When I call the following example I receive a pretty report, but with confidence equal to 95% specifically:
julia> OneSampleTTest([1,2,3], 2)
One sample t-test
-----------------
Population details:
parameter of interest: Mean
value under h_0: 2
point estimate: 2.0
95% confidence interval: (-0.48413771175033027,4.48413771175033)
Test summary:
outcome with 95% confidence: fail to reject h_0
two-sided p-value: 1.0 (not significant)
Details:
number of observations: 3
t-statistic: 0.0
degrees of freedom: 2
empirical standard error: 0.5773502691896258
I'd like to receive similar report but for confidence equal to 99%.
The docs state that OneSampleTest implements method confint, which does receive parameter alpha, but it does not give me a full report as shown above.
Related
I'm currently struggling with how to report, following APA-6 recommendations, the output of rstanarm::stan_lmer().
First, I'll fit a mixed model within the frequentist approach, then will try to do the same using the bayesian framework.
Here's the reproducible code to get the data:
library(tidyverse)
library(neuropsychology)
library(rstanarm)
library(lmerTest)
df <- neuropsychology::personality %>%
select(Study_Level, Sex, Negative_Affect) %>%
mutate(Study_Level=as.factor(Study_Level),
Negative_Affect=scale(Negative_Affect)) # I understood that scaling variables is important
Now, let's fit a linear mixed model in the "traditional" way to test the impact of Sex (male/female) on Negative Affect (negative mood) with the study level (years of education) as random factor.
fit <- lmer(Negative_Affect ~ Sex + (1|Study_Level), df)
summary(fit)
The output is the following:
Linear mixed model fit by REML t-tests use Satterthwaite approximations to degrees of
freedom [lmerMod]
Formula: Negative_Affect ~ Sex + (1 | Study_Level)
Data: df
REML criterion at convergence: 3709
Scaled residuals:
Min 1Q Median 3Q Max
-2.58199 -0.72973 0.02254 0.68668 2.92841
Random effects:
Groups Name Variance Std.Dev.
Study_Level (Intercept) 0.04096 0.2024
Residual 0.94555 0.9724
Number of obs: 1327, groups: Study_Level, 8
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 0.01564 0.08908 4.70000 0.176 0.868
SexM -0.46667 0.06607 1321.20000 -7.064 2.62e-12 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Correlation of Fixed Effects:
(Intr)
SexM -0.149
To report it, I would say that "we fitted a linear mixed model with negative affect as outcome variable, sex as predictor and study level was entered as a random effect. Within this model, the male level led to a significant decrease of negative affect (beta = -0.47, t(1321)=-7.06, p < .001).
Is that correct?
Then, let's try to fit the model within a bayesian framework using rstanarm:
fitB <- stan_lmer(Negative_Affect ~ Sex + (1|Study_Level),
data=df,
prior=normal(location=0, scale=1),
prior_intercept=normal(location=0, scale=1),
prior_PD=F)
print(fitB, digits=2)
This returns:
stan_lmer
family: gaussian [identity]
formula: Negative_Affect ~ Sex + (1 | Study_Level)
------
Estimates:
Median MAD_SD
(Intercept) 0.02 0.10
SexM -0.47 0.07
sigma 0.97 0.02
Error terms:
Groups Name Std.Dev.
Study_Level (Intercept) 0.278
Residual 0.973
Num. levels: Study_Level 8
Sample avg. posterior predictive
distribution of y (X = xbar):
Median MAD_SD
mean_PPD 0.00 0.04
------
For info on the priors used see help('prior_summary.stanreg').
I think than median is the median of the posterior distribution of the coefficient and mad_sd the equivalent of standart deviation. These parameters are close to the beta and standart error of the frequentist model, which is reassuring. However, I do not know how to formalize and put the output in words.
Moreover, if I do the summary of the model (summary(fitB, probs=c(.025, .975), digits=2)), I get other features of the posterior distribution:
...
Estimates:
mean sd 2.5% 97.5%
(Intercept) 0.02 0.11 -0.19 0.23
SexM -0.47 0.07 -0.59 -0.34
...
Is something like the following good?
