Left-Tailed One-Proportion Test Example

Testing a Claim That a Proportion Is Lower

A question such as “Has the defect rate fallen below 5%?” calls for a left-tailed one-proportion test. The test asks whether a sample result this low, or lower, would be unusual if the population proportion were still 5%.

As in Choosing One-Sided or Two-Sided Alternatives, the research question sets the direction of the alternative hypothesis before you examine the sample results. As in Calculating a One-Proportion z-Test by Hand, the null proportion \(p_0\) is used to calculate the standard error. The key left-tail decision is to measure probability in the direction of values below the null benchmark.

Definition: A left-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. Its p-value is the probability, assuming \(H_0\) is true, of obtaining a test statistic at least as small as the observed statistic.

The test statistic compares the observed sample proportion \(\hat{p}=x/n\) with the null value \(p_0\), in standard-error units. The formula is the same one used for a right-tailed one-proportion test; the alternative hypothesis determines which tail supplies the p-value.

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

For a left-tailed test, the p-value is the area to the left of the observed \(z\) statistic under the standard Normal curve. A very negative \(z\) means the sample proportion is well below \(p_0\), which can produce a small left-tail p-value. A positive \(z\) means the sample proportion is above \(p_0\); that result is not in the direction of the alternative and will not provide strong evidence for a lower proportion.

For example, if a problem gives \(z=-1.835\), the left-tail area is \(\operatorname{normalcdf}(-1\text{E}99,-1.835,0,1)\approx0.03325\), rounded. Do not double that area: the alternative is one-sided. When a test statistic comes from sample data, use the calculator’s unrounded statistic to find the p-value, even if you report \(z\) rounded in your written answer.

The Four Steps for a Left-Tailed Test

A complete test response connects the research question, the conditions, the calculation, and the conclusion. The four-step structure below keeps those parts together.

State.
Define \(p\) in context, write \(H_0:p=p_0\) and \(H_a:p<p_0\), 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, the \(z\) statistic, and the left-tail p-value. A 1-PropZTest can calculate the statistic and p-value if the inputs and less-than 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: 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, use 1-PropZTest, enter \(p_0\), the observed success count \(x\), and the sample size \(n\), and select the less-than alternative. Check that the calculator’s \(\hat{p}=x/n\), \(z\), and p-value fit the sample and the direction of the question. The calculator does not choose the hypotheses or verify the study conditions.

Worked Examples

Worked Example: Is a Defect Rate Below 5%?

A fictional factory randomly selects 400 components from a production run of 10,000. Twelve selected components are defective. Test whether the true proportion of components in this run that are defective is below 0.05. Use \(\alpha=0.05\).

State: Let \(p\) be the true proportion of components in this production run that are defective. The hypotheses are \(H_0:p=0.05\) and \(H_a:p<0.05\). The significance level is \(\alpha=0.05\).

Plan and check conditions: Use a one-proportion \(z\)-test. The components were randomly selected, so the Random condition is met. The sample was drawn without replacement from 10,000 components, and \(400\leq0.10(10{,}000)=1{,}000\), so the 10% condition is met. Under the null, the expected number of defective components is \(np_0=400(0.05)=20\), and the expected number of nondefective components is \(n(1-p_0)=400(0.95)=380\). 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{12}{400}=0.03

Calculate the null standard error and test statistic, keeping precision in the intermediate calculations:

SE_0=\sqrt{\frac{0.05(0.95)}{400}} =\sqrt{0.00011875} \approx0.01089725, \qquad z=\frac{0.03-0.05}{0.01089725} \approx-1.8353

The left-tail p-value using the unrounded statistic is \(\operatorname{normalcdf}(-1\text{E}99,-1.835325,0,1)\approx0.03323\), rounded. A TI-84 1-PropZTest with \(p_0=0.05\), \(x=12\), \(n=400\), and the less-than alternative should give approximately the same statistic and p-value.

Conclude: Assuming the true defect proportion is 0.05, the probability of obtaining a test statistic of about \(-1.8353\) or smaller is approximately 0.03323. Since \(0.03323<0.05\), reject \(H_0\). The sample provides convincing evidence that fewer than 5% of the components in this production run are defective.

Worked Example: A Larger Sample Gives Stronger Evidence

A second fictional production run contains 20,000 components. A random sample of 500 components includes 12 defective components. Test whether the true defect proportion in this run is below 0.05, using \(\alpha=0.01\).

State: Let \(p\) be the true proportion of components in this production run that are defective. The hypotheses are \(H_0:p=0.05\) and \(H_a:p<0.05\). Here, \(\alpha=0.01\).

Plan and check conditions: A one-proportion \(z\)-test is appropriate if its conditions hold. The components were randomly selected, meeting the Random condition. Because \(500\leq0.10(20{,}000)=2{,}000\), the 10% condition is met. Under the null, the expected success count is \(np_0=500(0.05)=25\), and the expected failure count is \(n(1-p_0)=500(0.95)=475\). Both are at least 10, so the Large Counts condition is met.

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

\hat{p}=\frac{12}{500}=0.024, \qquad SE_0=\sqrt{\frac{0.05(0.95)}{500}} =\sqrt{0.000095} \approx0.00974679

z=\frac{0.024-0.05}{0.00974679} =\frac{-0.026}{0.00974679} \approx-2.668

Using the unrounded value of \(z\), the left-tail p-value is approximately \(\operatorname{normalcdf}(-1\text{E}99,-2.66754,0,1)=0.00382\), rounded. The count-based calculation gives the same statistic: under the null there are 25 expected successes, so \((12-25)/\sqrt{500(0.05)(0.95)}=-13/\sqrt{23.75}\approx-2.668\). The calculator’s 1-PropZTest, with the less-than alternative, should agree.

