Running 1-PropZTest on the Calculator

Let the Calculator Carry Out the Test

In Finding a P-Value from a z Statistic with normalcdf, you used a calculated \(z\) statistic and the alternative hypothesis to find a p-value. The TI-84’s 1-PropZTest runs those calculations together: you enter the null proportion, the observed number of successes, the sample size, and the direction of the alternative.

The calculator is doing arithmetic, not making the statistical decisions for you. You still need to define the population proportion, choose hypotheses that fit the question, check the conditions, and interpret the result in context. As in Calculating a One-Proportion z-Test by Hand, the test statistic uses the null proportion \(p_0\) in its standard error.

Definition: The TI-84’s 1-PropZTest is a calculator procedure for a one-proportion \(z\)-test. It uses \(p_0\), the observed number of successes \(x\), and the sample size \(n\) to report the sample proportion \(\hat{p}\), test statistic \(z\), and p-value for the selected alternative.

Enter the Hypothesis Test Correctly

First write the hypotheses and define \(p\) in context. For the standard test, the null hypothesis is \(H_0:p=p_0\), and the alternative is one of \(H_a:p>p_0\), \(H_a:p<p_0\), or \(H_a:p\ne p_0\). The alternative is chosen from the research question, not from the direction that would make the observed data look more convincing.

On a TI-84, open STAT, move to TESTS, and select 1-PropZTest (usually option 5). Fill in the fields as follows:

p0: Enter the null proportion \(p_0\), as a decimal between 0 and 1.
x: Enter the observed number of successes, not the sample proportion or a percentage.
n: Enter the total sample size.
Alternative: Select \(p\ne p_0\), \(p<p_0\), or \(p>p_0\), matching \(H_a\).
Select Calculate to display the results.

The calculator typically displays \(z\), the p-value (often labeled \(p\)), and \(\hat{p}\). Here the output labeled \(p\) means the p-value, not the population proportion \(p\). The displayed \(\hat{p}\) should equal \(x/n\). The displayed \(z\) should agree with the one-proportion test statistic formula, and the p-value should agree with the tail or tails specified by the alternative.

\hat{p}=\frac{x}{n}, \qquad SE_0=\sqrt{\frac{p_0(1-p_0)}{n}}, \qquad z=\frac{\hat{p}-p_0}{SE_0}

After checking those outputs, compare the p-value with the significance level \(\alpha\), if one is given. A p-value at or below \(\alpha\) leads to rejecting \(H_0\); a p-value above \(\alpha\) leads to failing to reject \(H_0\). The output does not tell you whether the study’s sampling method supports the intended conclusion, so check conditions before relying on the test.

Conditions: Before using the 1-PropZTest result, check the Random condition, the 10% condition when sampling without replacement from a finite population, and the test’s Large Counts condition. For \(H_0:p=p_0\), that condition is \(np_0\geq10\) and \(n(1-p_0)\geq10\). These are expected counts under the null, not the observed counts.

A Reliable Calculator Workflow

State the test.
Define \(p\), write \(H_0\) and \(H_a\), and identify the significance level if one is provided.

Check conditions.
Use the study description to assess randomness and independence, then calculate the two expected counts using \(p_0\).

Enter the data.
Type \(p_0\), \(x\), and \(n\); select the alternative that matches \(H_a\); and calculate.

Verify and conclude.
Compare the calculator’s \(\hat{p}\), \(z\), and p-value with the hand work, then make a conclusion in context.

This order helps catch common input errors before they affect the conclusion. In particular, \(x\) is a count, while \(p_0\) is a proportion. If 72 of 120 sampled people meet the success criterion, enter \(x=72\), not \(0.60\) or \(60\). The calculator obtains \(\hat{p}=0.60\) from \(x/n\).

Worked Examples

Worked Example: Testing Whether a Rate Is Higher

A fictional random sample of 120 customers at a meal-delivery service includes 78 who say their latest delivery arrived within the promised time. The service’s benchmark is 60%. Test whether the true proportion of customers whose latest delivery arrives within the promised time is higher than 0.60, using \(\alpha=0.05\). The sample was drawn without replacement from 2,000 customers.

