Why a Simple Alternative Does Not Imply a Two-Tail Test

One Specified Alternative Determines One Direction

A test’s rejection region is determined by the hypotheses, not by a general rule that “mean tests use two tails.” In a Neyman–Pearson problem, the alternative is simple when it specifies one particular distribution. If that specified alternative has a larger normal mean than the null, the likelihood ratio orders observations from lower to higher sample means. If it has a smaller mean, the ordering reverses. In either case, the most powerful test at a fixed size uses one tail.

A two-sided alternative, such as “the mean differs from the null value,” includes values on both sides. It is composite: it does not specify one alternative distribution and therefore does not supply the single likelihood ratio used in the Neyman–Pearson lemma. A two-sided test may be a sensible choice when departures in either direction matter, but it is not the Neyman–Pearson test against one specified alternative.

Definition: A simple hypothesis specifies one probability distribution, including all its parameter values. A composite hypothesis allows more than one distribution. A test has size \(\alpha\) when its probability of rejecting the null is \(\alpha\) under the null distribution (or no greater than \(\alpha\), for a test whose size is bounded by \(\alpha\)). For simple \(H_0\) and simple \(H_1\), the likelihood ratio is \(f_1(\mathbf{x})/f_0(\mathbf{x})\), where \(f_1\) and \(f_0\) are the joint densities under the two hypotheses.

The Neyman–Pearson lemma says that, among tests with a specified size, a most powerful test rejects for sufficiently large values of \(f_1(\mathbf{x})/f_0(\mathbf{x})\). “Large” means that the observed data are relatively more likely under the specified alternative than under the null. It does not mean “far from the null in either direction.” The form of the rejection region follows from the likelihood-ratio ordering.

The Normal-Mean Likelihood Ratio

Suppose \(X_1,\ldots,X_n\) are independent observations from \(N(\mu,\sigma^2)\), with the variance \(\sigma^2\) known and the same under both hypotheses. Consider \(H_0:\mu=\mu_0\) against the single specified alternative \(H_1:\mu=\mu_1\), where \(\mu_1\ne\mu_0\). The sample mean is \(\bar X\). The joint density under a specified value \(\mu\) is

f_\mu(\mathbf{x})=(2\pi\sigma^2)^{-n/2} \exp\left\{-\frac{1}{2\sigma^2}\sum_{i=1}^n(x_i-\mu)^2\right\}.

Taking the ratio of the alternative density to the null density cancels the common normalizing factor. Expanding the squared terms gives a log-likelihood ratio that depends on the data only through \(\bar X\):

\begin{aligned} \log\frac{f_{\mu_1}(\mathbf{x})}{f_{\mu_0}(\mathbf{x})} &=-\frac{1}{2\sigma^2}\sum_{i=1}^n \left[(x_i-\mu_1)^2-(x_i-\mu_0)^2\right]\\ &=\frac{n(\mu_1-\mu_0)}{\sigma^2} \left(\bar X-\frac{\mu_0+\mu_1}{2}\right). \end{aligned}

This formula explains the direction of the test. If \(\mu_1>\mu_0\), the coefficient of \(\bar X\) is positive, so the likelihood ratio increases with \(\bar X\). The NP rejection region is therefore an upper tail. If \(\mu_1<\mu_0\), the coefficient is negative, so the ratio decreases with \(\bar X\), and the rejection region is a lower tail. This is the normal-mean instance of likelihood-ratio ordering.

Theorem: For independent \(N(\mu,\sigma^2)\) observations with known common variance, testing \(H_0:\mu=\mu_0\) against \(H_1:\mu=\mu_1\) gives a likelihood ratio monotone in \(\bar X\). For \(\mu_1>\mu_0\), reject for large \(\bar X\); for \(\mu_1<\mu_0\), reject for small \(\bar X\). A two-tail rule does not follow from this simple-versus-simple test.

To translate the ordering into a level-\(\alpha\) test, use the null distribution \(\bar X\sim N(\mu_0,\sigma^2/n)\). Write \(z_p\) for the \(p\)-quantile of the standard normal distribution. For a larger specified alternative, the upper-tail critical value is \(\mu_0+z_{1-\alpha}\sigma/\sqrt n\). For a smaller specified alternative, the lower-tail critical value is \(\mu_0-z_{1-\alpha}\sigma/\sqrt n\). These are one-tail cutoffs: the full size \(\alpha\) is allocated to the direction favored by the specified alternative.

Read the hypotheses.
Check whether the alternative gives one particular mean or allows a range of means. The NP ordering applies directly to two specified distributions.

Find the direction.
Compare \(\mu_1\) with \(\mu_0\). A larger alternative mean gives an upper-tail ordering; a smaller alternative mean gives a lower-tail ordering.

