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Critical value

A critical value fixes a hypothesis-test boundary. Compare one- and two-tailed cutoffs, normal and t references, and a return example at 5% and 1%.

A critical value is a boundary on a test statistic that separates rejection from non-rejection of a null hypothesis. It comes from the test's reference distribution and chosen significance level. The testing rule, including its tail direction and level, is specified before examining the results; the observed statistic is then compared with that boundary.

For a researcher testing a strategy's mean return, this turns “how far from zero is far enough?” into a defined rule. It does not turn statistical evidence into a guarantee of profitability.

Choose the tail before looking at the return

Write zq=Φ−1(q)z_q=\Phi^{-1}(q) for the qq quantile of the standard normal distribution: the probability to its left is qq. If ZZ has this distribution under the null, the following rules have false-rejection probability α\alpha:
AlternativeStrict rejection ruleCutoff at α=0.05\alpha=0.05
Positive departureZ>z1−αZ>z_{1-\alpha}Z>1.644854Z>1.644854
Negative departureZ<zαZ<z_{\alpha}Z<−1.644854Z<-1.644854
Departure in either direction∣Z∣>z1−α/2\lvert Z\rvert>z_{1-\alpha/2}∣Z∣>1.959964\lvert Z\rvert>1.959964
The two-sided rule allocates half the error budget to each tail. Using 1.645 in both tails would spend about 10%, rather than 5%. Choosing a positive one-sided test only after seeing a positive sample mean also changes the procedure; it does not preserve the advertised error rate. NIST's normal quantile table distinguishes the relevant quantiles.
Boundary convention. This article rejects only strictly beyond the critical boundary; equality means do not reject. For a continuous null distribution, equality has probability zero, so replacing a strict inequality with an inclusive one leaves the theoretical size unchanged. Still, software should document the rule and compare unrounded values. For these continuous tests, our convention corresponds to p<αp<\alpha.

A 0.5% average crosses one boundary but not the other

Consider a hypothetical, fixed study of 100 daily excess returns. Assume independent, identically distributed normal returns, and suppose the population standard deviation is known to be 2 percentage points per day. The observed average is 0.5 percentage points per day. Test H0:μ=0H_0:\mu=0 against H1:μ≠0H_1:\mu\ne0.
In decimal-return units, σ=0.02\sigma=0.02 and xˉ=0.005\bar x=0.005. The standard error and observed statistic are
SE⁡(Xˉ)=0.02100=0.002,zobs=0.005−00.002=2.5.\operatorname{SE}(\bar X)=\frac{0.02}{\sqrt{100}}=0.002, \qquad z_{\mathrm{obs}}=\frac{0.005-0}{0.002}=2.5.
The standard error is 0.2 percentage points, not 0.2% of the estimated mean. Under the stipulated null model, the statistic has a standard normal distribution exactly. NIST's one-mean testing guidance gives this known-standard-deviation construction.
Two-sided levelStatistic boundariesEquivalent sample-mean boundaries, percentage points per dayResult for the observed mean +0.5+0.5
5%±1.95996398\pm1.95996398±0.3919928\pm0.3919928Reject
1%±2.57582930\pm2.57582930±0.5151659\pm0.5151659Do not reject

These raw-mean boundaries are centered on the null mean, zero. They are not confidence-interval endpoints centered on the observed mean.

Standard normal null densities for the same observed z of 2.5. Two-sided 5% critical boundaries are minus and plus 1.960, so the result rejects; 1% boundaries are minus and plus 2.576, so it does not reject. Shaded tails total the stated level.
Standard normal null densities for the same observed z of 2.5. Two-sided 5% critical boundaries are minus and plus 1.960, so the result rejects; 1% boundaries are minus and plus 2.576, so it does not reject. Shaded tails total the stated level.
Original hypothetical normal-return calculation. Shaded rejection tails extend to infinity; the observed statistic stays fixed while the preselected significance level changes the boundaries.
The two-sided p-value is 2[1−Φ(2.5)]≈0.012419332[1-\Phi(2.5)]\approx0.01241933, consistent with both comparisons. This table compares two possible study designs; it is not permission to select the more convenient level after observing the result. Non-rejection at 1% does not establish that the true mean is zero.

Why 1.96 is not a universal hurdle

If the population standard deviation is estimated from these same normal observations, use

T=Xˉ−μ0S/n∼tn−1under H0.T=\frac{\bar X-\mu_0}{S/\sqrt n}\sim t_{n-1} \quad\text{under }H_0.
For n=100n=100, the two-sided 5% critical magnitude is t99,0.975≈1.984217t_{99,0.975}\approx1.984217, rather than 1.959964. The degrees of freedom matter; the numerical size of an estimated standard deviation does not make it a known population parameter. NIST's Student t table explains the degree-of-freedom and tail selections. Other statistics may require chi-squared, F, exact discrete or appropriately calibrated simulation reference laws.
In a discrete test, attainable tail probabilities jump. A nonrandomized rejection region may therefore have actual size below the nominal level; the boundary and whether its probability mass is included must be explicit. For a composite null, level control concerns the largest rejection probability over the null parameter set, not an arbitrary single null value.
A cutoff is only as credible as its reference model and research design. Dependent returns can invalidate the simple standard error; repeated backtest searches or repeated peeking can invalidate a single-test false-alarm claim. Tightening the boundary reduces false rejections but also reduces statistical power at a fixed alternative and sample size. Crossing it establishes neither economic value after costs nor the probability that the null is false.
For the derivation of the normal-family pivot and its matching confidence interval, read Confidence Intervals and Hypothesis Testing. A public preview is available; the full lesson requires membership.

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