"we fitted a linear mixed model within the bayesian framework with negative affect as outcome variable, sex as predictor and study level was entered as a random effect. Priors for the coefficient and the intercept were set to normal (mean=0, sd=1). Within this model, the features of the posterior distribution of the coefficient associated with the male level suggest a decrease of negative affect (mean = -0.47, sd = 0.11, 95% CI[-0.59, -0.34]).
Thanks for your help.
The following is personal opinion that may or may not be acceptable to a psychology journal.
To report it, I would say that "we fitted a linear mixed model with negative affect as outcome variable, sex as predictor and study level was entered as a random effect. Within this model, the male level led to a significant decrease of negative affect (beta = -0.47, t(1321)=-7.06, p < .001).
Is that correct?
That is considered correct from a frequentist perspective.
The key concepts from a Bayesian perspective are that (conditional on the model, of course)
There is a 0.5 probability that the true effect is less than the posterior median and a 0.5 probability that the true effect is greater than the posterior median. Frequentists tend to see a posterior median as being like a numerical optimum.
The posterior_interval function yields credible intervals around the median with a default probability of 0.9 (although a lower number produces more accurate estimates of the bounds). So, you can legitimately say that there is a probability of 0.9 that the true effect is between those bounds. Frequentists tend to see confidence intervals as being like credible intervals.
The as.data.frame function will give you access to the raw draws, so mean(as.data.frame(fitB)$male > 0) yields the probability that the expected difference in the outcome between men and women in the same study is positive. Frequentists tend to see these probabilities as being like p-values.
For a Bayesian approach, I would say
We fit a linear model using Markov Chain Monte Carlo with negative affect as the outcome variable, sex as predictor and the intercept was allowed to vary by study level.
And then talk about the estimates using the three concepts above.
I am having trouble interpreting the results of a logistic regression. My outcome variable is Decision and is binary (0 or 1, not take or take a product, respectively).
My predictor variable is Thoughts and is continuous, can be positive or negative, and is rounded up to the 2nd decimal point.
I want to know how the probability of taking the product changes as Thoughts changes.
The logistic regression equation is:
glm(Decision ~ Thoughts, family = binomial, data = data)
According to this model, Thoughts has a significant impact on probability of Decision (b = .72, p = .02). To determine the odds ratio of Decision as a function of Thoughts:
exp(coef(results))
Odds ratio = 2.07.
Questions:
How do I interpret the odds ratio?
Does an odds ratio of 2.07 imply that a .01 increase (or decrease) in Thoughts affect the odds of taking (or not taking) the product by 0.07 OR
Does it imply that as Thoughts increases (decreases) by .01, the odds of taking (not taking) the product increase (decrease) by approximately 2 units?
How do I convert odds ratio of Thoughts to an estimated probability of Decision?
Or can I only estimate the probability of Decision at a certain Thoughts score (i.e. calculate the estimated probability of taking the product when Thoughts == 1)?
The coefficient returned by a logistic regression in r is a logit, or the log of the odds. To convert logits to odds ratio, you can exponentiate it, as you've done above. To convert logits to probabilities, you can use the function exp(logit)/(1+exp(logit)). However, there are some things to note about this procedure.
First, I'll use some reproducible data to illustrate
library('MASS')
data("menarche")
m<-glm(cbind(Menarche, Total-Menarche) ~ Age, family=binomial, data=menarche)
summary(m)
This returns:
Call:
glm(formula = cbind(Menarche, Total - Menarche) ~ Age, family = binomial,
data = menarche)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.0363 -0.9953 -0.4900 0.7780 1.3675
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -21.22639 0.77068 -27.54 <2e-16 ***
Age 1.63197 0.05895 27.68 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 3693.884 on 24 degrees of freedom
Residual deviance: 26.703 on 23 degrees of freedom
AIC: 114.76
Number of Fisher Scoring iterations: 4
The coefficients displayed are for logits, just as in your example. If we plot these data and this model, we see the sigmoidal function that is characteristic of a logistic model fit to binomial data
#predict gives the predicted value in terms of logits
plot.dat <- data.frame(prob = menarche$Menarche/menarche$Total,
age = menarche$Age,
fit = predict(m, menarche))
#convert those logit values to probabilities
plot.dat$fit_prob <- exp(plot.dat$fit)/(1+exp(plot.dat$fit))
library(ggplot2)
ggplot(plot.dat, aes(x=age, y=prob)) +
geom_point() +
geom_line(aes(x=age, y=fit_prob))
Note that the change in probabilities is not constant - the curve rises slowly at first, then more quickly in the middle, then levels out at the end. The difference in probabilities between 10 and 12 is far less than the difference in probabilities between 12 and 14. This means that it's impossible to summarise the relationship of age and probabilities with one number without transforming probabilities.
