To make things a bit easier to manage, Iâll select the variables I want to use in the analysis and do some cleanup. Note that I conver the cond_new and stock_photo variables to logical first (using boolean comparisons) and also coerce them to be numeric values (using as.numeric()). This results in 1/0 values corresponding to the observations shown/described in Figure 9.13 and 9.14 on p. 365 of the textbook.
To make things a bit easier to manage, Iâll select the variables I want to use in the analysis and do some cleanup. Note that I convert the cond_new and stock_photo variables to logical TRUE/FALSE values first (using boolean comparisons) and then coerce them to be numeric values (using as.numeric()). This results in 1/0 values corresponding to the observations shown/described in Figure 9.13 and 9.14 on p. 365 of the textbook.
mariokart <- mariokart %>%
select(
price = total_pr, cond_new = cond, stock_photo, duration, wheels
@@ -1596,7 +1596,7 @@ mariokartNow letâs look at the variables in our model. Summary statistics for each, univariate density plots for the continuous measures, and some bivariate plots w each predictor and the outcome variable.
+Now letâs summarize and explore the variables in our model. Iâll calculate summary statistics for each, univariate density plots for the continuous measures, boxplots for the dichotomous indicators, and some bivariate plots with each predictor and the outcome variable.
summary(mariokart)## price cond_new stock_photo duration
## Min. : 28.98 Min. :0.0000 Min. :0.0000 Min. : 1.000
@@ -1620,6 +1620,8 @@ mariokart## [1] 0.8471829qplot(data = mariokart, x = price, geom = "density")
qplot(data = mariokart, y = price, geom = "boxplot") # check out the outliers!
qplot(data = mariokart, x = duration, geom = "density")
qplot(data = mariokart, x = wheels, geom = "density")The description of the model
+Based on the information in the textbook, Iâm assuming that the model is an ordinary least squares regression.
model <- lm(price ~ cond_new + stock_photo + duration + wheels, data = mariokart)
summary(model)##
@@ -1674,15 +1676,16 @@ summary(model)While Iâm looking at that, Iâll go ahead and calculate a confidence interval around the only parameter for which the model rejects the null hypothesis (wheels):
Huh, that doesnât quite look like whatâs in the textbook Figure 9.15â¦
+Iâll go ahead and calculate a confidence interval around the only parameter for which the model rejects the null hypothesis (wheels):
confint(model, "wheels")## 2.5 % 97.5 %
## wheels 4.572473 15.32278Overall, this resembles the model results in Figure 9.15, but notice that itâs different in a number of ways! Without more information from the authors of the textbook itâs very hard to determine exactly where or why the differences emerge.
+Without more information from the authors of the textbook itâs very hard to determine exactly where or why the differences emerge. My guess at this point is that it might have something to do with those outlying price values (take a look at that boxplot and density plot again). Maybe they did something to transform the variable? Remove the outliers? I have no idea.
Iâve already generated a bunch of univariate and bivariate summaries and plots. Letâs inspect the residuals more closely to see what else we can learn. Iâll use the autoplot() function from the ggfortify package to help me do this.
Iâve already generated a bunch of univariate and bivariate summaries and plots. Letâs inspect the residuals more closely to see what else we can learn. Iâll use the autoplot() function from the ggfortify package.
autoplot(model)## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
@@ -1690,9 +1693,13 @@ summary(model)
Overall, there are a number of issues with this model fit that Iâd want to mention/consider: * The distribution of the dependent variable (price) is very skewed, with two extreme outliers. Iâd recommend trying some transformations to see if it would look more appropriate for a linear regression and/or inspecting the cases that produced the outlying values more closely to understand whatâs happening there. * The plots of the residuals reveal those same two outlying points are also outliers with respect to the line of best fit. That said, they are not exerting huge amounts of leverage on the estimates, so itâs possible that the estimates from the fitted model wouldnât change much without those two points. Indeed, based on the degrees of freedom reported in Figure 9.15 (136) vs. the number reported in our version of the model (138) my best guess is that the textbook authors silently dropped those two outlying observations from their model!
