Corrigendum

Abstract

Jiang, Y., Boyle, D. K., Bott, M. J., Wick, J. A., Yu, Q., & Gajewski, B. J. (2014). Expediting Clinical and Translational Research via Bayesian Instrument Development. Applied Psychological Measurement, 38(4), 296-310. Original DOI: 10.1177/0146621613517165.

On page 298, section “Simple Illustration: Item to Domain Correlation,” the text “the true distribution of item to domain correlation, after Fisher transformation, is normal; that is, $g\left(\text{ρ}\right)~N(g\left(\text{ρ}\right),1/n)$ , where r is the sample …”

should be corrected as

“the true distribution of item to domain correlation, after Fisher transformation, is normal; that is, $g\left(r\right)~N(g\left(\text{ρ}\right),1/n)$ , where r is the sample …”

On page 299, the text “It is easy to show that using the MSE that BID is better than classic

when $\frac{n}{{(n+n_0)}^2}+\frac{n_0^2}{n+n_0}{(g\left({\rho }_0\right)-g\left(\theta \right))}^2<\frac{1}{n}$ , or ${(g\left({\rho }_0\right)-g\left(\theta \right))}^2<(\frac{1}{n}-\frac{n}{{(n+n_0)}^2})\frac{n_0^2}{n+n_0}$ .

Suppose the prior sample size is n₀ = 30 and data sample size is n = 77. Using the inequality, BID is better than classical method when ${(g\left({\rho }_0\right)-g\left(\theta \right))}^2<(\frac{1}{n}-\frac{n}{{(n+n_0)}^2})\frac{{n_0}^2}{n+n_0}=(\frac{1}{77}-\frac{77}{{(77+30)}^2})\frac{{30}^2}{77+30}$ = 0.0527.”

should be corrected as

“It is easy to show that using the MSE that BID is better than classic when $\frac{n}{{(n+n_0)}^2}+\frac{{n_0}^2}{{(n+n_0)}^2}{(g\left({\rho }_0\right)-g\left(\theta \right))}^2<\frac{1}{n}$ , or ${(g\left({\rho }_0\right)-g\left(\theta \right))}^2<(\frac{1}{n}-\frac{n}{{(n+n_0)}^2})\frac{{(n+n_0)}^2}{{n_0}^2}$ . Suppose the prior sample size is n₀ = 30 and data sample size is n = 77. Using the inequality, BID is better than classical method when ${(g\left({\rho }_0\right)-g\left(\theta \right))}^2<(\frac{1}{n}-\frac{n}{{(n+n_0)}^2})\frac{{(n+n_0)}^2}{{n_0}^2}=(\frac{1}{77}-\frac{77}{{(77+30)}^2})\frac{{(77+30)}^2}{{30}^2}$ = 0.079654.”

On page 303, section “Bias or Unbiased Priors,” the formula for $\mathrm{Bia}{s_j}^2$

$\mathrm{Bia}{s_j}^2=\sum _{j=1}^{1000}{(\widehat{{\rho }_j}(s)/1000-{\rho }_{\mathrm{TRUE}})}^2$

should be corrected as

$\mathrm{Bia}{s_j}^2=\sum _{j=1}^{1000}{(\widehat{{\rho }_j}\left(s\right)-{\rho }_{\mathrm{TRUE}})}^2/1000$