Tutorials › VEE Mathematical Statistics › Unequal Tail Allocation for Confidence Intervals

Home  /  Tutorials  /  VEE Mathematical Statistics / Confidence Intervals and Pivots

VEE Mathematical Statistics · Confidence Intervals and Pivots · Tutorial 526 of 880

Unequal Tail Allocation for Confidence Intervals

Use a pivot to construct two-sided confidence intervals when the lower- and upper-tail probabilities are unequal.

Intermediate 9 min read

What You'll Learn

  • Define separate lower- and upper-tail noncoverage probabilities whose sum is the total noncoverage.
  • Invert a standard normal pivot when the two tail allocations differ.
  • Match each pivot quantile to the correct confidence-limit endpoint.
  • Calculate and interpret unequal-tailed confidence intervals.
  • Identify endpoint reversals and tail-label errors.

Why Allocate the Tails Unequally?

A two-sided confidence interval does not have to divide its total noncoverage probability equally between its two tails. For a confidence level of 90%, for example, the total noncoverage probability is 10%; it may be allocated as 5% in each tail, or as 2% in one tail and 8% in the other. The resulting intervals have the same nominal coverage under the stated model, but their endpoints are located differently.

Unequal allocation can be appropriate when the two directions of error have different practical consequences, or when a problem explicitly specifies different tail probabilities. The main calculation is to choose the pivot quantiles that leave the required probability in each tail, then invert the pivot carefully. This tutorial is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let a pivot \(T(X,\theta)\) have a continuous distribution that does not depend on the unknown parameter \(\theta\). For total noncoverage probability \(\alpha\), choose tail probabilities \(\alpha_L\) and \(\alpha_U\) such that \(\alpha_L\geq 0\), \(\alpha_U\geq 0\), and \(\alpha_L+\alpha_U=\alpha\). If \(c_L\) and \(c_U\) satisfy \(P(T<c_L)=\alpha_L\) and \(P(T>c_U)=\alpha_U\), then \(P(c_L\leq T\leq c_U)=1-\alpha\). Inverting these pivot inequalities gives a confidence interval; the order of its endpoints depends on how the pivot relates to \(\theta\).

Unequal-Tailed Intervals From a Normal Pivot

We will use a normal sample with known standard deviation. Suppose \(X_1,\ldots,X_n\) are independent observations from \(N(\mu,\sigma^2)\), where \(\sigma\) is known and \(\mu\) is the parameter of interest. Then \(\bar X\) has distribution \(N(\mu,\sigma^2/n)\), so

$$ T=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\sim N(0,1). $$

Write \(z_p\) for the \(p\)-quantile of the standard normal distribution, so \(P(Z\leq z_p)=p\) for \(Z\sim N(0,1)\). The lower pivot cutoff is \(z_{\alpha_L}\). The upper cutoff is \(z_{1-\alpha_U}\), since the probability above it is \(\alpha_U\).

$$ P\left(z_{\alpha_L}\leq \frac{\bar X-\mu}{\sigma/\sqrt{n}} \leq z_{1-\alpha_U}\right)=1-\alpha. $$

The pivot increases as \(\bar X-\mu\) increases, but for fixed observed \(\bar X\), it decreases as \(\mu\) increases. Inverting therefore reverses the endpoint order: the upper pivot quantile determines the lower confidence limit, while the lower pivot quantile determines the upper confidence limit.

$$ \bar X-z_{1-\alpha_U}\frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar X-z_{\alpha_L}\frac{\sigma}{\sqrt{n}}. $$

This interval has coverage \(1-\alpha\) under the stated independent normal-sample model with known \(\sigma\). The allocation is not interchangeable: \(\alpha_L\) is the probability below the lower pivot cutoff, while \(\alpha_U\) is the probability above the upper pivot cutoff.

Worked Example: A 90% Interval With a 2% and 8% Split

A hypothetical laboratory takes a simple random sample of \(n=25\) measurements. Assume the measurements are independent and normally distributed with unknown mean \(\mu\) and known standard deviation \(\sigma=10\) units. The observed sample mean is \(\bar x=104\) units. Construct a 90% confidence interval, allocating \(\alpha_L=0.02\) and \(\alpha_U=0.08\).