There are those outliers again. At this point, there are a number of issues with this model fit that Iâd want to mention/consider: * The distribution of the dependent variable (price) is skewed, with two extreme outliers. Iâd recommend trying some transformations to see if it would look more appropriate for a linear regression and/or inspecting the cases that produced the outlying values more closely to understand whatâs happening there and identify reasons why theyâre so different. * The plots of the residuals reveal those same two outlying points are also outliers with respect to the line of best fit. That said, they are not exerting huge amounts of leverage on the estimates, so itâs possible that the estimates from the fitted model wouldnât change too much without those two points. Indeed, based on the degrees of freedom reported in Figure 9.15 (136) vs. the number reported in our version of the model (138) my best guess at this point is that the textbook authors silently dropped those two outlying observations from their model.
More out of curiosity than anything, Iâll create a version of the model that drops the two largest values of price. From the plots, I can see that those two are the only ones above $100, so Iâll use that information here:
-summary(lm(price ~ cond_new + stock_photo + duration + wheels, data = mariokart[mariokart$price < 100, ]))summary(
+ lm(price ~ cond_new + stock_photo + duration + wheels,
+ data = mariokart[mariokart$price < 100, ]
+ )
+)##
## Call:
## lm(formula = price ~ cond_new + stock_photo + duration + wheels,
@@ -1715,11 +1722,23 @@ summary(model)What do you know. That was it.
+What do you know. That was it. The difference in \(R^2\) is huge!
+A little further digging (by Nick Vincent) revealed that these two outliers come from auctions where the Mario kart game was being sold as part of a bundle along with other games. You can look this up in the title field from the original dataset using the following block of code:
data(mariokart)
+
+mariokart %>%
+ filter(total_pr > 100) %>%
+ select(id, total_pr, title)## # A tibble: 2 x 3
+## id total_pr title
+## <dbl> <dbl> <fct>
+## 1 110439174663 327. "Nintedo Wii Console Bundle Guitar Hero 5 Mario Kart "
+## 2 130335427560 118. "10 Nintendo Wii Games - MarioKart Wii, SpiderMan 3, etâ¦What do you make of the textbook authorsâ decision to drop the observations? Can you make a case for/against doing so? What seems like the right decision and the best way to handle this kind of situation?
The issues above notwithstanding, we can march ahead and interpret the model results. Here are some general comments and some specifically focused on the cond_new and stock_photo variables: * Overall, the linear model regressing total auction price on condition, stock photo, duration, and number of Wii wheels shows evidence of a positive, significant relationship between number of wheels and price. According to this model fit, an increase of 1 wheel is associated with a total auction price increase of $10 with the 95% confindence interval of ($4.57-$15.32). * The point estimate for selling a new condition game is positive, but with a large standard error. As a result, the model fails to reject the null of no association and provides no evidence of any relationship between the game condition and auction price. * The point estimate for including a stock photo is negative, but again, the standard error is very large and the model fails to reject the null hypothesis. There is no evidence of any relationship between including a stock photo and the final auction price.
The issues above notwithstanding, we can march ahead and interpret the results of the original model that I fit. Here are some general comments and some specifically focused on the cond_new and stock_photo variables: * Overall, the linear model regressing total auction price on condition, stock photo, duration, and number of Wii wheels shows evidence of a positive, significant relationship between number of wheels and price. According to this model fit, an increase of 1 wheel in a listing is associated with a total auction price increase of $10 on average (95% confindence interval: $4.57-$15.32). * The point estimate for selling a new condition game is positive, but with a large standard error. The model fails to reject the null of no association and provides no evidence of any relationship between the game condition and auction price. * The point estimate for including a stock photo is negative, but again, the standard error is very large and the model fails to reject the null hypothesis. There is no evidence of any relationship between including a stock photo and the final auction price.
