Problem set 8: Worked solutions
Statistics and statistical programming
Northwestern University
MTS 525
Aaron Shaw
November 23, 2020
Import libraries
I’ll start by loading some useful libraries…
library(openintro)
library(tidyverse)
library(ggfortify)
library(haven)Part I: Mario kart replication/extension
Import data and setup
data(mariokart)
mariokart## # A tibble: 143 x 12
## id duration n_bids cond start_pr ship_pr total_pr ship_sp seller_rate
## <dbl> <int> <int> <fct> <dbl> <dbl> <dbl> <fct> <int>
## 1 1.50e11 3 20 new 0.99 4 51.6 standa… 1580
## 2 2.60e11 7 13 used 0.99 3.99 37.0 firstC… 365
## 3 3.20e11 3 16 new 0.99 3.5 45.5 firstC… 998
## 4 2.80e11 3 18 new 0.99 0 44 standa… 7
## 5 1.70e11 1 20 new 0.01 0 71 media 820
## 6 3.60e11 3 19 new 0.99 4 45 standa… 270144
## 7 1.20e11 1 13 used 0.01 0 37.0 standa… 7284
## 8 3.00e11 1 15 new 1 2.99 54.0 upsGro… 4858
## 9 2.00e11 3 29 used 0.99 4 47 priori… 27
## 10 3.30e11 7 8 used 20.0 4 50 firstC… 201
## # … with 133 more rows, and 3 more variables: stock_photo <fct>, wheels <int>,
## # title <fct>To make things a bit easier to manage, I’ll 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.
mariokart <- mariokart %>%
select(
price = total_pr, cond_new = cond, stock_photo, duration, wheels
) %>%
mutate(
cond_new = as.numeric(cond_new == "new"),
stock_photo = as.numeric(stock_photo == "yes")
)
mariokart## # A tibble: 143 x 5
## price cond_new stock_photo duration wheels
## <dbl> <dbl> <dbl> <int> <int>
## 1 51.6 1 1 3 1
## 2 37.0 0 1 7 1
## 3 45.5 1 0 3 1
## 4 44 1 1 3 1
## 5 71 1 1 1 2
## 6 45 1 1 3 0
## 7 37.0 0 1 1 0
## 8 54.0 1 1 1 2
## 9 47 0 1 3 1
## 10 50 0 0 7 1
## # … with 133 more rowsNow let’s 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.
summary(mariokart)## price cond_new stock_photo duration
## Min. : 28.98 Min. :0.0000 Min. :0.0000 Min. : 1.000
## 1st Qu.: 41.17 1st Qu.:0.0000 1st Qu.:0.0000 1st Qu.: 1.000
## Median : 46.50 Median :0.0000 Median :1.0000 Median : 3.000
## Mean : 49.88 Mean :0.4126 Mean :0.7343 Mean : 3.769
## 3rd Qu.: 53.99 3rd Qu.:1.0000 3rd Qu.:1.0000 3rd Qu.: 7.000
## Max. :326.51 Max. :1.0000 Max. :1.0000 Max. :10.000
## wheels
## Min. :0.000
## 1st Qu.:0.000
## Median :1.000
## Mean :1.147
## 3rd Qu.:2.000
## Max. :4.000sd(mariokart$price)## [1] 25.68856sd(mariokart$duration)## [1] 2.585693sd(mariokart$wheels)## [1] 0.8471829qplot(data = mariokart, x = price, geom = "density")
qplot(data = mariokart, x = duration, geom = "density")
qplot(data = mariokart, x = wheels, geom = "density")
ggplot(data = mariokart, aes(cond_new, price)) +
geom_boxplot()## Warning: Continuous x aesthetic -- did you forget aes(group=...)?
ggplot(data = mariokart, aes(stock_photo, price)) +
geom_boxplot()## Warning: Continuous x aesthetic -- did you forget aes(group=...)?
