Two-Tailed One-Proportion Test Example

Testing Whether a Proportion Differs from a Benchmark

Some questions ask whether a proportion is higher or lower than a specified value. Others ask more generally whether it has changed or differs from that value, without specifying a direction. That second kind of question calls for a two-sided one-proportion test.

As in Choosing One-Sided or Two-Sided Alternatives, decide the direction from the research question, before examining the sample result. For a two-sided test, evidence against the null can come from a sample proportion noticeably above the benchmark or noticeably below it. As in Calculating a One-Proportion z-Test by Hand, calculate the test statistic using the null proportion in the standard error.

Definition: A two-sided one-proportion test evaluates \(H_0:p=p_0\) against \(H_a:p\ne p_0\), where \(p\) is the true proportion of a specified population with a specified characteristic. Its p-value is the probability, assuming \(H_0\) is true, of obtaining a test statistic at least as far from 0 as the observed statistic, in either direction.

The test statistic is calculated in the same way as for a one-sided one-proportion test. The difference is how the p-value is found: a two-sided test counts outcomes at least as extreme as the observed result in both tails of the null distribution.

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

Because the standard Normal curve is symmetric, the two-sided p-value is twice the area beyond the absolute value of the observed \(z\) statistic. Equivalently, find the smaller tail area beyond \(z_{\mathrm{obs}}\) in the direction of the observed difference, then double it. If the observed statistic is positive, that is the upper tail; if negative, it is the lower tail.

\text{Two-sided p-value} =2P(Z\geq |z_{\mathrm{obs}}|) =2P(Z\leq -|z_{\mathrm{obs}}|)

For instance, if \(z_{\mathrm{obs}}=2.00\), the two-sided p-value is twice the area to the right of 2.00. If \(z_{\mathrm{obs}}=-2.00\), it is twice the area to the left of \(-2.00\). In either case, the p-value is approximately 0.0455. The observed direction affects which tail you calculate first, but not the fact that both equally extreme tails count.

The Four Steps for a Two-Sided Test

A complete test response connects the research question, the conditions, the calculation, and the conclusion. This four-step structure also helps prevent a common error: finding only the tail in the observed direction when the alternative is two-sided.

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

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

Do.
Calculate \(\hat{p}\), the null standard error, and \(z_{\mathrm{obs}}\). Find the tail area beyond \(|z_{\mathrm{obs}}|\) and double it. A 1-PropZTest can calculate the statistic and p-value when the two-sided alternative is selected.

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 that the population proportion differs from the benchmark, in context.

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

On a TI-84, 1-PropZTest takes \(p_0\), the observed number of successes \(x\), the sample size \(n\), and the alternative. Select the not-equal alternative for a two-sided test. Check that the calculator’s \(\hat{p}=x/n\), \(z\), and p-value fit the data and research question. The calculator does not select the hypotheses or verify the study conditions for you.

Worked Examples

Worked Example: Does a Renewal Rate Differ from 60%?

A fictional service randomly selects 600 customers from a list of 12,000 customers whose subscriptions are due for renewal. Of those selected, 390 renew. Test whether the true proportion of these customers who renew differs from 0.60. Use \(\alpha=0.05\).

State: Let \(p\) be the true proportion of customers on this renewal list who renew their subscriptions. The hypotheses are \(H_0:p=0.60\) and \(H_a:p\ne0.60\). The significance level is \(\alpha=0.05\).

Plan and check conditions: Use a one-proportion \(z\)-test. The customers were randomly selected, so the Random condition is met. The sample was drawn without replacement from 12,000 customers, and \(0.10(12{,}000)=1{,}200\). Since \(600\leq1{,}200\), the 10% condition is met. Under the null, the expected number of renewals is \(np_0=600(0.60)=360\), and the expected number of nonrenewals is \(n(1-p_0)=600(0.40)=240\). Both expected counts are at least 10, so the Large Counts condition is met.