Use the correct size cutoff.
For a one-tail test, place probability \(\alpha\) in the appropriate tail of the null distribution. Do not split it between tails unless the testing problem calls for a two-sided procedure.

Worked Examples

Worked Example: A Specified Larger Mean

Suppose \(X_1,\ldots,X_4\) are independent observations from \(N(\mu,4)\), so the known variance is \(4\) and the standard deviation is \(2\). Test \(H_0:\mu=0\) against the specified alternative \(H_1:\mu=1\) at size \(\alpha=0.05\). For an observed sample mean \(\bar x=0.8\), determine the NP decision.

Method and assumptions. The hypotheses specify two normal distributions with the same known variance, so the NP likelihood-ratio test applies. Since \(1>0\), its likelihood ratio increases with \(\bar X\); use an upper-tail test.

Under the null, the sample mean has standard error \[ \frac{\sigma}{\sqrt n}=\frac{2}{\sqrt4}=1. \] The 95th standard normal quantile is approximately \(1.6449\), so the critical value for \(\bar X\) is \[ 0+1.6449(1)=1.6449. \] Reject \(H_0\) when \(\bar X>1.6449\). The observed value \(0.8\) does not exceed this cutoff, so the test does not reject.

The likelihood-ratio calculation confirms the ordering. Here \[ \log\frac{f_1(\mathbf{x})}{f_0(\mathbf{x})} =\frac{4(1-0)}{4}\left(\bar x-\frac{0+1}{2}\right) =0.8-0.5=0.3, \] so the observed ratio is \(e^{0.3}\approx1.3499\). At the critical boundary \(\bar x=1.6449\), the log ratio is \(1.6449-0.5=1.1449\), and the ratio is \(e^{1.1449}\approx3.142\). The observed ratio is below the boundary ratio, consistent with not rejecting. Common mistake avoided: using a two-tail cutoff for a single specified larger mean.

Worked Example: A Specified Smaller Mean

Let \(X_1,\ldots,X_4\) be independent observations from \(N(\mu,4)\). Test \(H_0:\mu=0\) against \(H_1:\mu=-1\) at size \(\alpha=0.10\). If \(\bar x=-1.5\), determine whether the NP test rejects.

Method and assumptions. This is again a simple null against a simple alternative, with known variance \(4\). Since \(-1<0\), the likelihood ratio decreases as \(\bar X\) increases, so the rejection region is a lower tail.

The null standard error is \(2/\sqrt4=1\). The 10th standard normal quantile is approximately \(-1.2816\), giving the lower-tail cutoff \[ 0-1.2816(1)=-1.2816. \] Reject when \(\bar X<-1.2816\). Since \(-1.5<-1.2816\), reject \(H_0\).

For a direct likelihood-ratio check, \[ \log\frac{f_{-1}(\mathbf{x})}{f_0(\mathbf{x})} =\frac{4(-1-0)}{4}\left(\bar x-\frac{0+(-1)}{2}\right) =-(\bar x+0.5). \] At \(\bar x=-1.5\), this is \(-(-1.5+0.5)=1\), so the ratio is \(e^1\approx2.7183\). At the critical boundary, the log ratio is \(-(-1.2816+0.5)=0.7816\), giving a boundary ratio of \(e^{0.7816}\approx2.185\). The observed ratio is larger, as required for rejection. Common mistake avoided: using an upper-tail rule merely because the test concerns a mean.

Worked Example: Why a Two-Tail Test Answers a Different Question

Suppose \(X_1,\ldots,X_9\) are independent observations from \(N(\mu,9)\), with known variance \(9\). The null is \(H_0:\mu=0\). Compare the simple alternative \(H_1:\mu=1\) with the composite alternative \(H_1:\mu\ne0\). At \(\alpha=0.05\), consider an observed sample mean \(\bar x=-2.2\).

Method and assumptions. Under the null, \(\bar X\sim N(0,9/9)=N(0,1)\). For the simple alternative \(\mu=1\), the NP test is upper-tailed, with critical value \(1.6449\). Since \(-2.2\) is not greater than \(1.6449\), this specified-alternative test does not reject.

A conventional symmetric two-sided test for \(H_0:\mu=0\) against departures in either direction rejects when \(|\bar X|>1.96\), allocating \(0.025\) to each tail. Here \(|-2.2|=2.2>1.96\), so that test rejects. Its decision differs because it addresses a different question: whether the mean departs from zero in either direction, rather than whether the mean is specifically \(1\).

The likelihood ratios explain the distinction. For the specified positive alternative \(\mu=1\), \[ \log\frac{f_1(\mathbf{x})}{f_0(\mathbf{x})} =\frac{9(1-0)}{9}\left(-2.2-\frac12\right)=-2.7, \] so the ratio is \(e^{-2.7}\approx0.0672\). The data are relatively unlikely under \(\mu=1\) compared with \(\mu=0\). For the alternative \(\mu=-1\), the log ratio is \[ \frac{9(-1-0)}{9}\left(-2.2+\frac12\right)=1.7, \] so the ratio is \(e^{1.7}\approx5.4739\). The same observation favors the negative alternative. Common mistake avoided: treating rejection in the lower tail as evidence for the specified positive alternative.