State: Let \(p\) be the true proportion of the service’s customers whose latest delivery arrives within the promised time. The hypotheses are \(H_0:p=0.60\) and \(H_a:p>0.60\).

Plan and check conditions: The customers were randomly sampled, so the Random condition is met. The sample was drawn without replacement from 2,000 customers, and \(120\leq0.10(2000)=200\), so the 10% condition is met. Under \(H_0\), the expected number of successes is \(np_0=120(0.60)=72\), and the expected number of failures is \(n(1-p_0)=120(0.40)=48\). Both are at least 10, so the Large Counts condition is met. A one-proportion \(z\)-test is appropriate.

Do by hand: The observed sample proportion is \(78/120=0.65\). The null standard error and test statistic are:

SE_0=\sqrt{\frac{0.60(0.40)}{120}}=\sqrt{0.002}\approx0.04472, \qquad z=\frac{0.65-0.60}{0.04472}\approx1.118

On the calculator, enter p0 = 0.60, x = 78, and n = 120. Select the alternative \(p>p_0\), then choose Calculate. The output should show \(\hat{p}=0.65\), \(z\approx1.118\), and a p-value of about 0.1318. As a hand check, this is the upper-tail area \(\operatorname{normalcdf}(1.118,1\text{E}99,0,1)\approx0.1318\).

Conclude: Assuming the true proportion of customers whose latest delivery arrives within the promised time is 0.60, the probability of obtaining a test statistic of 1.118 or greater is about 0.1318. Since \(0.1318>0.05\), fail to reject \(H_0\). The sample does not provide convincing evidence that more than 60% of the service’s customers receive their latest delivery within the promised time.

Worked Example: Testing Whether a Proportion Is Lower

A fictional random sample of 150 residents in a town includes 48 who report using a particular bus route at least once in the past week. A transit planner wants to know whether the true proportion of residents who used the route is below 0.40. The sample was drawn without replacement from 3,000 residents.

Let \(p\) be the true proportion of the town’s residents who used the bus route at least once in the past week. The hypotheses are \(H_0:p=0.40\) and \(H_a:p<0.40\).

The residents were randomly sampled, satisfying the Random condition. The sample size is no more than 10% of the source population because \(150\leq0.10(3000)=300\). Under the null, the expected counts are \(np_0=150(0.40)=60\) successes and \(n(1-p_0)=150(0.60)=90\) failures. Both counts are at least 10, so the Large Counts condition is met.

Enter p0 = 0.40, x = 48, and n = 150, and select the less-than alternative. The sample proportion is \(48/150=0.32\). The hand calculation gives:

SE_0=\sqrt{\frac{0.40(0.60)}{150}}=\sqrt{0.0016}=0.04, \qquad z=\frac{0.32-0.40}{0.04}=-2.00

The calculator should report \(\hat{p}=0.32\), \(z=-2.00\), and a p-value of about 0.0228. The alternative is lower, so the p-value is the area to the left of \(-2.00\): \(\operatorname{normalcdf}(-1\text{E}99,-2.00,0,1)\approx0.0228\). The negative statistic and lower-tail selection agree with the observed sample proportion being below 0.40.

If the significance level is \(\alpha=0.05\), reject \(H_0\) because \(0.0228<0.05\). The sample provides convincing evidence that less than 40% of the town’s residents used the bus route at least once in the past week.

Worked Example: Testing Whether a Proportion Differs

A fictional random sample of 100 students at a high school includes 61 who bring a reusable water bottle to school at least three days per week. A student group asks whether the true proportion differs from 0.50. The sample was drawn without replacement from 1,500 students.

Let \(p\) be the true proportion of students at the school who bring a reusable water bottle at least three days per week. The hypotheses are \(H_0:p=0.50\) and \(H_a:p\ne0.50\).