Method and assumptions. The total noncoverage is \(\alpha=1-0.90=0.10\), and the allocation is valid because \(0.02+0.08=0.10\). Under the specified normal model, use the standard normal pivot. The standard error is

$$ \frac{\sigma}{\sqrt{n}}=\frac{10}{\sqrt{25}}=\frac{10}{5}=2. $$

The required quantiles are \(z_{0.02}\approx -2.0537\) and \(z_{1-0.08}=z_{0.92}\approx 1.4051\). Substituting into the inverted-pivot interval gives

$$ \begin{aligned} L&=104-(1.4051)(2)=104-2.8102=101.1898,\\ U&=104-(-2.0537)(2)=104+4.1074=108.1074. \end{aligned} $$

Rounded to two decimal places, the interval is \([101.19,\ 108.11]\) units. As a check, its pivot cutoffs are approximately \(-2.0537\) and \(1.4051\), leaving 0.02 below and 0.08 above. The covered pivot probability is therefore \(1-0.02-0.08=0.90\).

Common mistake avoided: putting \(z_{0.02}\) in the lower-limit formula would produce \(104-(-2.0537)(2)=108.11\) as the lower limit. That is the upper endpoint, not the lower one; the quantiles must be matched to endpoints after accounting for the reversal during inversion.

Why the Endpoints Reverse

The endpoint order follows directly from solving the two inequalities, rather than from memorizing a formula. Let \(s=\sigma/\sqrt n\), which is positive. Starting from the pivot event,

$$ z_{\alpha_L}\leq \frac{\bar X-\mu}{s}\leq z_{1-\alpha_U}, $$

multiply all parts by \(s\). Then subtract \(\bar X\), and multiply by \(-1\). Multiplication by a negative number reverses each inequality:

$$ s z_{\alpha_L}\leq \bar X-\mu\leq s z_{1-\alpha_U} $$ $$ -s z_{\alpha_L}\geq \mu-\bar X\geq -s z_{1-\alpha_U} $$ $$ \bar X-s z_{1-\alpha_U}\leq \mu\leq \bar X-s z_{\alpha_L}. $$

This derivation explains both the endpoint formulas and their order. It also provides a reliable check: the final lower limit must use the upper pivot quantile \(z_{1-\alpha_U}\), and the final upper limit must use the lower pivot quantile \(z_{\alpha_L}\).

Worked Example: Seeing Both Limits Move When the Allocation Changes

A hypothetical calibration process produces independent normal measurements with unknown mean \(\mu\) and known standard deviation \(\sigma=8\) units. A sample of \(n=16\) has mean \(\bar x=52\) units. Compare the equal-tailed 90% interval with the interval that allocates \(\alpha_L=0.02\) and \(\alpha_U=0.08\).

Method and assumptions. The normal model and known standard deviation justify the same standard normal pivot. The standard error is

$$ \frac{8}{\sqrt{16}}=\frac{8}{4}=2. $$

For equal tails, \(\alpha_L=\alpha_U=0.05\). The cutoffs are \(z_{0.05}\approx -1.6449\) and \(z_{0.95}\approx 1.6449\), so

$$ \begin{aligned} L_{\mathrm{equal}}&=52-(1.6449)(2)=52-3.2898=48.7102,\\ U_{\mathrm{equal}}&=52-(-1.6449)(2)=52+3.2898=55.2898. \end{aligned} $$

Rounded to two decimal places, the equal-tailed interval is \([48.71,\ 55.29]\) units.

For the unequal allocation, use \(z_{0.02}\approx -2.0537\) and \(z_{0.92}\approx 1.4051\):

$$ \begin{aligned} L_{\mathrm{unequal}}&=52-(1.4051)(2)=52-2.8102=49.1898,\\ U_{\mathrm{unequal}}&=52-(-2.0537)(2)=52+4.1074=56.1074. \end{aligned} $$

The unequal-tailed interval is \([49.19,\ 56.11]\) units to two decimal places. Compared with the equal-tailed interval, both limits have moved upward. More specifically, the lower limit is higher because the probability assigned above the upper pivot cutoff is 0.08 rather than 0.05; the upper limit is higher because the probability below the lower pivot cutoff is only 0.02 rather than 0.05. Both intervals have 90% coverage under the model, but their tail risks differ.