Iâll start off by just importing things and summarizing the different variables we care about here:
grads <- readRDS(url("https://communitydata.science/~ads/teaching/2020/stats/data/week_11/grads.rds"))
@@ -1759,7 +1778,7 @@ summary(grads)ggplot(data = grads, aes(x = gpa, y = income)) +
geom_point()
Iâll also calculate some summary statistics and visual comparisons within cohorts:
+Iâll also calculate some summary statistics and visual comparisons within districts (cohorts):
tapply(grads$income, grads$cohort, summary)## $cohort_01
## Min. 1st Qu. Median Mean 3rd Qu. Max.
@@ -1874,7 +1893,58 @@ summary(grads)Huh. Those are remarkably similar values for the group means and the group standard deviationsâ¦
+Note that you could also do this pretty easily with a call to group_by and summarize in the tidyverse:
grads %>%
+ group_by(cohort) %>%
+ summarize(
+ n = n(),
+ min = min(income),
+ mean = mean(income),
+ max = max(income),
+ sd = sd(income)
+ )## # A tibble: 13 x 6
+## cohort n min mean max sd
+## <fct> <int> <dbl> <dbl> <dbl> <dbl>
+## 1 cohort_01 142 27.0 54.3 86.4 16.8
+## 2 cohort_02 142 25.4 54.3 78.0 16.8
+## 3 cohort_03 142 20.2 54.3 95.3 16.8
+## 4 cohort_04 142 22.0 54.3 98.3 16.8
+## 5 cohort_05 142 30.4 54.3 89.5 16.8
+## 6 cohort_06 142 17.9 54.3 96.1 16.8
+## 7 cohort_07 142 31.1 54.3 85.4 16.8
+## 8 cohort_08 142 22.3 54.3 98.2 16.8
+## 9 cohort_09 142 15.6 54.3 91.6 16.8
+## 10 cohort_10 142 27.4 54.3 77.9 16.8
+## 11 cohort_11 142 19.3 54.3 91.7 16.8
+## 12 cohort_12 142 21.9 54.3 85.7 16.8
+## 13 cohort_13 142 18.1 54.3 95.6 16.8grads %>%
+ group_by(cohort) %>%
+ summarize(
+ n = n(),
+ min = min(gpa),
+ mean = mean(gpa),
+ max = max(gpa),
+ sd = sd(gpa)
+ )## # A tibble: 13 x 6
+## cohort n min mean max sd
+## <fct> <int> <dbl> <dbl> <dbl> <dbl>
+## 1 cohort_01 142 14.4 47.8 92.2 26.9
+## 2 cohort_02 142 15.8 47.8 94.2 26.9
+## 3 cohort_03 142 5.65 47.8 99.6 26.9
+## 4 cohort_04 142 10.5 47.8 90.5 26.9
+## 5 cohort_05 142 2.73 47.8 99.7 26.9
+## 6 cohort_06 142 14.9 47.8 87.2 26.9
+## 7 cohort_07 142 4.58 47.8 97.8 26.9
+## 8 cohort_08 142 2.95 47.8 99.5 26.9
+## 9 cohort_09 142 0.0151 47.8 97.5 26.9
+## 10 cohort_10 142 0.217 47.8 99.3 26.9
+## 11 cohort_11 142 9.69 47.8 85.9 26.9
+## 12 cohort_12 142 16.3 47.8 85.6 26.9
+## 13 cohort_13 142 0.304 47.8 99.6 26.9Huh. Those are remarkably similar values for the group means and the group standard deviationsâ¦weird.
Onwards to plotting:
ggplot(data = grads, aes(x = cohort, y = income)) +
geom_boxplot()ggplot(data = grads, aes(x = gpa, y = income, color = cohort)) +
geom_point()
Those plots are also a little strange. I know this is just a simulated analysis, but it still seems weird that overall it just looks like a big mass of random points, but when I add the colors by cohort, I can see there are some lines and other regularities within groups. I wonder what happens when I plot each scatter within cohorts?
