Improving Power While Keeping Alpha Fixed
In “Calculating Power for a One-Proportion Test,” we used the null proportion to find a rejection region, then used a specified alternative proportion to estimate the probability of landing in that region. Here, we compare ways a study can make that probability larger while keeping the significance level \(\alpha\) unchanged.
The main possibilities are to collect a larger sample, study a situation where the true proportion is farther from the null value, or have less variability in the statistic. These ideas are related: power depends on how far the alternative sampling distribution lies from the rejection boundary, measured relative to its spread. A larger sample usually makes that spread smaller. A larger effect places the alternative mean farther from the null. Less variability can also make the rejection region easier to reach.
For a one-proportion test, the approximate standard deviation of \(\hat{p}\) when the true proportion is \(p\) is \(\sqrt{p(1-p)/n}\). This helps explain the strategies. Increasing \(n\) reduces the standard deviation. For the same \(n\), a value of \(p\) near 0 or 1 gives a smaller standard deviation than a value near 0.50. The alternative’s distance from \(p_0\) also matters: all else equal, an alternative farther in the direction of the test is more likely to fall in the rejection region.
The significance level controls the probability of a Type I error: rejecting \(H_0\) when it is true. Increasing \(\alpha\) can increase power, as discussed in “How Significance Level Affects Power and Type II Error,” but it also raises that Type I error probability. The strategies in this tutorial aim to improve power without making that tradeoff.
Strategy 1: Increase the Sample Size
When \(p_0\), \(\alpha\), and the specified alternative \(p_1\) stay fixed, a larger sample generally makes the sampling distributions narrower. The rejection cutoff still comes from the null model, but the alternative distribution becomes more concentrated. If \(p_1\) is in the direction of the alternative, a greater share of that distribution will usually fall in the rejection region. This is the practical reason researchers plan sample size before collecting data.
Worked Example: A Larger Survey to Detect Increased Transit Use
A city wants to test whether the proportion of residents who use a new bus route is greater than \(0.30\). The test is \(H_0:p=0.30\) versus \(H_a:p>0.30\), with \(\alpha=0.05\). Compare estimated power at \(p_1=0.40\) for samples of \(n=100\) and \(n=200\).
State: For each sample size, estimate the probability that the test rejects \(H_0\) if the true proportion of residents using the route is \(0.40\). We will keep the hypotheses, \(\alpha\), and \(p_1\) fixed.
Plan and conditions: Use the upper-tail normal approximation to \(\hat{p}\). Assume the residents are selected through an appropriate random process. If sampling without replacement, suppose the population contains at least 2,000 residents; even the larger sample satisfies \(200\leq0.10(2000)\), so the 10% condition is met. For \(n=100\), the Large Counts condition under \(p_0=0.30\) holds because \(np_0=30\) and \(n(1-p_0)=70\); under \(p_1=0.40\), the counts are 40 and 60. For \(n=200\), those counts are 60 and 140 under \(p_0\), and 80 and 120 under \(p_1\). All are at least 10.
Do: The upper-tail critical z-value for \(\alpha=0.05\) is \(1.6449\). With \(n=100\), the null standard deviation is \(\sqrt{0.30(0.70)/100}=\sqrt{0.0021}\approx0.04583\). The cutoff is \(0.30+1.6449(0.04583)\approx0.37538\). Under \(p_1=0.40\), the standard deviation is \(\sqrt{0.40(0.60)/100}=\sqrt{0.0024}\approx0.04899\). Thus,
For \(n=200\), the null standard deviation is \(\sqrt{0.30(0.70)/200}=\sqrt{0.00105}\approx0.03240\). The cutoff is \(0.30+1.6449(0.03240)\approx0.35330\). Under \(p_1=0.40\), the standard deviation is \(\sqrt{0.40(0.60)/200}=\sqrt{0.0012}\approx0.03464\). Therefore,
Conclude: Under the specified alternative, estimated power rises from about \(0.6924\) with 100 residents to about \(0.9112\) with 200 residents. The larger sample gives the test a better chance of detecting the increase without changing \(\alpha=0.05\). These are normal-approximation estimates, and the true power depends on the actual proportion.
