Assignment 2
JIaoyuan Huang
2019-07-31
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).
library(readr)
DLP_words <- read_csv("C:/Users/Emily/Desktop/540/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(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.
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.
model <- lm(rt ~ subtlex.frequency + length + POS + bigram.freq, data = DLP_words)
summary(model)##
## 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? subtlex.frequency, length, and bigram.freq are statistically significant.
- What do they suggest predicts response latency? (i.e., give the non-stats interpretation of the coefficients) The negative coefficent of subtlex.frequency shows that the words apears more frequent, it should be faster. The positive coefficent of bigram.freq shows that if the bigrams are frequent, then it will be slower.
- Which coefficients appear to predict the most variance? Length appears to predict the most variance. Calculate the \(pr^2\) values below:
a <- summary(model)$coefficients[-1 , 3]
pr <- a / sqrt(a^2 + model$df.residual)
pr^2## subtlex.frequency length POSAdjective POSVerb
## 5.908998e-03 1.879271e-02 1.601237e-04 2.647912e-05
## bigram.freq
## 9.759460e-04- What do the dummy coded POS values mean? It means that adj is faster than nouns, and nouns is fater than verbs. 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? The overall model is statistically significant.
- What is the practical importance of the overall model? R square is only 0.038, which means that only 3.8% of data can be explained. Even F-statistic is 95 and p value shows it is statistically significant. ## 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(model)
## 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.
summary(model, 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 1.00000000vif(model)## 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? Yes the model is linear - Include a plot and interpret the output.
plot(model, which = 2, pch = 16)
Normality
Are the errors normally distributed? Yes, normally distributed. - Include a plot and interpret the output.
hist(scale(residuals(model)))
Homoscedasticity/Homogeneity
Do the errors meet the assumptions of homoscedasticity and homogeneity? No the errors do not meet the assumptions of homoscedasticity and homogeneity - Include a plot and interpret the output (either plot option).
plot(model, 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? The estimates do not appear stable based on the 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
model_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)
summary(model_boot)
}