80$ . The p-value is $P (z \geq 1.80)$ , which is $1 - 0.9641 = 0.0359$ from the table, or 0.0361 from technology with the unrounded $z$ .
Conclude. Because the p-value of 0.036 is less than $α = 0.05$ , reject $H_{0}$ . There is convincing statistical evidence that the proportion of all teens in the state who say they are online almost constantly is greater than 0.45.

Answer.

Reject $H_{0}$ . The data give convincing statistical evidence that more than 45% of the state's teens are online almost constantly. The conclusion covers the population that was sampled, this state's teens, and nobody else.

Null distribution of the test statistic

If 45% of the state's teens were online almost constantly, $z$ would follow this standard normal curve. The shaded area at or above $z = 1.80$ is the p-value, about 0.036, and it is smaller than $α = 0.05$ .

Writing the decision and the conclusion
Comparison	Decision	Conclusion in context
p-value $\leq α$	Reject $H_{0}$	There is convincing statistical evidence that [the alternative hypothesis, in context].
p-value $> α$	Fail to reject $H_{0}$	There is not convincing statistical evidence that [the alternative hypothesis, in context].

Worked exampleWhen the evidence falls short

Last year, 62% of a university's incoming students said they planned to live on campus all four years. The housing office wants to know whether the proportion has changed for this year's 6,000 incoming students. In a random sample of 300 of them, 177 say they plan to, so $\hat{p} = 0.59$ . (Random sample; 300 is less than 10% of 6,000; $n p_{0} = 186$ and $n (1 - p_{0}) = 114$ .) Test at $α = 0.05$ .

Hypotheses. $H_{0} : p = 0.62$ and $H_{a} : p \neq 0.62$ , where $p$ is the proportion of all this year's incoming students who plan to live on campus all four years. The office asked about a change in either direction, so the alternative is two-sided.
Calculate. $z = \frac{0.59 - 0.62}{\sqrt{\frac{0.62 (0.38)}{300}}} \approx - 1.07$ . The p-value counts both tails: $2 P (z \leq - 1.07) \approx 0.285$ from the table, or 0.284 from technology with the unrounded $z$ .
Conclude. Because the p-value of 0.284 is greater than $α = 0.05$ , fail to reject $H_{0}$ . There is not convincing statistical evidence that the proportion of all this year's incoming students who plan to live on campus all four years is different from 0.62.

Answer.

Fail to reject $H_{0}$ . The office cannot conclude that the proportion changed, and it also cannot conclude that it stayed at 0.62.

Why you never accept the null hypothesis

In the housing example, a true proportion of 0.58 or 0.61 could easily have produced a sample like this one, too. The data are consistent with all of these values, so they cannot single out 0.62. Not having convincing evidence for $H_{a}$ is a statement about the evidence, not proof of $H_{0}$ .

That is why a conclusion is always written in terms of the alternative hypothesis, with non-definitive language: there is or there is not convincing statistical evidence that the proportion is different. Name the parameter and the population every time. The conclusion is the answer to the investigative question that started the study, and it applies only to the population that was sampled.

Check your understanding

A test of $H_{0} : p = 0.20$ against $H_{a} : p > 0.20$ uses a random sample of 400, in which 98 are successes. Find the value of the test statistic. Round to two decimal places.

Your answer

A smartwatch company claims that more than 30% of its buyers use the sleep tracker every night. In a random sample of 500 buyers, 165 do. The test of $H_{0} : p = 0.30$ against $H_{a} : p > 0.30$ gives $z = 1.46$ and a p-value of 0.072. At $α = 0.05$ , which is the correct decision and conclusion?

A researcher announces before collecting data that she will use a significance level of $α = 0.05$ . What does the 0.05 mean?

Two analysts test the same hypotheses with the same data and get a p-value of 0.03. One planned to use $α = 0.05$ and the other planned to use $α = 0.01$ . Which statement is correct?

A principal wants to know whether more than half of the students at her school want a later start time. She surveys a random sample of 150 of the school's 1,900 students, and the test gives convincing statistical evidence that more than half want a later start. To which group does this conclusion apply?

Practice

Practice until it is automatic

New numbers every time. Each one is checked the moment you answer, with the full working shown.

Significance tests for a proportion practice page

Course alignment, for teachers

AP Statistics topic 3.7, Unit 3: Inference for Categorical Data: Proportions.

Skill 3.E: Calculate appropriate statistical inference method results.
Skill 4.G: Justify a claim based on statistical inference method results.