The students were randomly sampled, so the Random condition is met. Also, \(100\leq0.10(1500)=150\), verifying the 10% condition. Under the null, the expected success count is \(100(0.50)=50\), and the expected failure count is \(100(0.50)=50\). Both are at least 10, so the Large Counts condition is met.

Enter p0 = 0.50, x = 61, and n = 100. Select the not-equal-to alternative. The hand calculation gives \(\hat{p}=61/100=0.61\) and:

SE_0=\sqrt{\frac{0.50(0.50)}{100}}=0.05, \qquad z=\frac{0.61-0.50}{0.05}=2.20

The calculator should display \(\hat{p}=0.61\), \(z=2.20\), and a two-sided p-value of about 0.0278. To verify it by hand, find the upper-tail area beyond 2.20 and double it:

2\operatorname{normalcdf}(2.20,1\text{E}99,0,1) =2(0.0139)\approx0.0278

At \(\alpha=0.05\), reject \(H_0\). The sample provides convincing evidence that the proportion of students at the school who bring a reusable water bottle at least three days per week differs from 0.50. The conclusion says “differs,” not “is greater,” because the alternative allows departures in either direction.

Common Mistakes and What Full Credit Says

A calculator result is useful only when its inputs and output match the problem. Before accepting the screen, check the entries against the written hypotheses and the observed data.

Entering \(\hat{p}\) instead of \(x\). The x field requires a count of successes. Entering 0.65 instead of 78 does not tell the calculator that 78 out of 120 observations were successes.
Entering a percentage as a whole number for p0. A null proportion of 60% must be entered as 0.60, not 60.
Selecting the alternative after seeing the sample result. The research question determines the alternative. A sample proportion above \(p_0\) does not by itself justify choosing \(p>p_0\).
Confusing the output labels. In the calculator output, \(\hat{p}\) is the sample proportion and \(p\) is commonly the p-value. Verify that \(\hat{p}=x/n\) before interpreting the p-value.
Reporting only calculator output. A complete inference response names the parameter and hypotheses, justifies conditions, reports the test result, and concludes in context. The calculator cannot supply those explanations.
Using observed counts for a test’s Large Counts check. Check \(np_0\) and \(n(1-p_0)\), the expected counts under \(H_0\), as in Checking the Success-Failure Condition for Tests.

AP Exam Tip: Show the entries or identify \(p_0\), \(x\), \(n\), and the selected alternative; report the calculator’s \(z\) and p-value; and connect them to the hand formulas. A full-credit conclusion compares the p-value with \(\alpha\), uses “reject” or “fail to reject,” and states whether the data provide convincing evidence for the alternative in context.

Key Takeaway

The 1-PropZTest can quickly calculate a one-proportion test statistic and p-value, but it cannot choose the question’s hypotheses or verify the study conditions for you. Enter the null proportion, success count, sample size, and alternative carefully; then compare the output with the hand work and explain what the test says about the population proportion.

Key takeaway: In 1-PropZTest, enter \(p_0\), \(x\), and \(n\), then select the alternative that matches \(H_a\). Check that the calculator’s \(\hat{p}=x/n\), its \(z\) agrees with the null-standard-error calculation, and its p-value uses the correct tail or tails.

Check Your Understanding

Answer each question by describing the correct entry or interpretation.

A sample has \(n=90\) observations, of which \(x=54\) are successes. What values should be entered for \(x\) and \(n\), and what should the calculator report for \(\hat{p}\)?
For \(H_0:p=0.35\) against \(H_a:p<0.35\), what values should be entered for p0, and which alternative should be selected?
A sample of 80 is taken without replacement from a population of 700. Does it meet the 10% condition? Show the comparison.
For a test with \(p_0=0.25\) and \(n=60\), calculate the two expected counts. Does the Large Counts condition hold?
The calculator reports a p-value of 0.08 for a test at \(\alpha=0.05\). What is the decision, and how should the conclusion be worded in context?