Choosing Alpha, Power, and Effect Targets

Planning Is a Set of Connected Choices

Before collecting data, a researcher has to decide what result would count as evidence, how much false-positive risk is acceptable, and how likely the study should be to detect an effect worth finding. Those choices cannot be made independently. For a fixed amount of variability and a fixed sample size, demanding stronger protection against false positives or greater power generally makes a study less able to detect small effects. To maintain the same power, the design may need more participants.

This tutorial uses a two-sided test for a population mean to make the trade-offs explicit. The same planning logic applies more broadly, but the formulas below are for the stated mean-test model. Values such as alpha and target power should be selected in light of the consequences of errors and the study’s purpose—not chosen merely because they are familiar conventions.

Definition: The significance level $\alpha$ is the probability, under the null hypothesis and the test’s assumptions, of rejecting $H_0$ when $H_0$ is true. This is the planned false-positive rate for the test. The Type II error probability $\beta$ is the probability of failing to reject $H_0$ under a specified alternative. The power at that alternative is $1-\beta$, the probability that the test rejects $H_0$ when that alternative is true.

Power is not a single property of a test in isolation. It depends on the alternative effect being considered, the sample size, the variability, the significance level, and whether the test is one-sided or two-sided. Consequently, a power target is meaningful only when paired with a specific effect target and a defined testing plan.

Effect Targets and the Mean-Test Planning Model

An effect target is the difference the study is designed to detect. It should be chosen for its scientific or practical importance, not simply because it produces an attractive sample-size calculation. For example, a difference in average recovery time of one day might matter to patients and clinicians, whereas a much smaller difference might not change decisions. The appropriate target depends on the context.

For the examples, suppose observations are independent and come from a population with standard deviation $\sigma$. Assume $\sigma$ is known for the test, or is a planning value that reasonably represents the variability expected in the study. The population is normal, or the sample is large enough for the sampling distribution of the sample mean to be approximately normal. Under these conditions, the standardized sample mean has a normal distribution. A planned study should also use a sampling or assignment design appropriate to its scientific question; a formula cannot repair biased selection, dependence, or poor measurement.

Planning model: For a two-sided test of $H_0:\mu=\mu_0$ against $H_a:\mu\ne\mu_0$, let $\delta>0$ be the absolute difference from $\mu_0$ that the study aims to detect. If the significance level is $\alpha$, the power target is $1-\beta$, and the planning standard deviation is $\sigma$, the usual normal-approximation sample-size calculation is $$ n\approx\left(\frac{(z_{1-\alpha/2}+z_{1-\beta})\sigma}{\delta}\right)^2. $$ Here $z_q$ is the $q$-quantile of the standard normal distribution. The calculated $n$ is rounded up to the next whole participant.

The formula makes several relationships visible. A smaller $\delta$ means the study seeks to detect a subtler difference, and requires a larger sample. A larger $\sigma$ means more variation obscures the mean difference, also requiring a larger sample. Raising the target power or lowering $\alpha$ increases the relevant normal quantile sum and usually increases the required sample size.

Why the Formula Has This Form

A short derivation connects the planning equation to the test’s rejection rule. Under the model, the standard error of the sample mean is $\sigma/\sqrt{n}$. For a two-sided test at level $\alpha$, reject $H_0$ when the standardized statistic falls below $-z_{1-\alpha/2}$ or above $z_{1-\alpha/2}$. If the true mean differs from $\mu_0$ by $\delta$, the standardized statistic has a normal distribution with standard deviation 1 and mean $\lambda=\delta\sqrt{n}/\sigma$, taking the positive direction without loss of generality.

\text{Power} =P(Z>z_{1-\alpha/2}-\lambda)+P(Z<-z_{1-\alpha/2}-\lambda), \qquad \lambda=\frac{\delta\sqrt{n}}{\sigma}.

The first term is the probability of crossing the upper rejection boundary when the true mean is above $\mu_0$; the second is the probability of crossing the lower boundary. For a useful approximation when the effect is in one direction, the second tail is very small. Setting the first-tail probability to the target power gives $z_{1-\alpha/2}-\lambda\approx -z_{1-\beta}$. Thus $\lambda\approx z_{1-\alpha/2}+z_{1-\beta}$. Substituting $\lambda=\delta\sqrt{n}/\sigma$, then squaring both sides, gives the planning formula. The formula is an approximation because it neglects the opposite tail under the alternative; the full two-tail power expression can be used for a more exact check.

Worked Planning Cases

Worked Example: Starting With a Meaningful Difference

Hypothetical planning case. A team plans a two-sided test of mean recovery time. It judges a difference of 4 days from the null value to be worth detecting. Planning assumptions are $\sigma=12$ days, $\alpha=0.05$, and target power $0.90$. The observations are assumed independent, the normal model is reasonable for the sample mean, and the planning standard deviation is considered credible.

For $\alpha=0.05$, $z_{1-\alpha/2}=z_{0.975}=1.960$. For 90% power, $z_{1-\beta}=z_{0.90}=1.282$. Substituting into the planning formula gives

n\approx\left(\frac{(1.960+1.282)(12\text{ days})}{4\text{ days}}\right)^2 =\left(\frac{(3.242)(12)}{4}\right)^2 =(9.726)^2 \approx94.60.