+Those plots are also a little strange. Even though this is just a simulated analysis, it still seems weird that overall the scatterplot just looks like a big mass of points, but when I color the points by district, I can see some regularities within groups. At this point, I might want to facet the scatterplots by district to see any patterns more clearly.
ggplot(data = grads, aes(x = gpa, y = income, color = cohort)) +
geom_point() +
facet_wrap(vars(cohort))
Hmmm. Thatâsâ¦absurd (in particular, cohort 8 looks like a sideways dinosaur). At this point, if I were really working as a consultant on this project, I would write to the client and start asking some uncomfortable questions about data quality (who collected this data? how did it get recorded/stored/etc.? what quality controls were in place?). I would also feel obligated to tell them that thereâs just no way the data correspond to the variables they think are here. If you did that and the client was honest, they might tell you where the data actually came from.
-In the event that you marched ahead with the analysis and are curious about what that could have looked like, Iâve provided some example code below. That said, this is a situation where the assumptions and conditions necessary to identify ANOVA, t-tests, or regression are all pretty broken because the data was generated programmatically in ways that undermine the kinds of interpretation youâve been asked to make. The best response here (IMHO) is to abandon these kinds of analysis once you discover that thereâs something systematically weird going on. The statistical procedures will âworkâ in the sense that they will return a result, but because those results arenât even close to meaningful, any relationships you do observe in the data reflect something different than the sorts of relationships the statistical procedures were designed to identify.
+Okay, thatâsâ¦a joke (in particular, cohort 8 looks like a sideways dinosaur). At this point, if I were really working as a consultant on this project, I would write to the client and start asking some probing questions about data quality (who collected this data? how did it get recorded/stored/etc.? what quality controls were in place?). I would also feel obligated to tell them that I suspect thereâs just no way the data correspond to the variables they think are here. If you did that and the client was honest, they might tell you where the data actually came from.
+In the event that you marched ahead with the analysis and are curious about what that could have looked like, Iâve provided some example code below. That said, this is a situation where the assumptions and conditions necessary to identify ANOVA, t-tests, or regression are all pretty broken because the data was generated programmatically in ways that undermine the kinds of interpretation youâve been asked to make. The best response here (IMHO) is to abandon these kinds of analysis once you discover that thereâs something systematically weird going on like this. While the experience of discovering a scatterplot dinosaur in your data isâ¦unlikely outside of the context of a problem set, there are many situations in which careful data exploration will bring a realization that you just donât understand some important things about the sources or qualities of your data. You have to learn to identify these moments and develop strategies for dealing with them! Often, the statistical procedures will âworkâ in the sense that they will return a result without any apparent errors, but because those results arenât even close to meaningful, any relationships you do observe in the data reflect something different than the sorts of relationships the statistical procedures were designed to identify.
+Okay, if you wanted example code to look at for this, here it is. Please just keep in mind that the results are not informative!
summary(aov(income ~ cohort, data = grads)) # no global differences of means across groups## Df Sum Sq Mean Sq F value Pr(>F)
## cohort 12 0 0.0 0 1
@@ -1927,7 +2001,7 @@ summary(grads)confint(grads.model, "gpa") # 95% confidence interval## 2.5 % 97.5 %
## gpa -0.06955974 -0.01263512Note that the failure to reject the null of any association between district and income in the ANOVA does not provide conclusive evidence that the relationship between GPA and income does not vary by cohort. There were several things you might have done here. One is to calculate correlation coefficients within groups. Hereâs some tidyverse code that does that:
+Note that the failure to reject the null of any association between district and income in the ANOVA would not (even in the event of more realistic data) provide very compelling evidence that the relationship between GPA and income does not vary by cohort. There were several things you might have done here. One is to calculate correlation coefficients within groups. Hereâs some tidyverse code that does that:
grads %>%
group_by(cohort) %>%
summarize(
@@ -1949,13 +2023,14 @@ summary(grads)Because these correlation coefficients are nearly identical, I would likely end my analysis here and conclude that the correlation between gpa and income appears to be consistently small and negative. If you wanted to go further, you could theoretically calculate an interaction term in the model (by including I(gpa*cohort) in the model formula), but the analysis up to this point gives no indication that youâd be likely to find much of anything (and we havenât really talked about interactions yet).
