Distribution-dependent and distribution-free confidence intervals for the variance

Abstract

Finding an interval estimation procedure for the variance of a population that achieves a specified confidence level can be problematic. If the distribution of the population is known, then a distribution-dependent interval for the variance can be obtained by considering a power transformation of the sample variance. Simulation results suggest that this method produces intervals for the variance that maintain the nominal probability of coverage for a wide variety of distributions. If the underlying distribution is unknown, then the power itself must be estimated prior to forming the endpoints of the interval. The result is a distribution-free confidence interval estimator of the population variance. Simulation studies indicate that the power transformation method compares favorably to the logarithmic transformation method and the nonparametric bias-corrected and accelerated bootstrap method for moderately sized samples. However, two applications, one in forestry and the other in health sciences, demonstrate that no single method is best for all scenarios.

Appendix

Rose and Smith (2013) developed computer software which can find unbiased estimators of products of central population moments. An unbiased estimator of, say, $\mu _2 \mu _6$, is computed in terms of power sums, $s_j = \sum _{i=1}^n X_i^j$. The power sums can be converted to functions of central sample moments and the coefficients of the central sample moments depend on the sample size (n).

$$\begin{aligned} a_1= & {} n^6-8 n^5+115 n^4-436 n^3+424 n^2+840 n-2016 \\ a_2= & {} -3n (n^5+2 n^4+13 n^3-400 n^2+1908 n-3024) \\ a_3= & {} -2n (3n^5-24n^4+113n^3-64n^2-1292n+3024) \\ a_4= & {} 2n^2 (n^4+18n^3-244n^2+981n-1386) \\ a_5= & {} 2 (9 n^5-90 n^4+515 n^3-1322 n^2+672 n+2016) \\ a_6= & {} 3 (n^5+11n^4-143n^3+537n^2-406n-840) \\ a_7= & {} -(n-1)^2 (n^3+7 n^2+6 n+72)\\ b_1= & {} 4 (n^5+n^4-31 n^3+143 n^2-378 n+504) \\ b_2= & {} -2 n (n^5-10 n^4+95 n^3-302 n^2-360 n+2016) \\ b_3= & {} -32 n (n^4-9 n^3+28 n^2-6 n-84) \\ b_4= & {} n^2 (n-4)(n^3-2 n^2+99 n-378) \\ b_5= & {} 8 (n-3) (n^4-6 n^3+19 n^2-14 n+168) \\ b_6= & {} n^6-17 n^5+120 n^4-337 n^3+395 n^2-882 n+2520 \\ b_7= & {} -(n-1) (n^4-6 n^3+35 n^2-30 n+72). \end{aligned}$$