How many exit should we give?
Look at the original dataset
## student_id plan_id name Grade Weekly_Assessments_Given
## 1 1 584 T1 Rational Counting PK 0
## 2 1 589 T1 Rhyme PK 1
## 3 1 1097 T1 Quantification PK 0
## 4 1 1173 T1 Syllables PK 0
## 5 1 1696 T1 Number Symbols PK 0
## 6 1 1908 T1 Phonemes PK 3
## Number.of.Exit.Exam preclass_exit_exam first_exit_exam second_exit_exam
## 1 1 NA 80 NA
## 2 1 NA 100 NA
## 3 1 NA NA 100
## 4 1 NA 100 NA
## 5 1 NA NA 30
## 6 2 NA 100 100
## third_exit_exam Maximum_Exit_Score Pass_Exam_or_Not
## 1 NA 80 Pass
## 2 NA 100 Pass
## 3 NA 100 Pass
## 4 NA 100 Pass
## 5 NA 30 Fail
## 6 NA 100 PassPlan Name
summary(factor(exit2$name))## T1 Addition Strategies T1 Classification/Sorting
## 159 19
## T1 Comparative Value T1 Compound Words
## 85 256
## T1 Conversations T1 Expression: Ask & Answer
## 36 73
## T1 Expression: Descriptive Language T1 Final Sounds
## 293 2
## T1 Initial Sounds T1 Number Symbols
## 30 423
## T1 One-to-One T1 Oral Comprehension
## 324 200
## T1 Ordering T1 Patterns
## 9 10
## T1 Phonemes T1 Quantification
## 268 567
## T1 Rational Counting T1 Rhyme
## 929 786
## T1 Rote Counting T1 Shape Composition
## 255 12
## T1 Shape Identification T1 Spatial Awareness
## 27 14
## T1 Subtraction Strategies T1 Syllables
## 37 336
## T1 Word Awareness T2 Rote Counting
## 534 13
## T2 Word Awareness
## 2library('ggplot2')
ggplot(exit2, aes(x=NumExit)) + geom_density()
exit2 = exit2[exit2$NumExit<=3,] Build logistic regression model to check the effect of the number of exit exam on Pass
model <- glm(factor(Pass) ~ factor(Grade)+factor(NumExit),family=binomial(link='logit'),data=exit2)
summary(model)##
## Call:
## glm(formula = factor(Pass) ~ factor(Grade) + factor(NumExit),
## family = binomial(link = "logit"), data = exit2)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.8967 0.6015 0.6015 0.7215 0.8463
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.61787 0.05424 29.829 < 2e-16 ***
## factor(Grade)PS -0.40486 0.06731 -6.015 1.8e-09 ***
## factor(NumExit)2 -0.01713 0.08234 -0.208 0.8352
## factor(NumExit)3 -0.37041 0.21603 -1.715 0.0864 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 5680.9 on 5677 degrees of freedom
## Residual deviance: 5642.0 on 5674 degrees of freedom
## AIC: 5650
##
## Number of Fisher Scoring iterations: 4In this case, I don’t consider the effect of different plan, the result shows that Grade have a grade impact on Pass. In addition, the number of exit exam is 3 also has a great impact on Pass.
Build logistic regression model for the following plans alone: T1 rhyme, T1 word awareness, T1 rote counting, and T1 one-to-one (those are the frequently used plans)
data1 = exit2[exit2$name %in% c('T1 Rational Counting','T1 Rythm','T1 Word Awareness','T1 One to One'),]
summary(factor(data1$Grade))## PK PS
## 692 758summary(factor(data1$NumExit))## 1 2 3
## 1060 358 32model1 <- glm(factor(Pass) ~ factor(Grade)+factor(NumExit),family=binomial(link='logit'), data = data1)
summary(model1)##
## Call:
## glm(formula = factor(Pass) ~ factor(Grade) + factor(NumExit),
## family = binomial(link = "logit"), data = data1)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.0411 0.5158 0.6672 0.8725 0.8725
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.3890 0.1039 13.363 < 2e-16 ***
## factor(Grade)PS -0.6195 0.1253 -4.944 7.66e-07 ***
## factor(NumExit)2 0.2076 0.1466 1.416 0.157
## factor(NumExit)3 0.5610 0.4953 1.133 0.257
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1629.7 on 1449 degrees of freedom
## Residual deviance: 1600.4 on 1446 degrees of freedom
## AIC: 1608.4
##
## Number of Fisher Scoring iterations: 4The effect of weekly Assessment given on Pass
model2 <- glm(factor(Pass) ~ factor(Weekly_Assessments_Given),family=binomial(link='logit'),data=data1)
summary(model2)##
## Call:
## glm(formula = factor(Pass) ~ factor(Weekly_Assessments_Given),
## family = binomial(link = "logit"), data = data1)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.0393 0.5168 0.7916 0.7916 0.8446
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.99977 0.07324 13.650 <2e-16 ***
## factor(Weekly_Assessments_Given)1 0.28588 0.18034 1.585 0.1129
## factor(Weekly_Assessments_Given)2 0.45061 0.20414 2.207 0.0273 *
## factor(Weekly_Assessments_Given)3 -0.15247 0.27091 -0.563 0.5736
## factor(Weekly_Assessments_Given)4 0.94614 0.53952 1.754 0.0795 .
## factor(Weekly_Assessments_Given)5 0.09885 1.15702 0.085 0.9319
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1629.7 on 1449 degrees of freedom
## Residual deviance: 1618.9 on 1444 degrees of freedom
## AIC: 1630.9
##
## Number of Fisher Scoring iterations: 4stratified on Grade
model3 <- glm(factor(Pass) ~ factor(NumExit),family=binomial(link='logit'),data=data1[data1$Grade=='PS',])
summary(model3)##
## Call:
## glm(formula = factor(Pass) ~ factor(NumExit), family = binomial(link = "logit"),
## data = data1[data1$Grade == "PS", ])
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6719 -1.5015 0.8849 0.8849 0.8849
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.73571 0.08966 8.206 2.3e-16 ***
## factor(NumExit)2 0.37794 0.19552 1.933 0.0532 .
## factor(NumExit)3 0.36291 0.67267 0.540 0.5895
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 932.09 on 757 degrees of freedom
## Residual deviance: 928.04 on 755 degrees of freedom
## AIC: 934.04
##
## Number of Fisher Scoring iterations: 4model4 <- glm(factor(Pass) ~ factor(NumExit),family=binomial(link='logit'),data=data1[data1$Grade=='PK',])
summary(model4)##
## Call:
## glm(formula = factor(Pass) ~ factor(NumExit), family = binomial(link = "logit"),
## data = data1[data1$Grade == "PK", ])
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.1460 0.6512 0.6512 0.6512 0.6576
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.44316 0.11468 12.585 <2e-16 ***
## factor(NumExit)2 -0.02177 0.22050 -0.099 0.921
## factor(NumExit)3 0.75407 0.75407 1.000 0.317
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 671.54 on 691 degrees of freedom
## Residual deviance: 670.29 on 689 degrees of freedom
## AIC: 676.29
##
## Number of Fisher Scoring iterations: 4In a word, after strafying on Grade, we can see that when the grade is PK, the number of exit exam has no effect on Pass, while when the grade is PS, the number of exit exam has some effect on Pass, which is exp(0.37794)=1.46. It means that NumExit=2 will increase the odd of Pass to 1.46 with other varialbes holds.