A larger sample costs more time and resources, and it does not fix poor sampling or measurement. Researchers should choose a sample size that is feasible and large enough for the effect they consider important, rather than assuming that collecting as many observations as possible is always the best design.
Strategy 2: Study an Effect Farther From the Null
The effect size for this one-proportion test can be described as the difference between the specified true proportion and the null value, \(p_1-p_0\). For an upper-tail test, a larger positive difference generally means greater power when \(n\) and \(\alpha\) are fixed. The alternative sampling distribution is centered farther beyond the null-based rejection cutoff. “How Effect Size Affects Power” introduced this relationship; the example below compares two alternatives using the same test design.
Worked Example: Comparing Increases in Recycling Participation
A town tests \(H_0:p=0.30\) versus \(H_a:p>0.30\), where \(p\) is the proportion of households that participate in a recycling program. The sample size is \(n=150\), and \(\alpha=0.05\). Compare estimated power if the true proportion is \(p_1=0.36\) or \(p_1=0.42\).
Plan and conditions: Assume an appropriate random sample of households. If sampling without replacement, suppose there are at least 1,500 households, so \(150\leq0.10(1500)\) and the 10% condition is met. Under \(p_0=0.30\), the Large Counts values are \(np_0=45\) and \(n(1-p_0)=105\). Under \(p_1=0.36\), they are 54 and 96; under \(p_1=0.42\), they are 63 and 87. All are at least 10.
Do: For either alternative, the null standard deviation is \(\sqrt{0.30(0.70)/150}=\sqrt{0.0014}\approx0.03742\). The critical z-value is \(1.6449\), so the upper rejection cutoff is \(0.30+1.6449(0.03742)\approx0.36154\). For \(p_1=0.36\), the alternative standard deviation is \(\sqrt{0.36(0.64)/150}=\sqrt{0.001536}\approx0.03919\). The power estimate is
For \(p_1=0.42\), the alternative standard deviation is \(\sqrt{0.42(0.58)/150}=\sqrt{0.001624}\approx0.04030\). The power estimate is
Conclude: With the same \(n\) and \(\alpha\), power is much higher for a true proportion of \(0.42\) than for \(0.36\). The first alternative is \(0.12\) above \(p_0\); the second is only \(0.06\) above it. This comparison does not mean a researcher can choose the true effect. It shows why a test may be better at detecting a substantial increase than a modest one.
A study can sometimes make a practically meaningful effect more detectable by choosing a focused question, a well-defined population, or a treatment that is expected to produce a meaningful change. But researchers should not redefine an effect after seeing the data or claim that an effect is larger than it is. In a planned power calculation, \(p_1\) should represent a plausible and meaningful alternative, not a value selected just to produce an appealing power estimate.
Strategy 3: Understand and, Where Possible, Reduce Variability
Lower variability can make a test more powerful because the alternative distribution is more concentrated. For a quantitative response, a carefully controlled design or more precise measurement can sometimes reduce unexplained variation. For a single binary outcome, the standard deviation of \(\hat{p}\) is tied to \(p(1-p)\) and \(n\). A researcher cannot independently lower that standard deviation while holding both the true proportion and sample size fixed.
The next comparison illustrates the mathematical role of variability. It holds the sample size and absolute increase constant, but considers different null proportions. Because \(p(1-p)\) is smaller near 0 or 1 than near 0.50, the spreads differ. This is a comparison of situations, not a prescription to manipulate the population’s true proportion.
Worked Example: Equal Increases at Different Starting Proportions
Consider two upper-tail tests, each with \(n=100\) and \(\alpha=0.05\). One tests \(H_0:p=0.10\) against \(H_a:p>0.10\), with specified alternative \(p_1=0.20\). The other tests \(H_0:p=0.50\) against \(H_a:p>0.50\), with \(p_1=0.60\). Both alternatives are an increase of \(0.10\).
Plan and conditions: For each case, assume data come from an appropriate random process and, if sampling without replacement, that the population is at least 1,000 units; then \(100\leq0.10(1000)\). For the first test, the Large Counts values are 10 and 90 under the null and 20 and 80 under the alternative. For the second test, they are 50 and 50 under the null and 60 and 40 under the alternative. All are at least 10, so the normal approximations are supported.
