Errata (excluding minor typos) for the book.

If you notice an error that is not mentioned in the errata below, submit an issue at https://github.com/avehtari/ROS-Examples/issues or send an email.

## 1st and 2nd (2021) printing

• p. 231, $$\log(-.595)$$ $$\rightarrow$$ $$\log(0.595)$$ (thanks to Daniel Timar)
• p. 268, offset=log(exposure) $$\rightarrow$$ offset=log(exposure2) (thanks to A. Solomon Kurz)
• p. 285, The numbers for the earnings model on p.285 should be the same as on p. 284 (thanks to David Galley)
• p. 295, $$n=(2.8* 0.49/0.1)^2=196$$ $$\rightarrow$$ $$n=(2.8* 0.5/0.1)^2=196$$, where $$0.5$$ is used as a conservative estimate of the standard deviation (thanks to Solomon A. Kurz)
• p. 299, $$0.5/1.15 = 0.43$$ $$\rightarrow$$ $$0.5/1.25 = 0.4$$ (thanks to Solomon A. Kurz)
• p. 300, The paragraph starting “We illustrate with the example of the survey earnings and height discussed in Chapter 4.” and the next two paragraphs have been edited to:
We illustrate with the survey of earnings and height discussed in Chapter 12. The coefficient for the sex-earnings interaction in model (12.2) is plausible (a positive interaction, implying that an extra inch of height is worth 2% more for men than for women), with a standard error of 1%.
Extracting another significant figure from the fitted regression yields an estimated interaction of 0.018 with standard error 0.015. How large a sample size would have been needed for the coefficient on the interaction to be “statistically significant” at the conventional 95% level? A simple calculation uses the fact that standard errors are proportional to $$1/\sqrt{n}$$. For a point estimate of 0.018 to be two standard errors from zero, it would need a standard error of 0.009, which would require the sample size to be increased by a factor of $$(0.015/0.009)^2$$. The original survey had a sample of 1629; this implies a required sample size of $$1629*(0.015/0.009)^2=4500$$.
To perform a power calculation for this hypothetical larger survey of 4500 people, we could suppose that the true $$\beta$$ for the interaction is equal to 0.018 and that the standard error is as we have just calculated. With a standard error of 0.009 the estimate from the regression would then be conventionally “statistically significant” only if $$\hat{\beta}>0.018$$ (or, in other direction, if $$\hat{\beta}< -0.018$$, but that latter possibility is highly unlikely given our assumptions). If the true coefficient $$\beta$$ is 0.018, then we would expect $$\hat{\beta}$$ to exceed 0.018, and thus achieve statistical significance, with a probability of $$\frac{1}{2}$$—that is, 50% power. To get 80% power, we need the true $$\beta$$ to be 2.8 standard errors from zero, so that there is an 80% probability that $$\hat{\beta}$$ is at least 2 standard errors from zero. If $$\beta=0.018\%$$, then its standard error would have to be no greater than $$0.018/2.8$$, so that the survey would need a sample size of $$1629*(2.8*0.015/0.018)^2=900$$. (thanks to comment by Solomon A. Kurz)
• p. 407, $$\sum_{k=1}^K(X_{ik}-X_{jk})^2$$ $$\rightarrow$$ $$(\sum_{k=1}^K(X_{ik}-X_{jk})^2)^{1/2}$$ (thanks to Stefan Gehrig)