Do: The observed sample proportion is:

\hat{p}=\frac{x}{n}=\frac{390}{600}=0.65

Calculate the null standard error and test statistic:

SE_0=\sqrt{\frac{0.60(0.40)}{600}} =\sqrt{0.0004} =0.02, \qquad z_{\mathrm{obs}}=\frac{0.65-0.60}{0.02} =2.50

The result is above the null value, so first find the upper-tail area. Then double it for the two-sided alternative:

\text{p-value} =2\operatorname{normalcdf}(2.50,1\text{E}99,0,1) =2(0.0062097) \approx0.0124

The value 0.0062097 is rounded; the p-value is also rounded. A TI-84 1-PropZTest with \(p_0=0.60\), \(x=390\), \(n=600\), and the not-equal alternative gives approximately the same statistic and p-value.

Conclude: Assuming the true renewal proportion is 0.60, the probability of obtaining a test statistic at least as far from 0 as 2.50, in either direction, is approximately 0.0124. Since \(0.0124<0.05\), reject \(H_0\). The sample provides convincing evidence that the true renewal proportion among customers on this list differs from 0.60.

Worked Example: A Sample Proportion Below 60%

A fictional community program randomly selects 300 people from a list of 6,000 eligible residents. In the sample, 165 say they would use a proposed weekend service. Test whether the true proportion of eligible residents who would use the service differs from 0.60. Use \(\alpha=0.05\).

State: Let \(p\) be the true proportion of eligible residents on this list who would use the proposed weekend service. The hypotheses are \(H_0:p=0.60\) and \(H_a:p\ne0.60\), with \(\alpha=0.05\).

Plan and check conditions: Use a one-proportion \(z\)-test. The residents were randomly selected, meeting the Random condition. The sample was drawn without replacement from 6,000 residents, and \(0.10(6{,}000)=600\). Since \(300\leq600\), the 10% condition is met. Under the null, the expected number who would use the service is \(300(0.60)=180\), and the expected number who would not is \(300(0.40)=120\). 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{165}{300}=0.55, \qquad SE_0=\sqrt{\frac{0.60(0.40)}{300}} =\sqrt{0.0008} \approx0.0282843

z_{\mathrm{obs}} =\frac{0.55-0.60}{0.0282843} \approx-1.7678

The observed statistic is negative, so find the lower-tail area. Double that area to include an equally extreme result in the upper tail:

\text{p-value} =2\operatorname{normalcdf}(-1\text{E}99,-1.76776695,0,1) \approx2(0.03855) \approx0.0771

Conclude: Since \(0.0771>0.05\), fail to reject \(H_0\). The sample does not provide convincing evidence that the true proportion of eligible residents who would use the proposed service differs from 0.60. The sample proportion is below 0.60, but a two-sided test considers departures in either direction, and this result is not sufficiently unusual at the 0.05 significance level.

Worked Example: A Small Difference from 60%

A fictional school district randomly selects 250 families from a list of 5,000 families and asks whether they would use a new online scheduling tool. Of those selected, 157 say yes. Test whether the true proportion of families on the list who would use the tool differs from 0.60. Use \(\alpha=0.10\).

State: Let \(p\) be the true proportion of families on this district list who would use the new online scheduling tool. The hypotheses are \(H_0:p=0.60\) and \(H_a:p\ne0.60\), with \(\alpha=0.10\).

Plan and check conditions: A one-proportion \(z\)-test is appropriate if its conditions hold. The families were randomly selected, meeting the Random condition. The sample was drawn without replacement from 5,000 families, and \(0.10(5{,}000)=500\). Since \(250\leq500\), the 10% condition is met. Under the null, the expected number of families who would use the tool is \(250(0.60)=150\), and the expected number who would not is \(250(0.40)=100\). Both expected counts are at least 10, so the Large Counts condition is met.

