Three Values to Read in 1-PropZTest Output
In Running 1-PropZTest on the Calculator, you learned how to enter the null proportion, the number of successes, the sample size, and the alternative hypothesis. This tutorial focuses on what to do with the results. A calculator can report several values at once, but the labels can be easy to confuse.
For a one-proportion \(z\)-test, the TI-84 commonly displays \(z\), \(p\), and \(\hat{p}\). These values answer different questions. The \(z\) statistic describes how far the sample result is from the null proportion, measured in null standard errors. The output’s \(p\) is the p-value for the alternative you selected. The \(\hat{p}\) is the sample proportion, calculated from the successes and sample size.
That last distinction matters. In a hypothesis test, \(p\) usually names the fixed, unknown population proportion. In the calculator’s output, the label \(p\) is shorthand for the p-value. Use the surrounding context to tell which meaning is intended. When reporting the calculator output, calling it “the p-value” avoids ambiguity.
The three output values should also make sense together. The sign of \(z\) shows whether \(\hat{p}\) is above or below \(p_0\): a positive \(z\) means \(\hat{p}>p_0\), and a negative \(z\) means \(\hat{p}<p_0\). The p-value then depends on the alternative hypothesis selected. A right-tailed test uses the upper-tail probability, a left-tailed test uses the lower-tail probability, and a two-sided test uses both tails. The same data and null proportion can therefore have different p-values for different alternatives.
A Quick Check Before Making a Decision
Do not treat the calculator as if it had decided whether the study question was answered. It calculates from the entries you provide. You still need to confirm that those entries represent the intended test and that the test’s conditions are met. As covered in Conditions Check for a Proportion Test Scenario and Running 1-PropZTest on the Calculator, check the data-collection method, the 10% condition when sampling without replacement, and the Large Counts condition using \(p_0\).
A useful output check is to compare \(\hat{p}\) with \(x/n\), and compare the sign of \(z\) with the direction of the sample’s departure from \(p_0\). If either does not agree, review the entries. Then verify that the selected calculator alternative matches \(H_a\). An incorrect alternative can produce a p-value for a different question, even if the \(z\) and \(\hat{p}\) look reasonable.
The decision rule is the one used in Comparing P-Value to the Significance Level: reject \(H_0\) when the p-value is less than or equal to \(\alpha\); otherwise, fail to reject \(H_0\). The decision is not one of the three calculator values. It comes from comparing the p-value with the significance level.
Worked Examples: Reading and Using the Output
Worked Example: A Right-Tailed Test
A fictional community garden coordinator wants to know whether more than 40% of the garden’s registered members compost food scraps at home. A random sample of 150 members is selected from a registry of 3,000 members; 72 report that they compost. Let \(p\) be the true proportion of registered members who compost food scraps at home. The test is conducted at \(\alpha=0.05\).
State: The hypotheses are \(H_0:p=0.40\) and \(H_a:p>0.40\). The alternative is right-tailed because the question asks whether the proportion is greater than 0.40.
Plan: The sample was randomly selected, so the Random condition is met. The sample is no more than 10% of the registry because \(150\leq0.10(3{,}000)=300\), so the 10% condition is met. Under \(H_0\), the expected number of members who compost is \(np_0=150(0.40)=60\), and the expected number who do not compost is \(n(1-p_0)=150(0.60)=90\). Both are at least 10, so the Large Counts condition is met. Use a one-proportion \(z\)-test with the greater-than alternative.
Do: Enter \(p_0=0.40\), \(x=72\), \(n=150\), and select \(p>p_0\) in 1-PropZTest. The output gives \(z\approx2.0000\), a p-value of approximately \(0.0228\), and \(\hat{p}=0.48\). Check the sample proportion: \(72/150=0.48\). The null standard error is \(\sqrt{0.40(0.60)/150}=0.04\), so \(z=(0.48-0.40)/0.04=2.0000\). The positive statistic agrees with the sample proportion being above 0.40. Compare the p-value with \(\alpha\): \(0.0228\leq0.05\), so reject \(H_0\).
Conclude: The sample provides convincing evidence that more than 40% of the garden’s registered members compost food scraps at home. The conclusion refers to the population proportion, not just the 48% observed in the sample.
Worked Example: A Left-Tailed Test
A fictional recreation center is reviewing whether less than 25% of its members use a particular indoor climbing wall during a typical month. Staff randomly select 120 members from the center’s list of 2,000 members; 24 say they used the wall. Let \(p\) be the true proportion of the center’s members who use the wall during a typical month. Use \(\alpha=0.05\).
State: The hypotheses are \(H_0:p=0.25\) and \(H_a:p<0.25\). The research question specifies a proportion below 0.25, so the test is left-tailed.
Plan: The random selection meets the Random condition. The 10% condition is met because \(120\leq0.10(2{,}000)=200\). Under the null hypothesis, the expected numbers of wall users and nonusers are \(120(0.25)=30\) and \(120(0.75)=90\). Both counts are at least 10, so the Large Counts condition is met. Use a one-proportion \(z\)-test with the less-than alternative.
Do: Enter \(p_0=0.25\), \(x=24\), \(n=120\), and select \(p<p_0\). The calculator reports \(z\approx-1.2649\), a p-value of approximately \(0.1030\), and \(\hat{p}=0.20\). Check that \(24/120=0.20\). The test statistic is
The negative \(z\) agrees with \(\hat{p}<p_0\). Because the selected alternative is left-tailed, the p-value is the lower-tail probability for this statistic. Compare \(0.1030\) with \(0.05\): \(0.1030>0.05\), so fail to reject \(H_0\).