ggplot(data = mariokart, aes(duration, price)) +
geom_point()
ggplot(data = mariokart, aes(wheels, price)) +
geom_point()
I’m also going to calculate correlation coefficients for all of the variables. We can discuss what to make of this in class:
cor(mariokart)## price cond_new stock_photo duration wheels
## price 1.00000000 0.1273624 -0.08987876 -0.04123545 0.32998375
## cond_new 0.12736236 1.0000000 0.37554707 -0.48174393 0.42629801
## stock_photo -0.08987876 0.3755471 1.00000000 -0.36722999 0.06714198
## duration -0.04123545 -0.4817439 -0.36722999 1.00000000 -0.29947345
## wheels 0.32998375 0.4262980 0.06714198 -0.29947345 1.00000000Replicate model results
The description of the model
model <- lm(price ~ cond_new + stock_photo + duration + wheels, data = mariokart)
summary(model)##
## Call:
## lm(formula = price ~ cond_new + stock_photo + duration + wheels,
## data = mariokart)
##
## Residuals:
## Min 1Q Median 3Q Max
## -19.485 -6.511 -2.530 1.836 263.025
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 40.9385 7.3607 5.562 1.34e-07 ***
## cond_new 2.5816 5.2272 0.494 0.622183
## stock_photo -6.7542 5.1729 -1.306 0.193836
## duration 0.3788 0.9388 0.403 0.687206
## wheels 9.9476 2.7184 3.659 0.000359 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 24.4 on 138 degrees of freedom
## Multiple R-squared: 0.1235, Adjusted R-squared: 0.09808
## F-statistic: 4.86 on 4 and 138 DF, p-value: 0.001069While I’m looking at that, I’ll 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 it’s different in a number of ways! Without more information from the authors of the textbook it’s very hard to determine exactly where or why the differences emerge.
Assess model fit/assumptions
I’ve already generated a bunch of univariate and bivariate summaries and plots. Let’s inspect the residuals more closely to see what else we can learn. I’ll use the autoplot() function from the ggfortify package to help me do this.
autoplot(model)## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.
Overall, there are a number of issues with this model fit that I’d want to mention/consider: * The distribution of the dependent variable (price) is very skewed, with two extreme outliers. I’d 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 what’s 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 it’s possible that the estimates from the fitted model wouldn’t 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!
More out of curiosity than anything, I’ll 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 $100, so I’ll use that information here:
summary(lm(price ~ cond_new + stock_photo + duration + wheels, data = mariokart[mariokart$price < 100, ]))##
## Call:
## lm(formula = price ~ cond_new + stock_photo + duration + wheels,
## data = mariokart[mariokart$price < 100, ])
##
## Residuals:
## Min 1Q Median 3Q Max
## -11.3788 -2.9854 -0.9654 2.6915 14.0346
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 36.21097 1.51401 23.917 < 2e-16 ***
## cond_new 5.13056 1.05112 4.881 2.91e-06 ***
## stock_photo 1.08031 1.05682 1.022 0.308
## duration -0.02681 0.19041 -0.141 0.888
## wheels 7.28518 0.55469 13.134 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.901 on 136 degrees of freedom
## Multiple R-squared: 0.719, Adjusted R-squared: 0.7108
## F-statistic: 87.01 on 4 and 136 DF, p-value: < 2.2e-16What do you know. That was it.
Interpret some results
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-$15.32). * The point estimate for selling a new condition game is positive, but with a large standard error. As a result, the model fails to reject the null of no association and provides no evidence of any relationship between the game condition and auction price. * The point estimate for including a stock photo is negative, but again, the standard error is very large and the model fails to reject the null hypothesis. There is no evidence of any relationship between including a stock photo and the final auction price.
Recommendations
Based on this model result, I’d recommend the prospective vendor of a used copy of the game not worry about it too much unless they can get their hands on some extra Wii wheels, since it seems like the number of Wii wheels explains variations in the auction price outcome more than anything else they can control (such as whether or not a stock photo is included).
