Right-Tailed One-Proportion Test Example

Testing a Claim That a Proportion Is Higher

A question such as “Is the vaccination rate greater than 80%?” calls for a right-tailed one-proportion test. The direction matters: the test measures how surprising the sample result would be if the population proportion were 80%, looking only for results above that benchmark.

As in Choosing One-Sided or Two-Sided Alternatives, choose the alternative from the research question before examining the sample results. As in Calculating a One-Proportion z-Test by Hand, use the null proportion \(p_0\) to calculate the standard error. The new task is to connect those choices and calculations in a complete test.

Definition: A right-tailed one-proportion test evaluates \(H_0:p=p_0\) against \(H_a:p>p_0\), where \(p\) is the true proportion of a specified population with a specified characteristic. The p-value is the upper-tail probability, assuming \(H_0\) is true, of obtaining a test statistic at least as large as the observed statistic.

The observed sample proportion is \(\hat{p}=x/n\), where \(x\) is the number of successes and \(n\) is the sample size. Under the null hypothesis, the test statistic is:

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

For a right-tailed alternative, the p-value is the area to the right of the observed \(z\) statistic under the standard Normal curve. A large positive \(z\) statistic gives a small p-value and may be evidence that \(p>p_0\). A negative statistic puts the observed sample proportion below \(p_0\), so it cannot produce a small right-tail p-value.

The Four Steps for a Right-Tailed Test

A calculator can report the test statistic and p-value, but a complete answer also explains what is being tested, why the procedure is appropriate, and what the result means in context. Use the four-step structure below.

State.
Define \(p\) in context, write \(H_0:p=p_0\) and \(H_a:p>p_0\), and identify the significance level \(\alpha\) if it is given.

Plan and check conditions.
Identify the one-proportion \(z\)-test. Check the Random condition, the 10% condition for sampling without replacement from a finite population, and Large Counts using \(np_0\) and \(n(1-p_0)\).

Do.
Calculate \(\hat{p}\), the null standard error, the \(z\) statistic, and the right-tail p-value. A 1-PropZTest can also calculate these values if its entries and alternative are correct.

Conclude.
Compare the p-value with \(\alpha\). Reject \(H_0\) when the p-value is at most \(\alpha\); otherwise, fail to reject \(H_0\). State whether the data provide convincing evidence for the alternative, in context.

Conditions: For a one-proportion \(z\)-test, the data should come from a random sample or an appropriate randomized process. If sampling without replacement from a finite population, check \(n\leq0.10N\). Under \(H_0:p=p_0\), check the Large Counts condition: \(np_0\geq10\) and \(n(1-p_0)\geq10\). The test condition uses expected counts under the null, not the observed counts.

For a right-tailed test, the TI-84’s 1-PropZTest should be set to the greater-than alternative. Enter \(p_0\), the success count \(x\), and the sample size \(n\), then check that the displayed \(\hat{p}\), \(z\), and p-value make sense. The calculator does not select the hypotheses or verify the conditions for you.

Worked Examples

Worked Example: Is the Vaccination Rate Above 80%?

A fictional health department takes a random sample of 150 adults from a community of 2,400 adults. In the sample, 129 report receiving a specified vaccination. Test whether the true proportion of adults in this community who received the vaccination exceeds 0.80. Use \(\alpha=0.05\).

State: Let \(p\) be the true proportion of adults in this community who received the specified vaccination. The hypotheses are \(H_0:p=0.80\) and \(H_a:p>0.80\). The significance level is \(\alpha=0.05\).

Plan and check conditions: This is a one-proportion \(z\)-test. The adults were randomly sampled, so the Random condition is met. The sample was drawn without replacement from 2,400 adults, and \(150\leq0.10(2400)=240\), so the 10% condition is met. Under the null, the expected success count is \(np_0=150(0.80)=120\), and the expected failure count is \(n(1-p_0)=150(0.20)=30\). Both are at least 10, so the Large Counts condition is met.

Do: The observed sample proportion is:

\hat{p}=\frac{x}{n}=\frac{129}{150}=0.86

Using \(p_0=0.80\), the null standard error and test statistic are:

SE_0=\sqrt{\frac{0.80(0.20)}{150}} =\sqrt{0.0010667}\approx0.03266, \qquad z=\frac{0.86-0.80}{0.03266}\approx1.837

On a TI-84, enter \(p_0=0.80\), \(x=129\), and \(n=150\), and select the greater-than alternative. The calculator should give \(\hat{p}=0.86\), \(z\approx1.837\), and a right-tail p-value of about 0.0331. This matches the upper-tail area \(\operatorname{normalcdf}(1.837,1\text{E}99,0,1)\approx0.0331\), rounded.

Conclude: Assuming the true proportion of adults in this community who received the vaccination is 0.80, the probability of obtaining a test statistic of 1.837 or greater is about 0.0331. Since \(0.0331<0.05\), reject \(H_0\). The sample provides convincing evidence that more than 80% of adults in this community received the specified vaccination.

Worked Example: A Sample Proportion Above the Benchmark Is Not Always Convincing Evidence

A fictional library system randomly samples 200 cardholders from a population of 5,000. Of those sampled, 150 say they borrowed at least one digital book during the past month. Test whether the true proportion of cardholders who did so is greater than 0.70. Use \(\alpha=0.05\).

State: Let \(p\) be the true proportion of the system’s cardholders who borrowed at least one digital book during the past month. The hypotheses are \(H_0:p=0.70\) and \(H_a:p>0.70\).

Plan and check conditions: A one-proportion \(z\)-test is appropriate if its conditions are met. The cardholders were randomly sampled, meeting the Random condition. Because \(200\leq0.10(5000)=500\), the 10% condition is met. The null expected counts are \(np_0=200(0.70)=140\) successes and \(n(1-p_0)=200(0.30)=60\) failures. Both are at least 10, so the Large Counts condition is met.

