CHILDES data for AoA paper

Option 1: Proportion of kids who produce word

For each age (12-144 months), calculate proportion of children who produce word.

• Currently not accounting for transcript length or number of times child said word in a transcript
  
• Word must be produced >3x within transcript to count as being said
  
• Age bins: 1-month (12mos, 13mos, 14mos, etc.) or 3-month (12-14mos, 15-18mos, 19-21mos, etc.)

Remaining questions:

• Do we need a cutoff for the minimum number of age bins where a word is produced? (e.g. words like “cauliflower” and “copper” are very low-frequency)
  
• Do we need to account for number of children per age bin? This varies a lot and is sparse after ~60mos:

DT::datatable(number_of_kids_1month, caption = "One-month age bins")

DT::datatable(number_of_kids_3month, caption = "Three-month age bins")

ggplot(nkids_3month_samplewords, aes(x=age_group, y=prop_say))+
    geom_point()+ 
    stat_smooth(method="glm", method.args=list(family="binomial"), se=TRUE)+
    scale_x_continuous(breaks=c(seq(0,144,12)))+
    theme_classic()+
    facet_wrap(~as.factor(word))

Option 2: Frequency of word production

For each age (12-144 months), calculate frequency of word production controlling for transcript length.
1. Group all data by transcript
2. Transcript length = number of child-produced words in transcript
3. Relative word frequency = N times word produced / N words in transcript
4. Calculated mean relative word frequency for each word in each age bin (1-month or 3-month)

• Currently not accounting for number of transcripts per child or number of children per age bin
  
• No minimum production count requirements for word inclusion at a given age

ggplot(frequency_3month_samplewords, aes(x=age_group, y=mean_proportion_frequency))+
    geom_point()+ 
    stat_smooth(method="glm", method.args=list(family="binomial"), se=TRUE)+
    scale_x_continuous(breaks=c(seq(0,144,12)))+
    theme_classic()+
    facet_wrap(~as.factor(word))

ggplot(frequency_3month_samplewords, aes(x=age_group, y=log_mean_freq))+
    geom_point()+ 
    scale_x_continuous(breaks=c(seq(0,144,12)))+
    theme_classic()+
    facet_wrap(~as.factor(word))

1. Get youngest age for each word where production >= .5, .6, .75 and rank

estimate_childes_aoa <- function(df, proportion){
  df %>%
    group_by(word) %>% 
    filter(prop_say >= proportion) %>% 
    filter(age_group == min(age_group)) %>% 
    ungroup() %>% 
    arrange(age_group) %>% 
    mutate(raw_rank_childes = seq(1, nrow(.))) %>% 
    group_by(age_group) %>% 
    mutate(averaged_rank_childes = mean(raw_rank_childes)) 
}

strictest_aoa_words <- estimate_childes_aoa(nkids_3months, .75)
middle_aoa_words <- estimate_childes_aoa(nkids_3months, .6)
lenient_aoa_words <- estimate_childes_aoa(nkids_3months, .5)

2. Spearman correlations for CHILDES, CDI, Kuperman

spearman_corr_df <- cdi %>%
  left_join(kuperman, by="word") %>% 
  left_join(lenient_aoa_words, by="word")

CDI vs. Kuperman

cor.test(spearman_corr_df$averaged_rank_cdi, spearman_corr_df$averaged_rank_kuperman, method="spearman")

## 
##  Spearman's rank correlation rho
## 
## data:  spearman_corr_df$averaged_rank_cdi and spearman_corr_df$averaged_rank_kuperman
## S = 4739399, p-value < 2.2e-16
## alternative hypothesis: true rho is not equal to 0
## sample estimates:
##       rho 
## 0.5813061

#rho = .581, p < .001

CHILDES vs. Kuperman

cor.test(spearman_corr_df$averaged_rank_childes, spearman_corr_df$averaged_rank_kuperman, method="spearman")

## 
##  Spearman's rank correlation rho
## 
## data:  spearman_corr_df$averaged_rank_childes and spearman_corr_df$averaged_rank_kuperman
## S = 69889, p-value = 8.222e-05
## alternative hypothesis: true rho is not equal to 0
## sample estimates:
##       rho 
## 0.4050951

#rho = .41, p < .001

CHILDES vs. CDI

cor.test(spearman_corr_df$averaged_rank_cdi, spearman_corr_df$averaged_rank_childes, method="spearman")

## 
##  Spearman's rank correlation rho
## 
## data:  spearman_corr_df$averaged_rank_cdi and spearman_corr_df$averaged_rank_childes
## S = 99201, p-value = 0.1454
## alternative hypothesis: true rho is not equal to 0
## sample estimates:
##       rho 
## 0.1555928

#rho = .15, p = .14