Part II: Hypothetical study
Import and explore
I’ll start off by just importing things and summarizing the different variables we care about here:
grads <- readRDS(url("https://communitydata.science/~ads/teaching/2020/stats/data/week_11/grads.rds"))
summary(grads)## id cohort gpa income
## Min. : 1.0 cohort_01:142 Min. : 0.01512 Min. :15.56
## 1st Qu.: 462.2 cohort_02:142 1st Qu.:22.56107 1st Qu.:41.07
## Median : 923.5 cohort_03:142 Median :47.59445 Median :52.59
## Mean : 923.5 cohort_04:142 Mean :47.83510 Mean :54.27
## 3rd Qu.:1384.8 cohort_05:142 3rd Qu.:71.81078 3rd Qu.:67.28
## Max. :1846.0 cohort_06:142 Max. :99.69468 Max. :98.29
## (Other) :994table(grads$cohort)##
## cohort_01 cohort_02 cohort_03 cohort_04 cohort_05 cohort_06 cohort_07 cohort_08
## 142 142 142 142 142 142 142 142
## cohort_09 cohort_10 cohort_11 cohort_12 cohort_13
## 142 142 142 142 142sd(grads$gpa)## [1] 26.84777sd(grads$income)## [1] 16.713qplot(data = grads, x = gpa, geom = "density")
qplot(data = grads, x = income, geom = "density")
ggplot(data = grads, aes(x = gpa, y = income)) +
geom_point()
I’ll also calculate some summary statistics and visual comparisons within cohorts:
tapply(grads$income, grads$cohort, summary)## $cohort_01
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 27.02 41.03 56.53 54.27 68.71 86.44
##
## $cohort_02
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 25.44 50.36 50.98 54.26 75.20 77.95
##
## $cohort_03
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 20.21 42.81 54.26 54.27 64.49 95.26
##
## $cohort_04
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 22.00 42.29 53.07 54.26 66.77 98.29
##
## $cohort_05
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 30.45 49.96 50.36 54.27 69.50 89.50
##
## $cohort_06
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 17.89 41.54 54.17 54.27 63.95 96.08
##
## $cohort_07
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 31.11 40.09 47.14 54.26 71.86 85.45
##
## $cohort_08
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 22.31 44.10 53.33 54.26 64.74 98.21
##
## $cohort_09
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 15.56 39.72 53.34 54.27 69.15 91.64
##
## $cohort_10
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 27.44 35.52 64.55 54.27 67.45 77.92
##
## $cohort_11
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 19.29 41.63 53.84 54.27 64.80 91.74
##
## $cohort_12
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 21.86 43.38 54.02 54.27 64.97 85.66
##
## $cohort_13
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 18.11 42.89 53.14 54.27 64.47 95.59tapply(grads$income, grads$cohort, sd)## cohort_01 cohort_02 cohort_03 cohort_04 cohort_05 cohort_06 cohort_07 cohort_08
## 16.76896 16.76774 16.76885 16.76590 16.76996 16.76670 16.76996 16.76514
## cohort_09 cohort_10 cohort_11 cohort_12 cohort_13
## 16.76982 16.77000 16.76924 16.76001 16.76676tapply(grads$gpa, grads$cohort, summary)## $cohort_01
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 14.37 20.37 50.11 47.84 63.55 92.21
##
## $cohort_02
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 15.77 17.11 51.30 47.84 82.88 94.25
##
## $cohort_03
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 5.646 24.756 45.292 47.831 70.856 99.580
##
## $cohort_04
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 10.46 30.48 50.47 47.83 70.35 90.46
##
## $cohort_05
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 2.735 22.753 47.114 47.837 65.845 99.695
##
## $cohort_06
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 14.91 22.92 32.50 47.84 75.94 87.15
##
## $cohort_07
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 4.578 23.471 39.876 47.840 73.610 97.838
##
## $cohort_08
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 2.949 25.288 46.026 47.832 68.526 99.487
##
## $cohort_09
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.01512 24.62589 47.53527 47.83472 71.80315 97.47577
##
## $cohort_10
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.217 24.347 46.279 47.832 67.568 99.284
##
## $cohort_11
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 9.692 26.245 47.383 47.831 72.533 85.876
##
## $cohort_12
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 16.33 18.35 51.03 47.84 77.78 85.58
##
## $cohort_13
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.3039 27.8409 46.4013 47.8359 68.4394 99.6442tapply(grads$gpa, grads$cohort, sd)## cohort_01 cohort_02 cohort_03 cohort_04 cohort_05 cohort_06 cohort_07 cohort_08
## 26.93027 26.93019 26.93861 26.93988 26.93768 26.94000 26.93000 26.93540
## cohort_09 cohort_10 cohort_11 cohort_12 cohort_13
## 26.93974 26.93790 26.93573 26.93004 26.93610Huh. Those are remarkably similar values for the group means and the group standard deviations…
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?