Why One Test Cannot Be Most Powerful in Both Directions

For a fixed size strictly between zero and one, the NP test against a specified larger mean rejects in the upper tail. Against a specified smaller mean, it rejects in the lower tail. In the normal model with known variance, the likelihood ratio is strictly monotone in \(\bar X\), and the null distribution is continuous. Consequently, the corresponding most powerful rejection regions are determined up to events of null probability zero; they cannot be the same region.

For example, at a common size \(\alpha=0.05\), testing \(\mu=0\) against \(\mu=1\) requires an upper-tail region, while testing \(\mu=0\) against \(\mu=-1\) requires a lower-tail region. A test that rejects in both tails cannot be most powerful against each of these two specified alternatives at that same size: probability spent on the tail disfavored by one alternative is not being used in the direction that maximizes power for that alternative.

This does not make two-sided tests incorrect. It means that the objective must be stated accurately. If departures in either direction are relevant, a symmetric two-sided procedure is a natural choice. If the alternative specifies one value above or below the null, the NP lemma gives the most powerful test by ordering outcomes according to that one alternative’s likelihood ratio.

Common Exam Trap

Suppose \(X_1,\ldots,X_9\) are independent \(N(\mu,9)\), and the task is to test \(H_0:\mu=0\) against the specified alternative \(H_1:\mu=1\) at size \(0.05\). A mistaken calculation uses the two-sided cutoff and states:

\text{Wrong: reject if }|\bar X|>1.96.

That rule does not use the NP ordering for \(\mu=1\). It rejects for sufficiently negative values as well, although negative sample means decrease the likelihood ratio for this alternative. The correct null standard error is \(3/\sqrt9=1\), and the one-sided upper cutoff is \(1.6449\):

\text{Correct: reject if }\bar X>1.6449.

For instance, \(\bar x=-2\) would trigger the mistaken two-tail rule because \(|-2|>1.96\), but it does not trigger the NP test against \(\mu=1\), because \(-2\) is not greater than \(1.6449\). Always identify the alternative before choosing the tail.

Key takeaway: “Simple alternative” means one specified distribution, not one of the two tails. In the known-variance normal-mean model, a specified larger mean gives an upper-tail NP test and a specified smaller mean gives a lower-tail NP test. A two-sided test is designed for a different, composite alternative.

Check Your Understanding

For each problem, identify the relevant direction from the hypotheses before applying a cutoff or interpreting a likelihood ratio.

Independent observations have distribution \(N(\mu,16)\), with known variance \(16\). For \(n=16\), test \(H_0:\mu=10\) against \(H_1:\mu=11.5\) at size \(0.05\). Give the critical value for \(\bar X\). (Answer: Upper-tail test; critical value \(10+1.6449(4/4)=11.6449\).)
Independent observations have distribution \(N(\mu,25)\), with known variance \(25\). For \(n=25\), test \(H_0:\mu=8\) against \(H_1:\mu=7\) at size \(0.10\). Give the rejection rule for \(\bar X\). (Answer: Lower-tail test; reject if \(\bar X<8-1.2816(5/5)=6.7184\).)
Independent observations have distribution \(N(\mu,4)\), with known variance \(4\) and \(n=4\). For \(H_0:\mu=5\) against \(H_1:\mu=4\), calculate the likelihood ratio at \(\bar x=4.2\) and state whether the likelihood ratio is above or below 1. (Answer: \(\log(f_4/f_5)=\frac{4(4-5)}4(4.2-4.5)=0.3\), so \(f_4/f_5=e^{0.3}\approx1.3499\), above 1.)
Independent observations have distribution \(N(\mu,9)\), with known variance \(9\) and \(n=9\). Compare the size-\(0.05\) NP test for \(H_0:\mu=0\) against \(H_1:\mu=1\) with the conventional two-sided test. What do they decide at \(\bar x=-2\)? (Answer: The NP test does not reject because \(-2\not>1.6449\); the two-sided test rejects because \(|-2|>1.96\).)
For independent \(N(\mu,1)\) observations with known variance \(1\), compare the likelihood-ratio ordering for \(H_0:\mu=0\) against \(H_1:\mu=0.5\) with the ordering against \(H_1:\mu=-0.5\). State the direction in each case and explain whether one rejection tail can be most powerful for both. (Answer: The \(\mu=0.5\) alternative orders toward large \(\bar X\); the \(\mu=-0.5\) alternative orders toward small \(\bar X\). The upper- and lower-tail regions differ, so one tail cannot be most powerful against both.)

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