## 1st (2020) printing

• p. 11, the summary of the treatment-control comparison says “the treated units were 4.8 points higher than the controls, $$\bar{y} = 31.7$$ under the treatment and $$\bar{y} = 25.5$$ for the controls.” The difference in shown means is 6.2 and not 4.8. Not that these values change when rerunning the simulation. (thanks Desislava Petkova)
• p. 15, “multiplicative factor; see Exercise 3.8.” $$\rightarrow$$ “multiplicative factor; see Exercise 4.8.” (thanks to Omri Har-Shemesh and Juliette Unwin)
• p. 39, in the second sentence of the first full paragraph, “Figure 3.4 displays data on log metabolic rate vs. body mass indicating…” $$\rightarrow$$ “log metabolic rate vs. log body mass” (thanks Ravi Shroff)
• p. 39, “The slope of the line is then 7.4/10 = 7.4.” $$\rightarrow$$ “The slope of the line is then 7.4/10 = 0.74.” (thanks to Jacob Warren)
• p. 42, the last line, in “Linear transformations” section “Exercise 3.5” $$\rightarrow$$ “Exercise 3.6”. (thanks Ed Berry)
• p. 43, “$$\sigma_a,\sigma_b$$$$\rightarrow$$ $$\sigma_u,\sigma_v$$ (thanks to Braden Scherting)
• p. 46, “exactly 10 000 votes” $$\rightarrow$$ “exactly 100 000 votes” (thanks to Gal Matijevic)
• p. 54, in “Comparisons, visual and numerical” subsection, “Figure 4.2” in the first sentence should be “Figure 4.3.” (thanks Ravi Shroff)
• p. 55, se_weighted_avg <- sqrt(sum(W*se)^2) $$\rightarrow$$ se_weighted_avg <- sqrt(sum((W*se)^2)) (thanks to Desislava Petkova)
• p. 57, last line: n_C+n_T, 2 $$\rightarrow$$ n_C+n_T-2 (thanks Justin Reppert)
• p. 61, “$$T(y;\phi(y))$$, for $$j = 1,\dots, J$$$$\rightarrow$$$$T(y;\phi_j(y))$$, for $$j = 1,\dots, J$$” (thanks to Kenneth Tay)
• p. 73, 1.483*median(abs(y - median(z))) $$\rightarrow$$ 1.483*median(abs(z - median(z)))
• p. 73, quantile(z, 0.25, 0.75) $$\rightarrow$$ quantile(z, c(0.25, 0.75)) and quantile(z, 0.025, 0.975) $$\rightarrow$$ quantile(z, c(0.025, 0.975)) (thanks to Justin Reppert)
• p. 83, Fig 6.1 “y = 0.49 + 0.27 * x” $$\rightarrow$$ “y = 0.4 + 0.27 * x”
• p. 83, Fig 6.2 Estimate for $$\sigma$$ should be 0.49 (not 0.43)
• p. 84, “which returns the value R^2 = 0.10, meaning that the linear model accounts for only 9% of the variance”. The “9%” should be “10%”.
• p. 105, just before Maximum likelihood section “see Exercise 7.1” $$\rightarrow$$ “see Exercise 8.1” (thanks to Jacob Warren)
• p. 116, the correct code should have y_linpred <- a + b*as.numeric(new) and y_pred <- a + b*as.numeric(new) + rnorm(n_sims, 0, sigma) (thanks to Nathan Hoffmann)
• p. 128, “Do Exercise 3.8” $$\rightarrow$$ “Do Exercise 4.8” (thanks to Omri Har-Shemesh and Juliette Unwin)
• p. 131, “the average or expected difference in outcome $$y_k$$$$\rightarrow$$ “… $$y$$” (thanks to Yoav Kessler)
• p. 143, the correct values for mean and sd in the last column of Figure 10.7 are 260.1 and 2.5 (thanks to Eam O’Brien)
• p. 143, 3rd line from bottom the correct values for mean and sd mentioned in the text are 260.1 and 2.5 (thanks to Eam O’Brien)
• p. 159, “as in Figure 11.2” $$\rightarrow$$ “as in Figure 11.3” (thanks to Eam O’Brien)
• p. 174, “and so the log score is $$-\frac{1}{2}\log\sigma$$$$\rightarrow$$ “and so the log score is $$-\log\sigma$$” (thanks to Matěj Grabovsky)