To answer your specific questions:
How do you interpret odds ratios?
The odds ratio for the value of the intercept is the odds of a "success" (in your data, this is the odds of taking the product) when x = 0 (i.e. zero thoughts). The odds ratio for your coefficient is the increase in odds above this value of the intercept when you add one whole x value (i.e. x=1; one thought). Using the menarche data:
exp(coef(m))
(Intercept) Age
6.046358e-10 5.113931e+00
We could interpret this as the odds of menarche occurring at age = 0 is .00000000006. Or, basically impossible. Exponentiating the age coefficient tells us the expected increase in the odds of menarche for each unit of age. In this case, it's just over a quintupling. An odds ratio of 1 indicates no change, whereas an odds ratio of 2 indicates a doubling, etc.
Your odds ratio of 2.07 implies that a 1 unit increase in 'Thoughts' increases the odds of taking the product by a factor of 2.07.
How do you convert odds ratios of thoughts to an estimated probability of decision?
You need to do this for selected values of thoughts, because, as you can see in the plot above, the change is not constant across the range of x values. If you want the probability of some value for thoughts, get the answer as follows:
exp(intercept + coef*THOUGHT_Value)/(1+(exp(intercept+coef*THOUGHT_Value))
Odds and probability are two different measures, both addressing the same aim of measuring the likeliness of an event to occur. They should not be compared to each other, only among themselves!
While odds of two predictor values (while holding others constant) are compared using "odds ratio" (odds1 / odds2), the same procedure for probability is called "risk ratio" (probability1 / probability2).
In general, odds are preferred against probability when it comes to ratios since probability is limited between 0 and 1 while odds are defined from -inf to +inf.
To easily calculate odds ratios including their confident intervals, see the oddsratio package:
library(oddsratio)
fit_glm <- glm(admit ~ gre + gpa + rank, data = data_glm, family = "binomial")
# Calculate OR for specific increment step of continuous variable
or_glm(data = data_glm, model = fit_glm,
incr = list(gre = 380, gpa = 5))
predictor oddsratio CI.low (2.5 %) CI.high (97.5 %) increment
1 gre 2.364 1.054 5.396 380
2 gpa 55.712 2.229 1511.282 5
3 rank2 0.509 0.272 0.945 Indicator variable
4 rank3 0.262 0.132 0.512 Indicator variable
5 rank4 0.212 0.091 0.471 Indicator variable
Here you can simply specify the increment of your continuous variables and see the resulting odds ratios. In this example, the response admit is 55 times more likely to occur when predictor gpa is increased by 5.
If you want to predict probabilities with your model, simply use type = response when predicting your model. This will automatically convert log odds to probability. You can then calculate risk ratios from the calculated probabilities. See ?predict.glm for more details.
I found this epiDisplay package, works fine! It might be useful for others but note that your confidence intervals or exact results will vary according to the package used so it is good to read the package details and chose the one that works well for your data.
Here is a sample code:
library(epiDisplay)
data(Wells, package="carData")
glm1 <- glm(switch~arsenic+distance+education+association,
family=binomial, data=Wells)
logistic.display(glm1)
Source website
The above formula to logits to probabilities, exp(logit)/(1+exp(logit)), may not have any meaning. This formula is normally used to convert odds to probabilities. However, in logistic regression an odds ratio is more like a ratio between two odds values (which happen to already be ratios). How would probability be defined using the above formula? Instead, it may be more correct to minus 1 from the odds ratio to find a percent value and then interpret the percentage as the odds of the outcome increase/decrease by x percent given the predictor.
