Assignment 2: Regression Analyses
Arun Kumar
09/06/19
Import the Data
Import the data from the Dutch Lexicon Project DLP_words.csv. All materials are from here: http://crr.ugent.be/programs-data/lexicon-projects
Variables we are going to use: - rt: Response Latency to the Lexical Decision Task - subtlex.frequency: The frequency of the word from the Dutch Subtitle Project. - length: Length of the word. - POS: part of speech. - bigram.freq: Summed frequency of the bigrams in the word (the sum of each two-letter combination frequency).
DLP_words <- readr::read_csv("DLP_words.csv")## Parsed with column specification:
## cols(
## spelling = col_character(),
## lexicality = col_character(),
## rt = col_double(),
## subtlex.frequency = col_double(),
## length = col_double(),
## POS = col_character(),
## bigram.freq = col_double()
## )Load the Libraries + Functions
Load all the libraries or functions that you will use to for the rest of the assignment. It is helpful to define your libraries and functions at the top of a report, so that others can know what they need for the report to compile correctly.
library(car)## Loading required package: carDatalibrary(readr)
library(boot)##
## Attaching package: 'boot'## The following object is masked from 'package:car':
##
## logitClean Up Part of Speech
Update the part of speech variable so that the Nouns are the comparison category. Here’s what the labels mean:
ADJ - Adjective N - Noun WW - Verbs
DLP_words$POS <- factor(DLP_words$POS, levels = c("N", "ADJ", "WW"), labels = c("Noun", "Adjective", "Verb"))
DLP_words## # A tibble: 12,026 x 7
## spelling lexicality rt subtlex.frequency length POS bigram.freq
## <chr> <chr> <dbl> <dbl> <dbl> <fct> <dbl>
## 1 aaide W 643. 11 5 Verb 211373
## 2 aaien W 614. 92 5 Verb 317860
## 3 aak W 561. 5 3 Noun 75056
## 4 aal W 669. 35 3 Noun 83336
## 5 aalmoes W 636. 42 7 Noun 166581
## 6 aam W NA 1 3 Adjective 71881
## 7 aambeeld W 691. 22 8 Noun 218551
## 8 aanblik W 614. 96 7 Noun 141520
## 9 aanbod W 568. 1174 6 Noun 139596
## 10 aanbouw W 652. 32 7 Noun 119589
## # ... with 12,016 more rowsDeal with Non-Normality
Since we are using frequencies, we should consider the non-normality of frequency. - Include a histogram of the original subtlex.frequency column. - Log-transform the subtlex.frequency column. - Include a histogram of bigram.freq - note that it does not appear extremely skewed.
## The histogram of the original `subtlex.frequency` column
hist(DLP_words$subtlex.frequency, breaks = 100)
## Log-transform the `subtlex.frequency` column
DLP_words$log_subtlex_freq <- log(DLP_words$subtlex.frequency)
## The histogram of `bigram.freq`
hist(DLP_words$bigram.freq, breaks = 100)
Create Your Linear Model
See if you can predict response latencies (DV) with the following IVs: subtitle frequency, length, POS, and bigram frequency.
lm_mod <- lm(rt ~ subtlex.frequency + length + POS + bigram.freq, data = DLP_words)
summary(lm_mod)##
## Call:
## lm(formula = rt ~ subtlex.frequency + length + POS + bigram.freq,
## data = DLP_words)
##
## Residuals:
## Min 1Q Median 3Q Max
## -258.31 -46.27 -9.90 36.87 651.07
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.894e+02 2.474e+00 238.250 < 2e-16 ***
## subtlex.frequency -4.084e-04 4.834e-05 -8.449 < 2e-16 ***
## length 6.261e+00 4.128e-01 15.166 < 2e-16 ***
## POSAdjective -2.536e+00 1.829e+00 -1.387 0.165526
## POSVerb -8.008e-01 1.420e+00 -0.564 0.572825
## bigram.freq 2.220e-05 6.481e-06 3.425 0.000617 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 62.72 on 12009 degrees of freedom
## (11 observations deleted due to missingness)
## Multiple R-squared: 0.03806, Adjusted R-squared: 0.03766
## F-statistic: 95.03 on 5 and 12009 DF, p-value: < 2.2e-16Interpret Your Model
Coefficients
- Which coefficients are statistically significant? Coefficients which are significan at 5% level: variable subtlex.frequency length bigram.freq
What do they suggest predicts response latency? (i.e., give the non-stats interpretation of the coefficients)
The coefficent on variable subtlex.frequency is negative, therefore frequent words are faster The coeffcient on length is positive: Thereforet longer words take longer to react The coefficient on the variable POS is not significant: Hence Adjectives, Verbs and Nouns have the same response latency The coeffcient on bigram.freq is positive: which implies that more frequent bigrams are slower
options(scipen = 999)
t <- summary(lm_mod)$coefficients[-1 , 3]
pr <- t / sqrt(t^2 + lm_mod$df.residual)
pr^2## subtlex.frequency length POSAdjective POSVerb
## 0.00590899844 0.01879270997 0.00016012374 0.00002647912
## bigram.freq
## 0.00097594604- What do the dummy coded POS values mean? Calculate the means of each group below to help you interpret:
tapply(DLP_words$rt, DLP_words$POS, mean, na.rm = TRUE)## Noun Adjective Verb
## 633.1077 628.9087 633.5708Overall Model
- Is the overall model statistically significant? Since the F-statistic is 95.03 and p-value is close to zero,overall model is deemed to be significant at the 5% level.