Conclude: Since \(0.00382<0.01\), reject \(H_0\). The sample provides convincing evidence, at the 0.01 significance level, that fewer than 5% of the components in this production run are defective. This example has a smaller observed sample proportion than 0.05 and a larger sample size than the first example; both facts contribute to a test statistic farther into the left tail.

Worked Example: A Low Sample Proportion Is Not Necessarily Convincing

A fictional city randomly samples 200 buildings from 3,000 buildings in a specified neighborhood. Eight sampled buildings have a particular type of outdated heating system. Test whether the true proportion of buildings in that neighborhood with this system is below 0.05. Use \(\alpha=0.05\).

State: Let \(p\) be the true proportion of buildings in the specified neighborhood with this type of outdated heating system. The hypotheses are \(H_0:p=0.05\) and \(H_a:p<0.05\).

Plan and check conditions: Use a one-proportion \(z\)-test. The buildings were randomly selected, meeting the Random condition. The sample was drawn without replacement, and \(200\leq0.10(3{,}000)=300\), so the 10% condition is met. Under the null, the expected number of buildings with the system is \(200(0.05)=10\), and the expected number without it is \(200(0.95)=190\). Both are at least 10, so the Large Counts condition is met.

Do: The sample proportion is \(8/200=0.04\). The null standard error and test statistic are:

SE_0=\sqrt{\frac{0.05(0.95)}{200}} =\sqrt{0.0002375} \approx0.01541104, \qquad z=\frac{0.04-0.05}{0.01541104} \approx-0.6489

The left-tail p-value is \(\operatorname{normalcdf}(-1\text{E}99,-0.6489,0,1)\approx0.2582\), rounded. Using the less-than alternative in 1-PropZTest should give a p-value of approximately 0.2582.

Conclude: Since \(0.2582>0.05\), fail to reject \(H_0\). Although the sample proportion, 0.04, is below 0.05, the difference is not sufficiently surprising under the null at the 0.05 significance level. The sample does not provide convincing evidence that fewer than 5% of the buildings in this neighborhood have the specified heating system.

Common Mistakes and What Full Credit Says

The alternative hypothesis determines the tail. In a left-tailed test, only outcomes at or below the observed statistic count as evidence against the null in the direction of the claim. A correct calculation with the wrong tail answers a different question.

Choosing the alternative after seeing the data. Write \(H_a:p<p_0\) when the research question asks whether the population proportion is lower. Do not choose a left-tailed alternative merely because the sample proportion happened to fall below \(p_0\).
Using the right tail or doubling the area. For \(H_a:p<p_0\), find the area to the left of the observed \(z\). Do not use the right-tail area and do not double the left-tail area; this is a one-sided test.
Rounding too early. Use the unrounded standard error and test statistic to calculate the p-value. Report rounded values consistently, and do not let a rounded \(z\) value create a small discrepancy in the reported p-value.
Checking observed counts for the test. The Large Counts condition for a one-proportion test uses \(np_0\) and \(n(1-p_0)\), not the observed counts \(x\) and \(n-x\). As discussed in Checking the Success-Failure Condition for Tests, the test checks expected counts under the null model.
Writing “accept the null.” If the p-value exceeds \(\alpha\), write “fail to reject \(H_0\).” This does not prove \(p=p_0\); it means the evidence is not strong enough to support the alternative at the stated significance level.
Leaving the conclusion out of context. Do not stop at “reject” or “the result is significant.” Identify the population characteristic and state whether the data provide convincing evidence that its proportion is below the benchmark.

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 left-tail p-value, compares the p-value with \(\alpha\), and concludes in context. Use “convincing evidence that the population proportion is less than the benchmark” when you reject \(H_0\); use “does not provide convincing evidence” when you fail to reject.

Key Takeaway

A left-tailed one-proportion test evaluates whether a population proportion is below a benchmark. Write \(H_a:p<p_0\) from the research question, use the null standard error, and find the area to the left of the observed \(z\) statistic. Check the conditions and compare the p-value with \(\alpha\) before describing the evidence in context.

Key takeaway: For a left-tailed test of \(H_0:p=p_0\) against \(H_a:p<p_0\), use the null standard error and the lower-tail p-value. Reject \(H_0\) only when the p-value is at most \(\alpha\).

Check Your Understanding

For each question, identify the direction, check the relevant test reasoning, or interpret the result.

A researcher asks whether fewer than 30% of residents use a public transit pass. Write the alternative hypothesis using \(p\), the true proportion of residents who use the pass.
For a test of \(H_0:p=0.08\) with \(n=200\), calculate both expected counts for the Large Counts condition. Does the condition hold?
A left-tailed test has observed \(z=-1.40\). Is the p-value the area to the left or right of \(-1.40\)? Should that area be doubled?
A left-tailed test at \(\alpha=0.05\) gives a p-value of 0.09. State the decision and interpret it for the claim that \(p<p_0\).
Why does the test’s Large Counts condition use \(p_0\) rather than the observed sample proportion \(\hat{p}\)?