Because these correlation coefficients are nearly identical, I would likely end an analysis here and conclude that the correlation between gpa and income appears to be consistently small and negative. If you wanted to go further, you could theoretically calculate an interaction term in the model (by including I(gpa*cohort) in the model formula), but the analysis up to this point gives no indication that youâd be likely to find much of anything (and we havenât really talked about interactions yet). On top of that, thereâs a literal dinosaur lurking in your dataâ¦just give up!
Revisit the text and worked solutions for problem set 7 for more details about the study design, data collection and more.
## reminder that the "read_dta()" function requires the "haven" library
df <- read_dta(url("https://communitydata.science/~ads/teaching/2020/stats/data/week_07/Halloween2012-2014-2015_PLOS.dta"))
@@ -1984,7 +2059,7 @@ dfLetâs fit and summarize the model:
+That looks consistent with what we want here. Letâs fit and summarize the model:
f <- formula(fruit ~ obama + age + male + year)
fit <- glm(f, data = df, family = binomial("logit"))
@@ -2017,8 +2092,9 @@ summary(fit)Interesting. Looks like adjusting for these other variables in a regression setting can impact the results.
-Onwards to generating more interpretable results:
+Interesting. Looks like adjusting for these other variables in a regression setting allows us to uncover some different the results.
+Onwards to generating more interpretable results. You might recall that the big problem with interpreting logistic regression is that the results are given to you in âlog-odds.â Not only is it difficult to have intuitions about odds, but intuitions about the natural logarithms of odds are just intractable (for most of us).
+To make things easier, the typical first step is to calculare odds-ratios instead of log-odds. This is done by exponentiating the coefficients (as well as the corresponding 95% confidence intervals):
## Odds ratios (exponentiated log-odds!)
exp(coef(fit))## (Intercept) obamaTRUE age maleTRUE year2014 year2015
@@ -2031,9 +2107,10 @@ exp(coef(fit))## model-predicted probabilities for prototypical observations:
-fake.kids <- data.frame(
- obama = rep(c(FALSE, TRUE), 2),
+You can use these to construct statements about the change in odds of the dependent variable flipping from 0 to 1 (or FALSE to TRUE) predicted by a 1-unit change in the corresponding predictor (where an odds ratio of 1 corresponds to unchanged odds). Weâll interpret the obamaTRUE odds ratio below.
+Now, model-predicted probabilities for prototypical observations. Recall that itâs necessary to create synthetic (âfakeâ), hypothetical individuals to generate predicted probabilities like these. In this case, Iâll create two versions of each fake kid: one assigned to the treatment condition and one assigned to the control. Then Iâll use the predict() function to generate fitted values for each of the fake kids.
+fake.kids <- data.frame(
+ obama = c(FALSE, FALSE, TRUE, TRUE),
year = factor(rep(c("2015", "2012"), 2)),
age = rep(c(9, 7), 2),
male = rep(c(FALSE, TRUE), 2)
@@ -2044,10 +2121,10 @@ fake.kids.pred <- cbind(fake.kids, pred.prob = predict(fit, fake.kids, type =
fake.kids.pred
## obama year age male pred.prob
## 1 FALSE 2015 9 FALSE 0.2906409
-## 2 TRUE 2012 7 TRUE 0.2537326
-## 3 FALSE 2015 9 FALSE 0.2906409
+## 2 FALSE 2012 7 TRUE 0.2117531
+## 3 TRUE 2015 9 FALSE 0.3414844
## 4 TRUE 2012 7 TRUE 0.2537326
-Note that this UCLA logit regression tutorial also contains example code to help extract standard errors and confidence intervals around these predicted probabilities. You were not asked to produce them here, but if youâd like an example here you go (I can try to clarify in class):
+Note that this UCLA logit regression tutorial also contains example code to help extract standard errors and confidence intervals around these predicted probabilities. You were not asked to create them here, but if youâd like an example here you go. The workhorse is the plogis() function, which essentially does the inverse logit transformation detailed in the textbook chapter 8:
fake.kids.more.pred <- cbind(
fake.kids,
predict(fit, fake.kids, type = "link", se = TRUE)
@@ -2060,101 +2137,68 @@ within(fake.kids.more.pred, {
})
## obama year age male fit se.fit residual.scale upper lower
## 1 FALSE 2015 9 FALSE -0.8922736 0.2243091 1 0.3887361 0.2088421
-## 2 TRUE 2012 7 TRUE -1.0788031 0.2594135 1 0.3611555 0.1697706
-## 3 FALSE 2015 9 FALSE -0.8922736 0.2243091 1 0.3887361 0.2088421
+## 2 FALSE 2012 7 TRUE -1.3143903 0.2682218 1 0.3124531 0.1370389
+## 3 TRUE 2015 9 FALSE -0.6566864 0.2363758 1 0.4518027 0.2460144
## 4 TRUE 2012 7 TRUE -1.0788031 0.2594135 1 0.3611555 0.1697706
## pred.prob
## 1 0.2906409
-## 2 0.2537326
-## 3 0.2906409
+## 2 0.2117531
+## 3 0.3414844
## 4 0.2537326
To do this, weâll need to create a slightly new model formula that drops the year term since weâre going to restrict each subset of the data along that dimension.
