R: Logistic regression statistics on logit binomial data with repeated-measures










0















I would like to know how to do a logistic regression of accuracy with logit binomial distribution in a dataset with repeated-measures.



I tried to do a logist regression like the one above, but it doesn't seem to consider the "repeated-measures" by subject, such as I can do with RT in an AOV with "Error(subject)" factor.



df <- read.table(text = "subject condition acc rt
1 A TRUE 254
1 B TRUE 645
2 A FALSE 243
2 B TRUE 656
3 A FALSE 234
3 B TRUE 456", header= TRUE)

acc <- with(df, aggregate(acc, list(subject, condition), sum))
colnames(acc) <- c("subject", "condition", "true")
acc$false <- 2-acc$true

summary(glm(cbind(acc$true, acc$false) ~ condition, data = acc, family = "binomial"))
anova(glm(cbind(acc$true, acc$false) ~ condition, data = acc, family = "binomial"), test = "Chisq")

summary(aov(df$rt ~ df$condition + Error(df$subject)))`


Please, could somebody help me? Thank you a lot!










share|improve this question
























  • When considering dependent data like this, its common to used a random/mixed effects model (aka multilevel model) that considers the clustering that occurs due to your subjects. Its a different model than the RM-ANOVA, but might be worth considering. Take a look at a lme4 package in R

    – Simon
    Nov 13 '18 at 1:51















0















I would like to know how to do a logistic regression of accuracy with logit binomial distribution in a dataset with repeated-measures.



I tried to do a logist regression like the one above, but it doesn't seem to consider the "repeated-measures" by subject, such as I can do with RT in an AOV with "Error(subject)" factor.



df <- read.table(text = "subject condition acc rt
1 A TRUE 254
1 B TRUE 645
2 A FALSE 243
2 B TRUE 656
3 A FALSE 234
3 B TRUE 456", header= TRUE)

acc <- with(df, aggregate(acc, list(subject, condition), sum))
colnames(acc) <- c("subject", "condition", "true")
acc$false <- 2-acc$true

summary(glm(cbind(acc$true, acc$false) ~ condition, data = acc, family = "binomial"))
anova(glm(cbind(acc$true, acc$false) ~ condition, data = acc, family = "binomial"), test = "Chisq")

summary(aov(df$rt ~ df$condition + Error(df$subject)))`


Please, could somebody help me? Thank you a lot!










share|improve this question
























  • When considering dependent data like this, its common to used a random/mixed effects model (aka multilevel model) that considers the clustering that occurs due to your subjects. Its a different model than the RM-ANOVA, but might be worth considering. Take a look at a lme4 package in R

    – Simon
    Nov 13 '18 at 1:51













0












0








0








I would like to know how to do a logistic regression of accuracy with logit binomial distribution in a dataset with repeated-measures.



I tried to do a logist regression like the one above, but it doesn't seem to consider the "repeated-measures" by subject, such as I can do with RT in an AOV with "Error(subject)" factor.



df <- read.table(text = "subject condition acc rt
1 A TRUE 254
1 B TRUE 645
2 A FALSE 243
2 B TRUE 656
3 A FALSE 234
3 B TRUE 456", header= TRUE)

acc <- with(df, aggregate(acc, list(subject, condition), sum))
colnames(acc) <- c("subject", "condition", "true")
acc$false <- 2-acc$true

summary(glm(cbind(acc$true, acc$false) ~ condition, data = acc, family = "binomial"))
anova(glm(cbind(acc$true, acc$false) ~ condition, data = acc, family = "binomial"), test = "Chisq")

summary(aov(df$rt ~ df$condition + Error(df$subject)))`


Please, could somebody help me? Thank you a lot!










share|improve this question
















I would like to know how to do a logistic regression of accuracy with logit binomial distribution in a dataset with repeated-measures.



I tried to do a logist regression like the one above, but it doesn't seem to consider the "repeated-measures" by subject, such as I can do with RT in an AOV with "Error(subject)" factor.



df <- read.table(text = "subject condition acc rt
1 A TRUE 254
1 B TRUE 645
2 A FALSE 243
2 B TRUE 656
3 A FALSE 234
3 B TRUE 456", header= TRUE)

acc <- with(df, aggregate(acc, list(subject, condition), sum))
colnames(acc) <- c("subject", "condition", "true")
acc$false <- 2-acc$true

summary(glm(cbind(acc$true, acc$false) ~ condition, data = acc, family = "binomial"))
anova(glm(cbind(acc$true, acc$false) ~ condition, data = acc, family = "binomial"), test = "Chisq")

summary(aov(df$rt ~ df$condition + Error(df$subject)))`


Please, could somebody help me? Thank you a lot!







r statistics regression logistic-regression






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edited Nov 12 '18 at 18:01







Gustavo

















asked Nov 12 '18 at 17:01









GustavoGustavo

12




12












  • When considering dependent data like this, its common to used a random/mixed effects model (aka multilevel model) that considers the clustering that occurs due to your subjects. Its a different model than the RM-ANOVA, but might be worth considering. Take a look at a lme4 package in R

    – Simon
    Nov 13 '18 at 1:51

















  • When considering dependent data like this, its common to used a random/mixed effects model (aka multilevel model) that considers the clustering that occurs due to your subjects. Its a different model than the RM-ANOVA, but might be worth considering. Take a look at a lme4 package in R

    – Simon
    Nov 13 '18 at 1:51
















When considering dependent data like this, its common to used a random/mixed effects model (aka multilevel model) that considers the clustering that occurs due to your subjects. Its a different model than the RM-ANOVA, but might be worth considering. Take a look at a lme4 package in R

– Simon
Nov 13 '18 at 1:51





When considering dependent data like this, its common to used a random/mixed effects model (aka multilevel model) that considers the clustering that occurs due to your subjects. Its a different model than the RM-ANOVA, but might be worth considering. Take a look at a lme4 package in R

– Simon
Nov 13 '18 at 1:51












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