Common mistake avoided: changing one tail probability while leaving the other at 0.05 would give a total noncoverage of 0.07 or 0.13, not the required 0.10. Always verify \(\alpha_L+\alpha_U=\alpha\) before calculating the limits.

Reading the Allocation in Context

The subscripts \(L\) and \(U\) refer to the lower and upper tails of the pivot distribution, not directly to the lower and upper confidence limits. In the normal-mean setup above, the pivot is \((\bar X-\mu)/s\). Because this pivot decreases as \(\mu\) increases for fixed data, the pivot’s upper tail corresponds to missing the parameter below the interval’s lower limit, and the pivot’s lower tail corresponds to missing it above the interval’s upper limit.

This distinction matters when a problem describes the allocation in terms of which side of the parameter interval should have more noncoverage. Translate the requested parameter-side error into a pivot-tail probability before choosing quantiles. In the normal-mean construction:

  • The probability of the interval lying entirely above \(\mu\) is the pivot upper-tail probability \(\alpha_U\).
  • The probability of the interval lying entirely below \(\mu\) is the pivot lower-tail probability \(\alpha_L\).

The interval formulas in this tutorial rely on the stated normal model, known \(\sigma\), and a continuous standard normal pivot. They should not be applied automatically to a different pivot or model: first establish the pivot’s distribution, identify its tail quantiles, and solve the inequalities for the parameter. A discrete pivot, an estimated standard deviation, or a pivot with a different relationship to the parameter can require a different construction.

Key takeaway: Select \(\alpha_L\) and \(\alpha_U\) so their sum is the total noncoverage \(\alpha\). For the pivot \(T=(\bar X-\mu)/(\sigma/\sqrt n)\), use the cutoffs \(z_{\alpha_L}\) and \(z_{1-\alpha_U}\), then invert: the upper pivot cutoff gives the lower limit, and the lower pivot cutoff gives the upper limit.

Check Your Understanding

For each problem, state the tail quantiles you use and check that the two tail probabilities sum to the requested noncoverage. Unless specified otherwise, assume independent normal observations and a known population standard deviation.

  1. A sample of \(n=25\) has mean \(\bar x=30\), with known \(\sigma=5\). Construct a 90% interval using \(\alpha_L=0.04\) and \(\alpha_U=0.06\). Use \(z_{0.04}\approx-1.7507\) and \(z_{0.94}\approx1.5548\). (Answer: \([28.45,\ 31.75]\), to two decimal places.)
  2. A sample of \(n=64\) has mean \(\bar x=12.3\), with known \(\sigma=4\). Construct a 95% interval using \(\alpha_L=0.01\) and \(\alpha_U=0.04\). Use \(z_{0.01}\approx-2.3263\) and \(z_{0.96}\approx1.7507\). (Answer: \([11.42,\ 13.46]\), to two decimal places.)
  3. A 92% interval is required, and the probability of the interval lying entirely above the true mean must be 0.05. For a sample with \(n=36\), \(\bar x=100\), and known \(\sigma=12\), determine the tail allocation and construct the interval. Use \(z_{0.03}\approx-1.8808\) and \(z_{0.95}\approx1.6449\). (Answer: \(\alpha_L=0.03,\ \alpha_U=0.05\); interval \([96.71,\ 103.76]\), to two decimal places.)
  4. A sample has \(n=9\), \(\bar x=7.5\), and known \(\sigma=3\). Construct a 90% interval with \(\alpha_L=0.08\) and \(\alpha_U=0.02\). Use \(z_{0.08}\approx-1.4051\) and \(z_{0.98}\approx2.0537\). (Answer: \([5.45,\ 8.91]\), to two decimal places.)
  5. For a standard normal pivot with a 90% confidence level, a candidate chooses \(\alpha_L=0.03\) and \(\alpha_U=0.05\). Is the allocation valid for a 90% interval? If not, what total confidence level does it imply? (Answer: No; the total noncoverage is \(0.03+0.05=0.08\), so it implies 92% confidence.)

Previous

Next

Back to Tutorials
Clinical Biostats