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Two-sided test

A test for departures in either direction: calculate both normal tails, separate a negative significant result from a positive edge, and read the matching confidence interval.

A two-sided test is a hypothesis test whose alternative allows departures from a specified parameter value in either direction. For a mean return, it asks whether the mean differs from a benchmark, rather than specifically whether it exceeds that benchmark.

That distinction matters when reviewing a strategy. A statistically detectable negative mean can reject “zero expected excess return” while providing no evidence for the positive edge the researcher hoped to find.

The question determines the tails

Let μ\mu be the strategy's mean daily excess return after specified costs and relative to a specified cash benchmark. Consider

H0:μ=0,H1:μ≠0.H_0:\mu=0, \qquad H_1:\mu\ne0.
Both negative and positive means belong to the alternative hypothesis. In the symmetric normal-mean test below, departures of equal standardized magnitude count as equally extreme. A 5% significance level is divided into two rejection tails of 2.5% each. It is a total error allowance, not 5% per tail.
Fix the hypotheses, sample window and decision rule before looking at the data. A one-sided test chosen because the observed average happens to be positive or negative is a different procedure from one whose direction was fixed in advance.

A negative result can be significant

Consider a hypothetical sample of 64 daily excess returns. Stipulate independent, identically distributed normal observations with a known population standard deviation of 1 percentage point (pp). The observed sample mean is −0.25 pp per day. These assumptions make the calculation exact; they are not claims about real return data.
The standard error is σ/n=1/64=0.125\sigma/\sqrt n=1/\sqrt{64}=0.125 pp. The observed test statistic is
zobs=xˉ−0σ/n=−0.250.125=−2.z_{\mathrm{obs}}=\frac{\bar x-0}{\sigma/\sqrt n} =\frac{-0.25}{0.125}=-2.
Under the null, ZZ has the standard normal distribution. Writing Φ\Phi for its cumulative distribution function, the two-sided p-value is
p=PH0 ⁣(∣Z∣≥∣zobs∣)=Φ(−2)+1−Φ(2)=2[1−Φ(2)]≈0.04550026.\begin{aligned} p&=\mathbb{P}_{H_0}\!\left(\lvert Z\rvert\ge\lvert z_{\mathrm{obs}}\rvert\right)\\ &=\Phi(-2)+1-\Phi(2)\\ &=2[1-\Phi(2)]\approx0.04550026. \end{aligned}
Each observed tail contributes about 2.275%, for a total of 4.550%. The positive tail is counted even though the observed result is negative: the predeclared alternative regards either direction as evidence against zero.

The 5% rule rejects when

∣zobs∣>Φ−1(0.975)≈1.95996,\lvert z_{\mathrm{obs}}\rvert>\Phi^{-1}(0.975) \approx1.95996,
equivalently when p<0.05p<0.05. Here it rejects. The critical values are fixed by the chosen level; the p-value uses the observed magnitude. They need not delimit the same tail area.
Two standard normal panels: the fixed rejection tails beyond ±1.960 each have area 2.5%; the observed p-value tails beyond ±2 each have area 2.275%, totaling 4.550%.
Two standard normal panels: the fixed rejection tails beyond ±1.960 each have area 2.5%; the observed p-value tails beyond ±2 each have area 2.275%, totaling 4.550%.
Original exact normal-model calculation for the hypothetical sample. The first panel shows the fixed 5% rejection region; the second shows the observed p-value. Shaded tails extend beyond the displayed axis.

Keep the sign in the report

“Significant” alone loses the economic meaning. The estimated mean is negative. Its matching 95% confidence interval is
xˉ±Φ−1(0.975)σn=−0.25±1.95996398(0.125)≈[−0.49500,−0.00500] pp per day.\bar x\pm\Phi^{-1}(0.975)\frac{\sigma}{\sqrt n} =-0.25\pm1.95996398(0.125) \approx[-0.49500,-0.00500]\text{ pp per day}.

Zero falls outside the interval, agreeing with the test. The interval also shows how close the upper endpoint is to zero. Its 95% coverage describes repeated sampling under the model, not a posterior probability assigned to this realized interval. With strict rejection and a closed interval, a null value exactly at an endpoint is not rejected.

For the explicitly positive alternative μ>0\mu>0, this same negative observation would instead have right-tailed p-value 1−Φ(−2)≈0.977249871-\Phi(-2)\approx0.97724987. Evidence against zero in either direction is therefore not a positive-edge certificate. Nor does it establish that a negative edge will persist, justify reversing the strategy, or show that the effect survives execution costs and a search across many strategies.

The known-volatility assumption also matters. When the population standard deviation is estimated from this normal sample, the usual exact mean test uses Student's tt distribution with n−1n-1 degrees of freedom. Confidence Intervals and Hypothesis Testing derives that change and the matching interval (public preview; full lesson requires membership). Dependent returns require an appropriate sampling-variance treatment; adding a second tail does not repair an invalid standard error.

What “two-sided” does not settle

Doubling one tail works here because the normal reference law is symmetric and extremeness is defined by absolute standardized distance. It is not a universal definition for asymmetric or discrete tests: specify the test's ordering of outcomes and its two-sided p-value convention.

Finally, failure to reject equality does not establish practical equivalence. A noisy estimate can be compatible with zero and with economically large positive or negative means. Demonstrating that an effect lies within a tolerable range requires a separately designed equivalence analysis.

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  • hypothesis testing
  • two tailed test
  • normal mean
  • confidence intervals
  • research design
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