Conclude: The sample does not provide convincing evidence that less than 25% of the recreation center’s members use the climbing wall during a typical month. This conclusion does not establish that the true proportion is 25%; it reports that the sample does not provide sufficient evidence for the stated alternative at the 0.05 level.
Worked Example: A Two-Sided Test
A fictional neighborhood association wants to know whether the proportion of residents who use its community tool library differs from 0.50. A random sample of 100 residents is taken from a list of 1,600 residents, and 58 report using the library. Let \(p\) be the true proportion of residents in the neighborhood who use the tool library. The significance level is \(\alpha=0.10\).
State: The hypotheses are \(H_0:p=0.50\) and \(H_a:p\ne0.50\). The question asks whether the proportion differs in either direction, so this is a two-sided test.
Plan: The residents were randomly selected, meeting the Random condition. The 10% condition is met because \(100\leq0.10(1{,}600)=160\). Under \(H_0\), the expected counts of users and nonusers are \(100(0.50)=50\) and \(100(0.50)=50\). Both are at least 10, so the Large Counts condition is met. Use a one-proportion \(z\)-test with the not-equal alternative.
Do: Enter \(p_0=0.50\), \(x=58\), \(n=100\), and select \(p\ne p_0\). The output gives \(z=1.6000\), a p-value of approximately \(0.1096\), and \(\hat{p}=0.58\). The sample proportion checks because \(58/100=0.58\). The null standard error is \(\sqrt{0.50(0.50)/100}=0.05\), giving \(z=(0.58-0.50)/0.05=1.6000\). For the two-sided alternative, the calculator counts results at least this far from zero in either direction. Since \(0.1096>0.10\), fail to reject \(H_0\).
Conclude: At the 0.10 significance level, the sample does not provide convincing evidence that the proportion of neighborhood residents who use the community tool library differs from 0.50. Although the sample proportion is above 0.50, the test does not provide sufficient evidence of a difference in either direction at this significance level.
What the Output Can—and Cannot—Tell You
The value \(\hat{p}\) describes the sample: it is the observed fraction of sampled individuals with the characteristic. It does not describe the null hypothesis, and it is not itself a decision. In the examples, the sample proportions were 0.48, 0.20, and 0.58. Those values can be reported as 48%, 20%, and 58%, but the test conclusions concern the corresponding population proportions.
The \(z\) statistic expresses the difference between \(\hat{p}\) and \(p_0\) in standard-error units, using the null model. A value near zero means the sample proportion is close to the null value relative to the expected sampling variation. A positive or negative sign identifies the direction of the observed difference. The alternative hypothesis—not the sign by itself—determines which tail or tails are used for the p-value.
The p-value measures how unusual the observed result, or a more extreme result, would be if \(H_0\) were true, according to the selected alternative. It is not the probability that \(H_0\) is true. As emphasized in Interpreting a P-Value in Context, give its meaning in terms of sample results under the null model rather than treating it as a probability about the hypothesis.
Finally, 1-PropZTest does not verify whether the sample was random, whether the 10% condition applies, or whether the Large Counts condition is satisfied. Nor can it decide whether the hypotheses match the research question. Those checks and choices belong to the reasoning in your solution. A correct-looking output is not enough if the wrong \(p_0\), counts, or alternative were entered.
Common Mistakes and AP Exam Tips
- Calling the calculator’s \(p\) the population proportion: In the output, \(p\) means p-value. Say “the p-value is…” when reporting it; use \(p\) for the population parameter in the hypotheses.
- Confusing \(p\) with \(\hat{p}\): \(\hat{p}\) is the sample proportion \(x/n\). Check it directly from the entered counts rather than interpreting it as a population value.
- Using the wrong tail: A calculator can return a p-value for the alternative selected, even if that alternative does not match the research question. State \(H_a\) first and confirm the matching calculator choice.
- Using only the sign of \(z\) to choose the alternative: The research question determines \(H_a\). Do not switch the alternative after seeing whether \(z\) is positive or negative.
- Reporting a decision without comparing the p-value to \(\alpha\): Show the comparison, such as \(0.0228\leq0.05\), and then state “reject \(H_0\).”
- Writing “accept \(H_0\)” when the p-value is large: Write “fail to reject \(H_0\)” and explain that the data do not provide convincing evidence for the alternative.
- Ending with a conclusion about the sample alone: The sample proportion is a statistic; the test’s conclusion addresses the population proportion defined in context.
Key Takeaway
Interpreting 1-PropZTest output means more than copying three values. Identify \(z\) as the test statistic, the calculator’s \(p\) as the p-value, and \(\hat{p}\) as the sample proportion. Check that the entries and alternative are correct, then compare the p-value with \(\alpha\) and state what the evidence says about the population proportion.
Check Your Understanding
Use the meanings of \(z\), the calculator’s \(p\), and \(\hat{p}\) to answer each question.
- A 1-PropZTest output shows \(z=-1.45\), \(p=0.0735\), and \(\hat{p}=0.31\). What does each value represent? Which one is the p-value?
- A test has \(x=45\) successes in a sample of \(n=150\). What value should the calculator report for \(\hat{p}\)?
- A right-tailed test has a positive \(z\) statistic and a p-value of 0.032. At \(\alpha=0.05\), state the decision. What should the conclusion say about the alternative?
- A two-sided test has a p-value of 0.084 and uses \(\alpha=0.05\). State the decision and a cautious contextual conclusion.
- Why should you verify the selected alternative before interpreting the p-value, even if the displayed \(\hat{p}\) seems reasonable?