Do: The sample proportion and null standard error are:

\hat{p}=\frac{157}{250}=0.628, \qquad SE_0=\sqrt{\frac{0.60(0.40)}{250}} =\sqrt{0.00096} \approx0.0309839

The test statistic is:

z_{\mathrm{obs}} =\frac{0.628-0.60}{0.0309839} \approx0.9037

The statistic is positive, so calculate the upper-tail area and double it:

\text{p-value} =2\operatorname{normalcdf}(0.903696,1\text{E}99,0,1) \approx2(0.1831) \approx0.3662

Conclude: Since \(0.3662>0.10\), fail to reject \(H_0\). The sample does not provide convincing evidence that the true proportion of families on this district list who would use the online scheduling tool differs from 0.60. A sample proportion above the benchmark is not, by itself, strong evidence of a difference; the size of the standardized difference and the two-sided p-value matter.

Common Mistakes and What Full Credit Says

The alternative hypothesis determines how extreme results are counted. In a two-sided test, both unusually high and unusually low sample proportions count as evidence against \(H_0\). The doubled tail area is part of the p-value, not an optional adjustment.

Writing a one-sided alternative for a “differs” question. If the question does not specify higher or lower, use \(H_a:p\ne p_0\). Do not choose a direction just because \(\hat{p}\) happens to be above or below \(p_0\).
Reporting only one tail. For a two-sided test, the p-value is twice the tail area beyond \(|z_{\mathrm{obs}}|\). Reporting just the area on the observed side answers a one-sided question instead.
Doubling the wrong area. Double the smaller tail beyond the observed statistic, not the larger area between the statistic and the center of the curve. For a negative statistic, find the area to its left; for a positive statistic, find the area to its right.
Using the wrong standard error. In a one-proportion test, use \(p_0\) in \(SE_0\), because the test models sampling variation under the null hypothesis. Do not substitute \(\hat{p}\).
Using observed counts for the test’s Large Counts check. Check \(np_0\) and \(n(1-p_0)\), the expected counts under the null, rather than \(x\) and \(n-x\).
Interpreting a large p-value as proof of equality. When the p-value is greater than \(\alpha\), say “fail to reject \(H_0\)” and explain that the data do not provide convincing evidence of a difference. Do not say that the null has been proved or accepted.

AP Exam Tip: A complete response defines \(p\), states both hypotheses, checks the Random, 10%, and Large Counts conditions with evidence, reports the test statistic and doubled p-value, compares that p-value with \(\alpha\), and concludes in context. For a two-sided claim, the conclusion should say whether there is convincing evidence that the population proportion “differs from” the benchmark—not that it is specifically higher or lower unless the question asks that.

Key Takeaway

A two-sided one-proportion test asks whether the population proportion differs from a benchmark in either direction. Calculate the test statistic with the null standard error, find the area beyond its absolute value, and double that area. Then compare the p-value with \(\alpha\) and describe the evidence in context.

Key takeaway: For \(H_0:p=p_0\) against \(H_a:p\ne p_0\), the p-value is twice the area in one tail beyond \(|z_{\mathrm{obs}}|\). Reject \(H_0\) only when the p-value is at most \(\alpha\).

Check Your Understanding

For each question, consider the hypotheses, conditions, tail areas, or conclusion for a two-sided one-proportion test.

A researcher asks whether the proportion of households using a particular heating source differs from 0.35. Write the null and alternative hypotheses using \(p\).
For a test of \(H_0:p=0.60\) with \(n=250\), calculate both expected counts for the Large Counts condition. Does the condition hold?
A two-sided test has observed \(z=-1.25\). Which tail area should you find first, and what should you do with that area to obtain the p-value?
A two-sided test at \(\alpha=0.05\) gives a p-value of 0.032. State the decision and interpret it in context for a claim that the proportion differs from its null value.
Why does a two-sided test count results in both tails, even when the observed sample proportion is above the null proportion?