Timeline for Introducing random slopes in nested model improves model fit but residuals variances become unequal
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 27, 2019 at 8:53 | history | edited | Ferdi |
Qqplot is more specific than linear-model
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| Dec 10, 2018 at 7:33 | vote | accept | BP86 | ||
| Dec 9, 2018 at 17:51 | comment | added | Michael Hardy | $\ldots\,$errors all have equal variances. $\qquad$ | |
| Dec 9, 2018 at 17:50 | comment | added | Michael Hardy | $\ldots\,$slightly more complicated model, assume $Y_i\sim \operatorname N(a+bx_i, \sigma^2)$ are independent and $\widehat a, \widehat{\,b\,}$ are the least-squares estimators of $a,b.$ Then $\varepsilon_i = Y_i - (a+bx_i)$ is the $i$th error and $\widehat\varepsilon_i =Y_i-\left(\widehat a+\widehat{\,b\,}x_i\right)$ is the $i$th residual. The residuals again must add up to e$0$ (whereas the errors don't) and must also satisfy $\sum_i \widehat\varepsilon_i x_i = 0.$ The variance of the $i$th residual gets bigger as $x_i$ gets farther from the mean $\overline x,$ whereas the$\,\ldots\qquad$ | |
| Dec 9, 2018 at 17:45 | comment | added | Michael Hardy | Variances of errors (not to be confused with residuals) are typically assumed equal in simple linear regression, but variances of residuals (not to be confused with errors) are typically not equal. If $Y_1,\ldots, Y_n \sim \text{i.i.d.} \operatorname N(\mu,\sigma^2)$ and $\overline Y = (Y_1+\cdots+Y_n)/n$ is the sample mean, then $Y_i - \mu$ is the $i$th _error_ and $Y_i - \overline Y$ is the $i$th residual. Note that the errors are independent but the residuals are negatively correlated; in particular since the sum of the residuals must be $0$, they can't be independent. In a$\,\ldots\qquad$ | |
| Dec 9, 2018 at 17:19 | answer | added | Isabella Ghement | timeline score: 2 | |
| S Dec 9, 2018 at 13:25 | history | edited | gung - Reinstate Monica | CC BY-SA 4.0 |
code formats
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| S Dec 9, 2018 at 13:25 | history | suggested | Benjamin Christoffersen | CC BY-SA 4.0 |
code formats
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| Dec 9, 2018 at 13:16 | comment | added | BP86 | Thanks @Benjamin Christoffersen. Yes, the boldness in both models are boxcox transformed. I found the boxcox parameter lambda from the maximum likelihood method using the boxcox function from the library MASS. The latter model as in, you mean the intercept only model? | |
| Dec 9, 2018 at 12:32 | comment | added | Benjamin Christoffersen |
The latter residual plot may suggest a non-linear effect of Trial.
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| Dec 9, 2018 at 12:27 | comment | added | Benjamin Christoffersen |
So Boldness in lmer(Boldness~Trial+(1|ID)+(Trial|colony),data=mydata) is boxcox transformed, right? How did you find the parameter in the boxcox transformation? You are comparing the models lmer(Boldness~Trial+(1|ID)+(Trial|colony),data=mydata), and lmer(Boldness~Trial+(1|ID)+(1|colony),data=mydata), right?
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| Dec 9, 2018 at 12:21 | review | Suggested edits | |||
| S Dec 9, 2018 at 13:25 | |||||
| Dec 9, 2018 at 10:26 | history | asked | BP86 | CC BY-SA 4.0 |