Once I fit the models, Iâll use the stargazer package to create a reasonably pretty regression table that incorporates all three summaries.
f2 <- formula(fruit ~ obama + age + male)
-summary(glm(f2, data = df[df$year == "2012", ], family = binomial("logit")))##
-## Call:
-## glm(formula = f2, family = binomial("logit"), data = df[df$year ==
-## "2012", ])
-##
-## Deviance Residuals:
-## Min 1Q Median 3Q Max
-## -1.0041 -0.7564 -0.6804 -0.5434 1.9929
-##
-## Coefficients:
-## Estimate Std. Error z value Pr(>|z|)
-## (Intercept) -1.53780 0.37711 -4.078 4.55e-05 ***
-## obamaTRUE 0.32568 0.38680 0.842 0.400
-## age 0.08532 0.06699 1.273 0.203
-## maleTRUE 0.32220 0.38802 0.830 0.406
-## ---
-## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
-##
-## (Dispersion parameter for binomial family taken to be 1)
-##
-## Null deviance: 177.56 on 163 degrees of freedom
-## Residual deviance: 174.90 on 160 degrees of freedom
-## (1 observation deleted due to missingness)
-## AIC: 182.9
-##
-## Number of Fisher Scoring iterations: 4summary(glm(f2, data = df[df$year == "2014", ], family = binomial("logit")))##
-## Call:
-## glm(formula = f2, family = binomial("logit"), data = df[df$year ==
-## "2014", ])
-##
-## Deviance Residuals:
-## Min 1Q Median 3Q Max
-## -0.8016 -0.7373 -0.6846 -0.6594 1.8071
-##
-## Coefficients:
-## Estimate Std. Error z value Pr(>|z|)
-## (Intercept) -1.13351 0.19798 -5.725 1.03e-08 ***
-## obamaTRUE 0.07348 0.24158 0.304 0.761
-## age 0.01198 0.03673 0.326 0.744
-## maleTRUE -0.23980 0.23499 -1.020 0.308
-## ---
-## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
-##
-## (Dispersion parameter for binomial family taken to be 1)
-##
-## Null deviance: 450.11 on 421 degrees of freedom
-## Residual deviance: 448.85 on 418 degrees of freedom
-## AIC: 456.85
-##
-## Number of Fisher Scoring iterations: 4summary(glm(f2, data = df[df$year == "2015", ], family = binomial("logit")))##
-## Call:
-## glm(formula = f2, family = binomial("logit"), data = df[df$year ==
-## "2015", ])
-##
-## Deviance Residuals:
-## Min 1Q Median 3Q Max
-## -0.9067 -0.7931 -0.7338 1.4761 1.7012
-##
-## Coefficients:
-## Estimate Std. Error z value Pr(>|z|)
-## (Intercept) -0.9971079 0.1383417 -7.208 5.7e-13 ***
-## obamaTRUE 0.3170892 0.1898404 1.670 0.0949 .
