Carrying Out a Proportion Test — AP® Statistics

Part 1 · Hook

Why this matters

A school district starts a mentoring program, and this year's random sample shows 14% of students chronically absent, down from 18% last year. The superintendent wants to announce that absence has dropped. A complete significance test says whether the data back her up, and it also says what she still cannot claim.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. For $H_0: p = 0.18$ , $H_a: p < 0.18$ , $n = 450$ and $\hat{p} = 0.14$ , which is the test statistic?

$\frac{0.14 - 0.18}{\sqrt{0.18(0.82)/450}}$
$\frac{0.14 - 0.18}{\sqrt{0.14(0.86)/450}}$
$\frac{0.18 - 0.14}{\sqrt{0.18(0.82)/450}}$
$\frac{0.14 - 0.18}{0.18(0.82)/450}$

Show the answer

Statistic minus null value, over the standard deviation computed with $p_0 = 0.18$ .

Correct: $\frac{0.14 - 0.18}{\sqrt{0.18(0.82)/450}}$ :
$\frac{0.14 - 0.18}{\sqrt{0.14(0.86)/450}}$ :
$\frac{0.18 - 0.14}{\sqrt{0.18(0.82)/450}}$ :
$\frac{0.14 - 0.18}{0.18(0.82)/450}$ :

2. Which hypotheses fit "has the proportion gone down from 18%?"

$H_0: p = 0.18$ , $H_a: p < 0.18$
$H_0: p = 0.18$ , $H_a: p \ne 0.18$
$H_0: p < 0.18$ , $H_a: p = 0.18$
$H_0: \hat{p} = 0.18$ , $H_a: \hat{p} < 0.18$

Show the answer

The equals sign goes in $H_0$ ; "gone down" gives a one-sided alternative about $p$ .

Correct: $H_0: p = 0.18$ , $H_a: p < 0.18$ :
$H_0: p = 0.18$ , $H_a: p \ne 0.18$ :
$H_0: p < 0.18$ , $H_a: p = 0.18$ :
$H_0: \hat{p} = 0.18$ , $H_a: \hat{p} < 0.18$ :

Part 4 · See it

See it first

Four boxes in a column joined by arrows, for a test about chronic absence in a school district. State: the hypotheses H0: p = 0.18 and Ha: p < 0.18, where p is the true proportion of all district students who are chronically absent, with alpha = 0.05. Plan: a one-sample z test for a proportion; random sample; 450 is less than 10% of the students; 450(0.18) = 81 and 450(0.82) = 369 are at least 10. Do: z = (0.14 − 0.18) / sqrt(0.18(0.82)/450) = −2.21 and p-value = 0.014. Conclude: because 0.014 < 0.05, reject H0; there is convincing evidence that the true proportion is less than 0.18. — State, plan, do, conclude: a complete one-sample z test for a proportion, with every part written out. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The p-value is less than $\alpha$ the result is too unusual under $H_0$ : reject $H_0$ , and there is convincing evidence for $H_a$
The p-value is greater than $\alpha$ fail to reject $H_0$ : there is not convincing evidence for $H_a$ (which does not make $H_0$ true)
The data come from a random samplethe conclusion applies to the population that was sampled
There was no random assignment to treatmentsthe test can show a change but not what caused it

Part 6 · Key ideas

Key ideas

Four steps: State ( $H_0$ , $H_a$ , $\alpha$ , $p$ defined), Plan (procedure, conditions), Do ( $z$ , p-value), Conclude.
Conclusion: "Because the p-value of __ is less than $\alpha$ = __, we reject $H_0$ . There is convincing evidence that the true proportion of all __ is __ than __."
Fail to reject: "there is not convincing evidence that ...", still naming the direction of $H_a$ . Never "accept $H_0$ " or "prove".
A two-sided test at $\alpha = 0.05$ almost always rejects when $p_0$ is outside the 95% interval, and fails to reject when it is inside.
Random sampling lets you generalize; only random assignment lets you claim cause.

Part 7 · Misconception

A common mistake

The wrong idea: The p-value of 0.38 is greater than 0.05, so we accept $H_0$ : the proportion has not changed.

What actually happens: Fail to reject $H_0$ : there is not convincing evidence that the proportion has changed. The data are consistent with $H_0$ , but a test cannot show that $H_0$ is true.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A mentoring program and chronic absence

Last year, 18% of a large school district's students were chronically absent (missed 10% or more of school days). This year the district started a mentoring program and wants to know whether the proportion has gone down. A researcher selects a random sample of 450 of the district's 21,000 students; 63 of them are chronically absent so far. She tests $H_0: p = 0.18$ against $H_a: p < 0.18$ at $\alpha = 0.05$ , where $p$ is the true proportion of all the district's students who are chronically absent this year.

Sample results (invented data)
Students sampled	Chronically absent	Sample proportion
450	63	$\hat{p} = 0.14$

1. Calculate the test statistic. Give two decimal places.

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

$z$ $= \frac{0.14 - 0.18}{\sqrt{0.18(0.82)/450}}$ $= \frac{-0.04}{0.0181}$ $= -2.21$ . Conditions: random sample; 450 is less than 10% of 21,000; $450(0.18) = 81$ and $450(0.82) = 369$ are at least 10.