Do: In the first test, the null standard deviation is \(\sqrt{0.10(0.90)/100}=0.03000\). With critical z-value \(1.6449\), the cutoff is \(0.10+1.6449(0.03000)\approx0.14935\). Under \(p_1=0.20\), the standard deviation is \(\sqrt{0.20(0.80)/100}=0.04000\). Thus,
In the second test, the null standard deviation is \(\sqrt{0.50(0.50)/100}=0.05000\). Its cutoff is \(0.50+1.6449(0.05000)\approx0.58224\). Under \(p_1=0.60\), the standard deviation is \(\sqrt{0.60(0.40)/100}\approx0.04899\). Therefore,
Conclude: In these examples, the test with \(p_0=0.10\) has greater estimated power than the test with \(p_0=0.50\), despite both alternatives being \(0.10\) above their null values. The standard deviations differ, and the normal distributions’ positions relative to their cutoffs determine the probabilities. The result is specific to these values; it does not mean that every test near a boundary automatically has higher power.
In practice, the most useful way to address variability is often to improve data quality: define “success” clearly, apply the same measurement rules to everyone, and avoid preventable recording errors. These steps make the data more trustworthy. For a binary response, though, they do not provide a separate mathematical reduction in \(\sqrt{p(1-p)/n}\) if the true \(p\) and \(n\) remain unchanged. Poor classification may even distort the observed proportion or blur a real effect.
Choosing a Strategy Without Raising Alpha
The options differ in how much control a researcher has over them. Sample size is a design choice, though budget and time constrain it. The effect size is a feature of the real situation; researchers can select a meaningful alternative for planning, but cannot guarantee that it is true. Variability may sometimes be reduced through design or measurement, but for a binary response its formula is tied to the proportion and sample size.
- Plan for an adequate sample: Use a plausible, meaningful \(p_1\) and the chosen \(\alpha\) to estimate power before collecting data. If power is too low, consider a larger feasible \(n\).
- Keep the outcome and population well defined: Clear definitions support consistent data collection and an interpretable effect.
- Do not promise power for every alternative: State the specified \(p_1\). A test may have high power for a large increase and low power for a small increase.
- Keep \(\alpha\) tied to error consequences: Do not quietly raise \(\alpha\) just to improve power. That changes the probability of a Type I error.
These strategies do not guarantee a significant result. They improve the chance of rejecting \(H_0\) when a specified alternative is true. If a test fails to reject \(H_0\), low power for a modest effect may be one reason; it does not prove that the null hypothesis is true.
Common Mistakes and AP Exam Tip
- Claiming that a larger sample changes alpha: With the test designed to use the same significance level, increasing \(n\) generally increases power but does not raise the Type I error probability above the chosen \(\alpha\).
- Treating a planned alternative as a known truth: Power at \(p_1\) is conditional on \(p=p_1\). Say “if the true proportion is \(p_1\),” not “the probability that \(p_1\) is true.”
- Calling effect size a study setting: Researchers can plan around a meaningful effect, but cannot choose the population’s actual response just to obtain higher power.
- Assuming variability is always controllable: For a one-proportion test, \(\sqrt{p(1-p)/n}\) depends on the true proportion and \(n\). Better measurement helps data quality, but it does not independently alter this formula at fixed \(p\) and \(n\).
- Ignoring the direction of the alternative: For an upper-tail test, power concerns the area above the cutoff; a change in the opposite direction does not help detect an increase.
For a full-credit comparison, name what stays fixed, identify the strategy being considered, and explain how it changes the alternative distribution’s position or spread relative to the rejection region. If giving a power value, state the specified alternative and interpret power conditionally in context.
Check Your Understanding
Answer each question using the ideas in this tutorial.
- With \(p_0\), \(\alpha\), and \(p_1\) fixed, why does increasing \(n\) generally increase power?
- For an upper-tail test, which alternative would generally have greater power with the same \(n\): \(p_1=0.34\) or \(p_1=0.40\), if \(p_0=0.30\)? Explain.
- For a binary outcome, what determines the approximate standard deviation of \(\hat{p}\)?
- Does improving measurement quality allow a researcher to independently lower \(\sqrt{p(1-p)/n}\) while \(p\) and \(n\) stay fixed? Explain.
- Why might raising \(\alpha\) increase power, and what error probability does that tradeoff affect?