-## age 0.0004615 0.0290552 0.016 0.9873
-## maleTRUE -0.1791721 0.1791828 -1.000 0.3173
-## ---
-## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
-##
-## (Dispersion parameter for binomial family taken to be 1)
-##
-## Null deviance: 743.8 on 634 degrees of freedom
-## Residual deviance: 740.0 on 631 degrees of freedom
-## (1 observation deleted due to missingness)
-## AIC: 748
-##
-## Number of Fisher Scoring iterations: 4Interesting. The treatment effect seems to emerge overwhelmingly within a single year of the data.
+## =============================================== +## Dependent variable: +## ----------------------------- +## fruit +## 2012 2014 2015 +## (1) (2) (3) +## ----------------------------------------------- +## obama 0.326 0.073 0.317* +## (0.387) (0.242) (0.190) +## +## age 0.085 0.012 0.0005 +## (0.067) (0.037) (0.029) +## +## male 0.322 -0.240 -0.179 +## (0.388) (0.235) (0.179) +## +## Constant -1.538*** -1.134*** -0.997*** +## (0.377) (0.198) (0.138) +## +## ----------------------------------------------- +## Observations 164 422 635 +## Log Likelihood -87.451 -224.427 -370.002 +## Akaike Inf. Crit. 182.902 456.853 748.005 +## =============================================== +## Note: *p<0.1; **p<0.05; ***p<0.01 +Well, for starters, the model providing a âpooledâ estimate of treatment effects while adjusting for age, gender, and study year suggests that the point estimate is âmarginallyâ statistically significant (\(p <0.1\)) indicating some evidence that the data support the alternative hypothesis (being shown a picture of Michelle Obama causes trick-or-treaters to be more likely to pick up fruit than the control condition). In more concrete terms, the trick-or-treaters shown the Obama picture were, on average, about 26% more likely to pick up fruit than those exposed to the control (95% CI: \(-4\%~-~+66\%\)).1 In even more concrete terms, the estimated probability that a 9 year-old girl in 2015 and a 7 year-old boy in 2012 would take fruit increase about 17% and 19% respectively on average (from 29% to 34% in the case of the 9 year-old and from 21% to 25% in the case of the 7 year-old). These findings are sort of remarkable given the simplicity of the intervention and the fairly strong norm that Halloween is all about candy.
+All of that said, the t-test results from Problem set 5 and the âunpooledâ results reported in the sub-group analysis point to some potential concerns and limitations. For starters, the fact that the experiment was run iteratively over multiple years and that the sample size grew each year raises some concerns that the study design may not have anticipated the small effect sizes eventually observed and/or was adapted on the fly. This would undermine confidence in some of the test statistics and procedures. Furthermore, because the experiment occurred in sequential years, thereâs a very real possibility that the significance of a picture of Michelle Obama shifted during that time period and/or the house in question developed a reputation for being âthat weird place where they show you pictures of Michelle Obama and offer you fruit.â Whatever the case, my confidence in the findings here is not so great and I have some lingering suspicions that the results might not replicate.
+On a more nuanced/advanced statistical note, I also have some concerns about the standard errors. This goes beyond the content of our course, but basically, a randomized controlled trial introduces clustering into the data by-design (you can think of it as analogous to the observations coming from the treatment âclusterâ and the control âclusterâ). In this regard, the normal standard error formulas can be biased. Luckily, thereâs a fix for this: compute ârobustâ standard errors as a result and re-calculate the corresponding confidence intervals. Indeed, robust standard errors are often considered to be the best choice even when you donât know about potential latent clustering or heteroskedastic error structures in your data. This a short pdf provides a little more explanation, citations, as well as example R code for how you might calculate robust standard errors.
+Remember when I said we would use those odds ratios to interpret the parameter on obamaTRUE? Here we are. The parameter value is approximately 1.26, which means that the odds of picking fruit are, on average, 1.26 times as large for a trick-or-treater exposed to the picture of Michelle Obama versus a trick-or-treater in the control condition. In other words, the odds go up by about 26% (\(= \frac{1.26-1}{1}\)).â©ï¸