ggplot(data = grads, aes(x = gpa, y = income, color = cohort)) +
geom_point() +
facet_wrap(vars(cohort))
Hmmm. That’s…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 there’s 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, I’ve 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 you’ve been asked to make. The best response here (IMHO) is to abandon these kinds of analysis once you discover that there’s something systematically weird going on. The statistical procedures will “work” in the sense that they will return a result, but because those results aren’t 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.
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
## Residuals 1833 515354 281.1summary(grads.model <- lm(income ~ gpa + cohort, data = grads)) # gpa percentile has a small, negative association with income.##
## Call:
## lm(formula = income ~ gpa + cohort, data = grads)
##
## Residuals:
## Min 1Q Median 3Q Max
## -37.381 -13.259 -1.536 13.000 43.362
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.623e+01 1.567e+00 35.894 < 2e-16 ***
## gpa -4.110e-02 1.451e-02 -2.832 0.00468 **
## cohortcohort_02 -7.026e-03 1.986e+00 -0.004 0.99718
## cohortcohort_03 -1.790e-03 1.986e+00 -0.001 0.99928
## cohortcohort_04 -6.281e-03 1.986e+00 -0.003 0.99748
## cohortcohort_05 2.481e-03 1.986e+00 0.001 0.99900
## cohortcohort_06 1.296e-03 1.986e+00 0.001 0.99948
## cohortcohort_07 -7.184e-03 1.986e+00 -0.004 0.99711
## cohortcohort_08 -4.368e-03 1.986e+00 -0.002 0.99825
## cohortcohort_09 -1.440e-03 1.986e+00 -0.001 0.99942
## cohortcohort_10 -7.512e-04 1.986e+00 0.000 0.99970
## cohortcohort_11 1.030e-03 1.986e+00 0.001 0.99959
## cohortcohort_12 -9.652e-05 1.986e+00 0.000 0.99996
## cohortcohort_13 3.578e-04 1.986e+00 0.000 0.99986
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 16.74 on 1832 degrees of freedom
## Multiple R-squared: 0.004359, Adjusted R-squared: -0.002707
## F-statistic: 0.6169 on 13 and 1832 DF, p-value: 0.8419confint(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. Here’s some tidyverse code that does that:
grads %>%
group_by(cohort) %>%
summarize(
correlation = cor(income, gpa)
)## # A tibble: 13 x 2
## cohort correlation
## <fct> <dbl>
## 1 cohort_01 -0.0630
## 2 cohort_02 -0.0603
## 3 cohort_03 -0.0686
## 4 cohort_04 -0.0617
## 5 cohort_05 -0.0694
## 6 cohort_06 -0.0685
## 7 cohort_07 -0.0656
## 8 cohort_08 -0.0645
## 9 cohort_09 -0.0641
## 10 cohort_10 -0.0666
## 11 cohort_11 -0.0686
## 12 cohort_12 -0.0683
## 13 cohort_13 -0.0690Because 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 you’d be likely to find much of anything (and we haven’t really talked about interactions yet).