Answer: -2.21

2. Find the p-value. Give four decimal places.

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

$H_a: p < 0.18$ , so the p-value is the left tail: $P(Z \le -2.21) = 0.0136$ . Technology (a 1-proportion z test) gives $z = -2.21$ and p-value 0.0136.

Answer: 0.0136

3. State the conclusion of the test at $\alpha = 0.05$ .

Write the conclusion in context.

Show the answer

Because $0.0136 < 0.05$ , reject $H_0$ . The conclusion is about the parameter, in context, with evidence language and the direction of $H_a$ .

A full-credit answer: Because the p-value of 0.0136 is less than α = 0.05, we reject H0. There is convincing evidence that the true proportion of all the district's students who are chronically absent this year is less than 0.18.

4. Suppose the researcher had chosen $\alpha = 0.01$ before collecting the data. What would she conclude?

Reject $H_0$ , because the p-value is small
Fail to reject $H_0$ : not convincing evidence that $p < 0.18$
Accept $H_0$ : the true proportion is 0.18
Change to $\alpha = 0.05$ so the result is significant

Show the answer

Because $0.0136 > 0.01$ , she would fail to reject $H_0$ : there would not be convincing evidence at the 1% level that the true proportion is less than 0.18. That does not mean the proportion is 0.18.

Reject $H_0$ , because the p-value is small: At $\alpha = 0.01$ , the p-value 0.0136 is not small enough: it is greater than 0.01.
Correct: Fail to reject $H_0$ : not convincing evidence that $p < 0.18$ : Right: $0.0136 > 0.01$ .
Accept $H_0$ : the true proportion is 0.18: A test never accepts $H_0$ ; failing to reject means the evidence is not strong enough.
Change to $\alpha = 0.05$ so the result is significant: The significance level is fixed before the data; changing it afterward to get a result is not allowed.

Table

Computer output: a four-day school week

A polling firm selects a random sample of 1,000 adults in a state and asks whether they would support a four-day school week. The firm wants to know whether more than half of the state's adults support it. Its software prints the output below.

Test and confidence interval for one proportion (invented data)
Test	X	N	Sample p	Z	P-Value
p = 0.5 vs p > 0.5	532	1000	0.532	2.02	0.0215

5. The firm had chosen $\alpha = 0.01$ . State the conclusion.

Write the conclusion in context.

Show the answer

Because $0.0215 > 0.01$ , fail to reject $H_0$ . The data do not give convincing evidence, at the 1% level, that more than half of the state's adults support a four-day week. This does not show that the true proportion is 0.5.

A full-credit answer: Because the p-value of 0.0215 is greater than α = 0.01, we fail to reject H0. There is not convincing evidence that the true proportion of all the state's adults who would support a four-day school week is greater than 0.5.

6. A test of $H_0: p = 0.25$ against $H_a: p \ne 0.25$ gives a p-value of 0.38. Which conclusion is worded correctly?

We accept $H_0$ : the true proportion is 0.25
We fail to reject $H_0$ : not convincing evidence that $p \ne 0.25$
There is a 38% chance that the true proportion is 0.25
The data prove that the proportion is still 0.25, just as it was before

Show the answer

Because $0.38 > 0.05$ , fail to reject $H_0$ , and say the evidence is not convincing for $H_a$ . Never "accept" $H_0$ or say it is true.

We accept $H_0$ : the true proportion is 0.25: A test never accepts $H_0$ ; a large p-value means the data are consistent with it, not that it is true.
Correct: We fail to reject $H_0$ : not convincing evidence that $p \ne 0.25$ : Right: the decision, then evidence language about $H_a$ .
There is a 38% chance that the true proportion is 0.25: A p-value is not the probability that $H_0$ is true.
The data prove that the proportion is still 0.25, just as it was before: Tests never prove; failing to reject only means the evidence was not strong enough.

7. Put the four steps of a significance test in order.

State: hypotheses, $\alpha$ , and the parameter in context
Plan: name the procedure and check the conditions
Do: compute the test statistic and the p-value
Conclude: compare the p-value with $\alpha$ and state the conclusion in context

Show the answer

State what you are testing, plan how (procedure and conditions), do the calculation, then conclude in context.

Correct order: 1. State: hypotheses, $\alpha$ , and the parameter in context 2. Plan: name the procedure and check the conditions 3. Do: compute the test statistic and the p-value 4. Conclude: compare the p-value with $\alpha$ and state the conclusion in context

Part 9 · Summary

Summary

A complete test runs state, plan, do, conclude. It states hypotheses about

p

with

\alpha

, names the one-sample z test for a proportion and checks random, 10% and large counts with

p_0

, computes

z

and the p-value, then concludes: compare the p-value with

\alpha

, reject or fail to reject

H_0

, and say whether there is convincing evidence for

H_a

about the true proportion of all ..., in context. Never accept

H_0

, and claim cause only from a randomized experiment.

Part 10 · Up next

What comes next

Part 11 · Connections