What is the practical importance of the overall model? The R-squared is 0.038
Diagnostic Tests
Outliers
Create an influence plot of the model using the car library. - Which data points appear to have the most influence on the model?
influencePlot(lm_mod)
## StudRes Hat CookD
## 4184 7.974493 0.5263562855 11.717236871
## 8552 10.428743 0.0002960303 0.005319831
## 11783 2.157855 0.0704942705 0.058838675Additivity
Do we have additivity in our model? - Show that the correlations between predictors is less than .9. - Show the VIF values.
## correlation
summary(lm_mod, correlation= TRUE)$correlation[-1, -1]## subtlex.frequency length POSAdjective POSVerb
## subtlex.frequency 1.00000000 0.09789214 -0.01679744 -0.09437705
## length 0.09789214 1.00000000 0.02513935 0.09441722
## POSAdjective -0.01679744 0.02513935 1.00000000 0.19608345
## POSVerb -0.09437705 0.09441722 0.19608345 1.00000000
## bigram.freq 0.02157752 -0.43240037 0.01337525 -0.30343414
## bigram.freq
## subtlex.frequency 0.02157752
## length -0.43240037
## POSAdjective 0.01337525
## POSVerb -0.30343414
## bigram.freq 1.00000000## VIF
vif(lm_mod)## GVIF Df GVIF^(1/(2*Df))
## subtlex.frequency 1.022716 1 1.011294
## length 1.251375 1 1.118649
## POS 1.121023 2 1.028972
## bigram.freq 1.358407 1 1.165507Linearity
Is the model linear? - Include a plot and interpret the output.
plot(lm_mod, which = 2, pch = 16)
Normality
Are the errors normally distributed? - Include a plot and interpret the output.
hist(scale(residuals(lm_mod)))
Homoscedasticity/Homogeneity
Do the errors meet the assumptions of homoscedasticity and homogeneity? - Include a plot and interpret the output (either plot option).
plot(lm_mod, which = 1)
Bootstrapping
Use the function provided from class (included below) and the boot library to bootstrap the model you created 1000 times. - Include the estimates of the coefficients from the bootstrapping. - Include the confidence intervals for at least one of the predictors (not the intercept). - Do our estimates appear stable, given the bootstrapping results?
Use the following to randomly sample 500 rows of data - generally, you have to have more bootstraps than rows of data, so this code will speed up your assignment. In the boot function use: data = DF[sample(1:nrow(DF), 500, replace=FALSE),] for the data argument changing DF to the name of your data frame.
bootcoef = function(formula, data, indices){
d = data[indices, ] #randomize the data by row
model = lm(formula, data = d) #run our model
return(coef(model)) #give back coefficients
}
# The estimates of the coefficients from the bootstrapping
lm_mod_boot <- boot(formula = rt ~ subtlex.frequency + length + POS + bigram.freq,
data = DLP_words[sample(1:nrow(DLP_words), size = 500, replace = FALSE), ],
statistic = bootcoef,
R = 1000)
lm_mod_boot##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = DLP_words[sample(1:nrow(DLP_words), size = 500, replace = FALSE),
## ], statistic = bootcoef, R = 1000, formula = rt ~ subtlex.frequency +
## length + POS + bigram.freq)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 591.21979418172 0.7635156646024 10.27444850290
## t2* -0.00349408177 -0.0006581106600 0.00171904484
## t3* 5.13021312371 -0.0866320837204 1.75365845974
## t4* 0.79332710091 0.0524178813671 8.62253008182
## t5* 9.62914463832 0.2501683854199 6.29971240592
## t6* 0.00004920488 0.0000009384283 0.00003167061