Part III: Trick or treating again
Import and update data
## 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"))
df <- df %>%
mutate(
obama = as.logical(obama),
fruit = as.logical(fruit),
year = as.factor(year),
male = as.logical(male),
age = age - mean(age, na.rm = T)
)
df## # A tibble: 1,223 x 7
## obama fruit year age male neob treat_year
## <lgl> <lgl> <fct> <dbl> <lgl> <dbl> <dbl>
## 1 FALSE FALSE 2014 -2.52 FALSE 1 4
## 2 FALSE TRUE 2014 -3.52 FALSE 1 4
## 3 FALSE FALSE 2014 0.480 TRUE 1 4
## 4 FALSE FALSE 2014 -3.52 TRUE 1 4
## 5 FALSE FALSE 2014 -1.52 FALSE 1 4
## 6 FALSE FALSE 2014 0.480 FALSE 1 4
## 7 FALSE FALSE 2014 1.48 TRUE 1 4
## 8 FALSE TRUE 2014 -3.52 FALSE 1 4
## 9 FALSE FALSE 2014 -0.520 TRUE 1 4
## 10 FALSE TRUE 2014 -1.52 FALSE 1 4
## # … with 1,213 more rowsLet’s fit and summarize the model:
f <- formula(fruit ~ obama + age + male + year)
fit <- glm(f, data = df, family = binomial("logit"))
summary(fit)##
## Call:
## glm(formula = f, family = binomial("logit"), data = df)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.9079 -0.7853 -0.7319 1.4685 1.8134
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.2509167 0.2180417 -5.737 9.63e-09 ***
## obamaTRUE 0.2355872 0.1385821 1.700 0.0891 .
## age 0.0109508 0.0214673 0.510 0.6100
## maleTRUE -0.1401293 0.1329172 -1.054 0.2918
## year2014 0.0002101 0.2215401 0.001 0.9992
## year2015 0.2600857 0.2107893 1.234 0.2173
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1374.9 on 1220 degrees of freedom
## Residual deviance: 1367.2 on 1215 degrees of freedom
## (2 observations deleted due to missingness)
## AIC: 1379.2
##
## Number of Fisher Scoring iterations: 4Interesting. Looks like adjusting for these other variables in a regression setting can impact the results.
Onwards to generating more interpretable results:
## Odds ratios (exponentiated log-odds!)
exp(coef(fit))## (Intercept) obamaTRUE age maleTRUE year2014 year2015
## 0.2862423 1.2656518 1.0110110 0.8692458 1.0002101 1.2970413exp(confint(fit))## 2.5 % 97.5 %
## (Intercept) 0.1846712 0.4348622
## obamaTRUE 0.9635277 1.6593989
## age 0.9691402 1.0543000
## maleTRUE 0.6697587 1.1280707
## year2014 0.6518923 1.5564945
## year2015 0.8649706 1.9799543## model-predicted probabilities for prototypical observations:
fake.kids <- data.frame(
obama = rep(c(FALSE, TRUE), 2),
year = factor(rep(c("2015", "2012"), 2)),
age = rep(c(9, 7), 2),
male = rep(c(FALSE, TRUE), 2)
)
fake.kids.pred <- cbind(fake.kids, pred.prob = predict(fit, fake.kids, type = "response"))
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
## 4 TRUE 2012 7 TRUE 0.2537326Note 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 you’d like an example here you go (I can try to clarify in class):
fake.kids.more.pred <- cbind(
fake.kids,
predict(fit, fake.kids, type = "link", se = TRUE)
)
within(fake.kids.more.pred, {
pred.prob <- plogis(fit)
lower <- plogis(fit - (1.96 * se.fit))
upper <- plogis(fit + (1.96 * se.fit))
})## 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
## 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
## 4 0.2537326Sub-group analysis
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.