Do: Calculate the sample proportion, null standard error, and test statistic:

\hat{p}=\frac{150}{200}=0.75, \qquad SE_0=\sqrt{\frac{0.70(0.30)}{200}} =\sqrt{0.00105}\approx0.03240, \qquad z=\frac{0.75-0.70}{0.03240}\approx1.543

For the right-tailed alternative, the p-value is the area above \(z=1.543\). Using a calculator gives \(\operatorname{normalcdf}(1.543,1\text{E}99,0,1)\approx0.0614\), rounded. The TI-84 1-PropZTest with \(p_0=0.70\), \(x=150\), and \(n=200\), using the greater-than alternative, should give the same result.

Conclude: Since \(0.0614>0.05\), fail to reject \(H_0\). The sample proportion is 0.75, above the 0.70 benchmark, but the difference is not sufficiently surprising under the null at the 0.05 significance level. The sample does not provide convincing evidence that more than 70% of the system’s cardholders borrowed at least one digital book during the past month.

Worked Example: Evidence Depends on the Significance Level

A fictional software company randomly samples 100 of its 3,000 active users. Seventy-nine sampled users say they enabled a new security feature. Test whether the true proportion of active users who enabled the feature exceeds 0.70, using \(\alpha=0.01\).

State: Let \(p\) be the true proportion of the company’s active users who enabled the new security feature. The hypotheses are \(H_0:p=0.70\) and \(H_a:p>0.70\). Here \(\alpha=0.01\).

Plan and check conditions: The users were randomly sampled, meeting the Random condition. The sample was drawn without replacement from 3,000 users, and \(100\leq0.10(3000)=300\), so the 10% condition is met. Under the null, the expected success count is \(100(0.70)=70\), and the expected failure count is \(100(0.30)=30\). Both are at least 10, so the Large Counts condition is met.

Do: The sample proportion is \(79/100=0.79\). The test statistic and upper-tail p-value are:

SE_0=\sqrt{\frac{0.70(0.30)}{100}} =\sqrt{0.0021}\approx0.04583, \qquad z=\frac{0.79-0.70}{0.04583}\approx1.964

For the right-tailed test, \(\operatorname{normalcdf}(1.964,1\text{E}99,0,1)\approx0.0248\), rounded. Entering \(p_0=0.70\), \(x=79\), and \(n=100\) in 1-PropZTest with the greater-than alternative should produce a p-value of about 0.0248.

Conclude: Since \(0.0248>0.01\), fail to reject \(H_0\) at the 0.01 significance level. The sample does not provide convincing evidence, at this significance level, that more than 70% of the company’s active users enabled the feature. Failing to reject does not prove that the true proportion is 0.70; it means this test did not reach the stated evidence threshold.

Common Mistakes and What Full Credit Says

The alternative determines which outcomes count as evidence against \(H_0\). For \(H_a:p>p_0\), use the right tail. A different direction changes the p-value, so choosing the tail correctly is part of the statistical reasoning, not just a calculator setting.

Choosing the direction after seeing the data. Write \(H_a:p>p_0\) only when the question asks whether the population proportion is higher. A sample proportion above \(p_0\) is not a reason to switch to a greater-than alternative after the fact.
Using the wrong tail. For a positive observed \(z\), the right-tail p-value is the area above \(z\), not the area below it and not twice the right-tail area. A right-tailed test is one-sided.
Calling the p-value the probability that \(H_0\) is true. The p-value is calculated assuming \(H_0\) is true. It describes how unusual the observed statistic, or one still farther in the direction of \(H_a\), would be under that assumption.
Checking the wrong counts. For the test’s Large Counts condition, calculate \(np_0\) and \(n(1-p_0)\). The observed counts \(x\) and \(n-x\) are not the counts used for this test condition.
Writing “accept the null.” A p-value above \(\alpha\) leads to “fail to reject \(H_0\),” not “accept” or “prove” \(H_0\). State that the data do not provide convincing evidence for the alternative at the chosen significance level.
Leaving out the context. A conclusion should name the population characteristic, not stop at “reject” or “the result is significant.”

AP Exam Tip: A full-credit response identifies \(p\) and the hypotheses, checks all relevant conditions with evidence, reports the test statistic and right-tail p-value, compares the p-value with \(\alpha\), and gives a conclusion in context. For example: “Because the p-value is less than \(\alpha\), reject \(H_0\). The data provide convincing evidence that [the population proportion] is greater than [the benchmark].”

Key Takeaway

A right-tailed one-proportion test asks whether the population proportion exceeds a specified benchmark. The research question sets \(H_a:p>p_0\), and the p-value is the area to the right of the observed \(z\) statistic. Check the conditions, compare the p-value with \(\alpha\), and describe the strength of evidence in context without claiming that the test proves either hypothesis.

Key takeaway: For a right-tailed test of \(H_0:p=p_0\) against \(H_a:p>p_0\), use the null standard error, find the upper-tail p-value, and reject \(H_0\) only when that p-value is at most \(\alpha\).

Check Your Understanding

For each question, explain the test direction or interpret the result as requested.

A researcher asks whether more than 65% of residents use a community garden. Write the alternative hypothesis using \(p\), the true proportion of residents who use the garden.
For \(H_0:p=0.40\) with \(n=100\), calculate both expected counts for the test’s Large Counts condition. Does the condition hold?
A right-tailed test has observed \(z=1.50\). Which area under the standard Normal curve is the p-value: to the left or to the right of 1.50?
A test at \(\alpha=0.05\) gives a p-value of 0.12. State the decision and what it means for a claim that the population proportion is greater than \(p_0\).
Why should a right-tailed alternative be chosen before examining the sample proportion?