I found the total change in deviance between 2 models to be 6.33 with 6 degrees of freedom.
How can I do a chisquared test to test the goodness of fit?
I tried dchisq(6,6.33) which gives me 0.11. Does it mean I can reject at 90%?
But when I look the chi-squared with 6 degrees of freedom, the 90% confidence interval is at 10.64, which is sure rejection. (I can also reject at 75% - 7.84, which contradicts the above which suggests rejecting at 89%)
I'm trying to understand the output from predict(), as well as understand whether this approach is appropriate for the problem I'm trying to solve. The prediction intervals don't make sense to me, but when I plot this on a scatterplot it looks like a good model:
I created a simple linear regression model of deal size ($) with a company's sales volume as a predictor variable. The data is faked, with deal size being a multiple of sales volume plus or minus some noise:
Call:
lm(formula = deal_size ~ sales_volume, data = accounts)
Residuals:
Min 1Q Median 3Q Max
-19123502 -3794671 -3426616 4838578 17328948
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.709e+06 1.727e+05 21.48 <2e-16 ***
sales_volume 1.898e-01 2.210e-03 85.88 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 6452000 on 1586 degrees of freedom
Multiple R-squared: 0.823, Adjusted R-squared: 0.8229
F-statistic: 7376 on 1 and 1586 DF, p-value: < 2.2e-16
The predictions were generated thusly:
d = data.frame(accounts, predict(fit, interval="prediction"))
When I plot sales_volume vs. deal_size on a scatterplot, and overlay the regression line with the prediction interval, it looks good, except for a few intervals that span negative values where sales is at or near zero.
I understand fit is the predicted value, but what are lwr and upr? Do they define the intervals in absolute terms (y coordinates)? The intervals seem to be extremely wide, wider than would make sense if my model was a good fit:
sales_volume deal_size fit lwr upr
0 0 3709276.494 -8950776.04 16369329.03
0 8586337.22 3709276.494 -8950776.04 16369329.03
110000 549458.6512 3730150.811 -8929897.298 16390198.92
When you use predict with an lm model, you can specify an interval. You have three choices: none will not return intervals, confidence and prediction. Both of those will return different values. The first column will be as you said the predicted values (column fit). You then have two other columns : lwr and upper which are the lower and upper levels of the confidence intervals.
What is the difference between confidence and prediction ?
confidence is a (by default 95%, use level if you wish to change that) confidence interval of the mean of the predicted value. It is the green interval on your plot. Whereas prediction is a (also 95%) confidence interval of all your values, meaning that should you repeat your experience/survey/... a huge number of times, you can expect that 95% of your values will fall in the yellow interval, thus making it a lot wider than the green one as the green one only evaluates the mean.
And as you an see on your plot, almost all values are in the yellow interval. R doesn't know that your values can only be positive so it explains why the yellow interval "begins" under 0.
Also, when you say "The intervals seem to be extremely wide, wider than would make sense if my model was a good fit", you can see in your plot that the interval is not that big, considering that you can expect 95% of the values to be in it, and you can clearly see a trend in your data. And your model is clearly a good fit as the adjusted R squared and the global p-value tells you.
Just a slight rephrasing of #etienne above, which is very good and accurate.
Confidence interval is the (1-alpha; eg 95%) interval for the mean prediction (or group response). IE if you have 10 new companies with sales volume of 2e+08 the predict(..., interval= "confidence") interval will give you the long-run average interval for your group mean.
With Var(\hat y|X= x*) = \sigma^2 (1/n + (x*-\bar x)^2 / SXX)
The prediction interval is the (1-alpha; eg 95%) interval for an individual response -- predict(..., interval= "predict"). IE for a single new company with sales volume of 2e+08...
With Var(\hat y|X= x*) = \sigma^2 (1 + 1/n + (x*-\bar x)^2 / SXX)
(Sorry that LaTeX isn't supported)
I'm trying to evaluate the output from a negative binomial mixed model using glmmadmb. To summarize the output I'm comparing the summary function with output forom the mcmc option. I have run this model:
pre1 <- glmmadmb(walleye~(1|year.center) + (1|Site) ,data=pre,
family="nbinom2",link="log",
mcmc=TRUE,mcmc.opts=mcmcControl(mcmc=1000))
I have two random intercepts: year and site. Year has 33 levels and site has 15.