Rounding up gives a target of 95 participants. The units cancel because the standard deviation and target difference are both measured in days. Under the planning assumptions, this sample size is designed to give approximately 90% power to detect a 4-day difference with a two-sided 5% significance level. It does not guarantee that the study will achieve that power if the actual variability, recruitment, or analysis differs from the assumptions.

Worked Example: How Stricter False-Positive Protection Changes the Plan

Hypothetical planning case. Keep the same recovery-time target of 4 days, planning standard deviation of 12 days, and 90% power, but now require $\alpha=0.01$. Use the same independent-observation and normal-sampling assumptions as in the previous case.

For a two-sided test, $z_{1-\alpha/2}=z_{0.995}=2.576$. The target-power quantile remains $z_{0.90}=1.282$. Therefore,

n\approx\left(\frac{(2.576+1.282)(12\text{ days})}{4\text{ days}}\right)^2 =\left(\frac{(3.858)(12)}{4}\right)^2 =(11.574)^2 \approx133.96.

Rounding up gives 134 participants, compared with 95 when $\alpha=0.05$. The stricter significance level lowers the planned false-positive probability under the null, but with the other targets held fixed it requires a larger sample to preserve 90% power for a 4-day difference. This is a real design trade-off, not evidence that either alpha value is universally best.

Worked Example: Checking Power for a Feasible Design

Hypothetical planning case. A team can recruit a fixed sample, and its assumptions imply that the target difference corresponds to a standardized shift of $\lambda=2.683$ under the alternative. It plans a two-sided test with $\alpha=0.05$. The independent-observation and normal-sampling assumptions are taken to be reasonable. Rather than assume a desired power can be met, calculate the power this design implies.

The two-sided critical value is $1.960$. Using the two-tail power expression, the probability of crossing the upper boundary is $P(Z>1.960-2.683)$, and the probability of crossing the lower boundary is $P(Z<-1.960-2.683)$. The arithmetic is

\begin{aligned} \text{Power} &=P(Z>1.960-2.683)+P(Z<-1.960-2.683)\\ &=P(Z>-0.723)+P(Z<-4.643)\\ &\approx0.7652+0.0000017\\ &\approx0.765. \end{aligned}

The design therefore has approximately 76.5% power for this target effect, below a hypothetical 90% goal. The team could recruit more participants, reconsider whether a different design can reduce variability, or accept a lower power target with a clear explanation. It should not redefine the meaningful effect merely to make the available sample appear adequate.

Choosing Targets Defensibly

The numerical calculations follow from the choices; they do not make those choices automatically. A defensible plan explains why each target suits the question and consequences of an error.

Choose alpha in light of false-positive consequences. A lower $\alpha$ can be important where a false claim would lead to harmful or costly decisions. It also tends to demand a larger sample to retain a chosen power for a given effect.
Set power for the effect that matters. A target such as 80% or 90% describes the desired probability of detecting a specified effect under the planning model. It is not the probability that the study’s conclusion will be correct.
Ground the effect target in context. Ask what difference would change a decision or matter to the people affected. A statistically detectable difference is not necessarily an important one.
Use a realistic variability assumption. If actual variability is larger than the planning value, power will be lower than planned at the same sample size. Sensitivity checks across plausible values can reveal whether the design is fragile.
Check feasibility honestly. If the required sample is impractical, report the trade-off. A feasible study with lower power may still be useful, but its ability to detect the target effect should not be overstated.

For a one-sided test, the direction must be specified before examining results, and the scientific question must justify treating evidence in the other direction differently. The critical value then comes from a one-sided rejection rule, rather than $z_{1-\alpha/2}$. Choosing a one-sided test only to reduce the required sample or obtain significance is not a defensible planning rationale.

Common Mistake: Treating a Conventional Target as a Guarantee

A common mistake is to state, “We used 80% power, so the study has an 80% chance of finding an effect,” without specifying which effect, alpha, variability, or model was used. That statement is incomplete. Power is calculated for an assumed alternative and design; it can differ for effects of different sizes, and it can be wrong if the assumptions are poor. Nor does 80% power mean there is an 80% probability that a reported result is true.

A better plan names the test, alpha, target power, meaningful effect, variability assumption, and resulting sample size. It then states the limitations: power refers to the specified effect under the assumed model. If the resulting sample is infeasible, the design discussion should make the consequences explicit rather than quietly changing the target after seeing the calculation.

Key takeaway: Alpha, power, effect size, variability, and sample size form a connected planning system. State the scientifically meaningful effect and the error-rate targets first, calculate the implied sample size, and then decide whether that design is feasible and adequate for its purpose.

Check Your Understanding

Answer these questions using the planning ideas above.

In a study plan, what does 90% power mean for a specified target effect, and what does it not mean?
With the effect, variability, and power target fixed, why does reducing alpha generally increase the required sample size?
A team doubles its target difference while holding the other planning quantities fixed. According to the mean-test formula, how does the approximate required sample size change?
Why should a meaningful effect target be chosen for its practical or scientific importance rather than selected only to fit the available sample?
What additional justification is needed before choosing a one-sided test instead of a two-sided test?