• p. 179, fit_2 <- update(fit_2, prior=hs()) $$\rightarrow$$ fit_2 <- update(fit_1, prior=hs()) (thanks to Omri Har-Shemesh)
• p. 183, $$\1600* 3.8=\63* 97=\101\,000\,000* 0.000\,060=\610$$ $$\rightarrow$$ $$\1600* 3.8=\63* 97=\101\,000\,000* 0.000\,060=\6100$$ (thanks to Ravi Shroff)
• p. 208, /sqrt(0.3*26) $$\rightarrow$$ *sqrt(0.3/26) (this is correct in the code in the web page, typo fix in the book thanks to Eugenia Migliavacca)
• p. 208, Figure 12.11 all top row subplots show the prior distribution for the regularized horseshoe prior. The correct subplot are produced by the R code in Student example and shown below (thanks to Zhengchen Cai for reporting the issue) • p. 217, First line in section 13.1 “The logistic function, logit(x)” $$\rightarrow$$ “The logit function, logit(x)” (thanks to VicentModesto)
• p. 219, “The inverse logistic function is curved” $$\rightarrow$$ “The inverse logit function is curved” (thanks to VicentModesto)
• p. 241-242, in paragraphs starting “The steps go…” and “Figure 14.2…”: $$(4.0/4.1)x_1$$ $$\rightarrow$$ $$(4.4/4.1)x_1$$ (thanks to Doug Davidson)
• p. 288, Chapter 15 Bibliographic note is missing the reference to the RiskyBehavior data: “The HIV risk data (RiskyBehavior) used in exercises comes fromcomes from El-Bassel et al (2003).” Full reference: El-Bassel, N., Witte, S. S., Gilbert, L., Wu, E., Chang, M., Hill, J., and Steinglass, P. (2003). The efficacy of a relationship-based HIV/STD prevention program for heterosexual couples. American Journal of Public Health, 93, 963–969.
• p. 314, Following Equation (17.1) “… did not include enough Republicans.” $$\rightarrow$$ “… included too many Democrats.” (thanks to Mark Fisher)
• p. 316, Figure 17.2 the values in the column $$N_j/N$$ should be $$0.16, 0.17, 0.19, 0.17, 0.16, 0.15$$ as in the code on the same page. (thanks to Noah Silbert)
• p. 344, “…Figure 18.1 yields a naive estimated treatment effect of -12.5.” $$\rightarrow$$ “…Figure 18.1 yields a naive estimated treatment effect of +12.5.” (thanks to Michael McLaren)
• p. 345, “difference in means, $$155-147.5$$$$\rightarrow$$ “difference in means, $$147.5 - 155$$” (thanks to Michael McLaren)
• p. 345, at the bottom “More to the point, the $$y_i^0$$’s are 140, 140, 150, and 170 in the control group and 150, 160, 160, and 170 in the treatment group.” $$\rightarrow$$ “More to the point, the $$y_i^0$$’s are 140, 140, 150, and 170 in the treatment group and 150, 160, 160, and 170 in the control group.” (thanks to Ravi Shroff)
• p. 371, “in Figure 19.4a” $$\rightarrow$$ “in Figure 19.4b” (thanks to Kenneth Tay)
• p. 385, equation number (20.1) should be for the second equation on the page $$y_i = \beta^*_0 + \beta^*_1z_i + \epsilon^*_i$$ (thanks to Matěj Grabovsky)
• p. 397, last line, in the standard error equation 1738 $$\rightarrow$$ 1820, 789 $$\rightarrow$$ 837, and 1.01 $$\rightarrow$$ 1.04 (thanks to Matěj Grabovsky)
• p. 435, equation 21.7: “$$\beta_2 *\mbox{math pre-test}_i$$$$\rightarrow$$$$\beta_3 *\mbox{math pre-test}_i$$” (thanks to Zhengchen Cai)
• p. 447, “populations homogeneous in $$w_i$$$$\rightarrow$$ “populations homogeneous in $$x_i$$” (thanks to Kenneth Tay)