The random effect parameter estimate for site and year from summary(pre1) do not seem to agree with the posterior distribution from the mcmc output. I am using the 50% confidence interval as the estimate that should coincide with the parameter estimate from the summary function. Is that incorrect? Is there a way to obtain an error around the random effect parameter using the summary function to gauge whether this is variance issue? I tried using postvar=T with ranef but that did not work. Also, Is there a way to format the mcmc output with informative row names to ensure I'm using the proper estimates?
summary output from glmmabmb:
summary(pre1)
Call:
glmmadmb(formula = walleye ~ (1 | year.center) + (1 | Site),
data = pre, family = "nbinom2", link = "log", mcmc = TRUE,
mcmc.opts = mcmcControl(mcmc = 1000))
AIC: 4199.8
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 3.226 0.154 21 <2e-16 ***
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Number of observations: total=495, year.center=33, Site=15
Random effect variance(s):
Group=year.center
Variance StdDev
(Intercept) 0.1085 0.3295
Group=Site
Variance StdDev
(Intercept) 0.2891 0.5377
Negative binomial dispersion parameter: 2.0553 (std. err.: 0.14419)
Log-likelihood: -2095.88
mcmc output:
m <- as.mcmc(pre1$mcmc)
CI <- t(apply(m,2,quantile,c(0.025,0.5,0.975)))
2.5% 50% 97.5%
(Intercept) 2.911667943 3.211775843 3.5537371345
tmpL.1 0.226614903 0.342206509 0.4600328729
tmpL.2 0.395353518 0.554211483 0.8619127547
alpha 1.789687691 2.050871824 2.3175742167
u.01 0.676758365 0.896844797 1.0726750539
u.02 0.424938481 0.588191585 0.7364795440
these estimates continue to u.48 to include year and site specific coefficients.
Thank you in advance for any thoughts on this issue.
Tiffany
The random effect parameter estimate for site and year from summary(pre1) do not seem to agree with the posterior distribution from the mcmc output. I am using the 50% confidence interval as the estimate that should coincide with the parameter estimate from the summary function. Is that incorrect?
It's not the 50% confidence interval, it's the 50% quantile (i.e. the median). The point estimates from the Laplace approximation of the among-year and among-site standard deviations respectively are {0.3295,0.5377}, which seem quite close to the MCMC median estimates {0.342206509,0.554211483} ... as discussed below, the MCMC tmpL parameters are the random-effects standard deviations, not the variances -- this might be the main cause of your confusion?
Is there a way to obtain an error around the random effect parameter using the summary function to gauge whether this is variance issue? I tried using postvar=T with ranef but that did not work.
The lme4 package (not the glmmadmb package) allows estimates of the variances of the conditional modes (i.e. the random effects associated with particular levels) via ranef(...,condVar=TRUE) (postVar=TRUE is now deprecated). The equivalent information on the uncertainty of the conditional modes is available via ranef(model,sd=TRUE) (see ?ranef.glmmadmb).
However, I think you might be looking for the $S (variance-covariance matrices) and $sd_S (Wald standard errors of the variance-covariance estimates) instead (although as stated above, I don't think there's really a problem).
Also, Is there a way to format the mcmc output with informative row names to ensure I'm using the proper estimates?
See p. 15 of vignette("glmmADMB",package="glmmADMB"):
The MCMC output in glmmADMB is not completely translated. It includes, in order:
pz zero-inflation parameter (raw)
fi xed eff ect parameters Named in the same way as the results of coef() or
fixef().
tmpL variances (standard-deviation scale)
tmpL1 correlation/off -diagonal elements of variance-covariance matrices (off -diagonal elements of the Cholesky factor of the correlation matrix'). (If you need to transform these to correlations, you will need to construct the relevant matrices with 1 on the diagonal and compute the cross-product, CC^T (see tcrossprod); if this makes no sense to you, contact the maintainers)
alpha overdispersion/scale parameter
u random eff ects (unscaled: these can be scaled using the estimated random-eff ects standard deviations from VarCorr())