Written by Seewoo Li
on 

Child Parent interaction in Korea

1. Introduction

  • The data is collected from PISA(Programme for International Student Assessment).
  • Country: Korea
  • Students’ age: HighSchool 1stGrade
  • A fosterer has responded to several questionnaire. And this report analyzes data obtained from those questionnaire.
  • Explanation of variables
    • Early : Child regularly attended prior to <grade 1 in ISCED 1>: Early childhood educational development […] 0: No, 1: Yes
    • Care : , participation hindered: I had no one to take care of my child/ children. 0: There was no one, 1: There was someone
    • Parent : Who will complete this questionnaire? 0: Father, 1: Mother
    • Talk : Activities with your child, how often: Spend time just talking to my child 1: Never or hardly ever 2: Once or twice a year 3: Once or twice a month 4: Once or twice a week 5: Every day or almost every day
    • Lastly, this report is to analyze the relationship among ‘Talk’, ‘Early’, ‘Care’, ‘Parent’.

2. Eploratory Data Analysis

## # A tibble: 11 x 5
##    Early  Care Parent  Talk     n
##    <int> <dbl>  <int> <dbl> <int>
##  1     0    NA     NA    NA  1011
##  2     1    NA     NA    NA  4382
##  3    NA     0     NA    NA   687
##  4    NA     1     NA    NA  4706
##  5    NA    NA      0    NA   828
##  6    NA    NA      1    NA  4565
##  7    NA    NA     NA     1    31
##  8    NA    NA     NA     2    62
##  9    NA    NA     NA     3   569
## 10    NA    NA     NA     4  2222
## 11    NA    NA     NA     5  2509

3. Selecting Model

Proportional and Unproportional

fit1 <- vglm(cbind(Never, Year, Month, Week, Day)~ Early + Care + Parent, 
             family = cumulative(parallel = TRUE), data = rdata)
fit2 <- vglm(cbind(Never, Year, Month, Week, Day)~ Early + Care + Parent, 
             family = cumulative(parallel = FALSE), data = rdata)

\(\Delta\chi^2=Deviance_{proportional}-Deviance_{unproportional}\) = 11061.27-11051.1=10.17

\(df=df_{proportional}-df_{unproportional}\) = 21565-21556=9

\(\chi^2_{.05}(9)\) = 16.9189776 Since 16.9189776 > 10.17, choose proportional model.

Dropping ‘Early’

fit_without_Early <- vglm(cbind(Never, Year, Month, Week, Day)~ Care + Parent,
                          family = cumulative(parallel = TRUE), data = rdata)

\(\Delta\chi^2=Deviance_{without Early}-Deviance_{proportional}\) = 11061.35-11061.27=0.08

\(df=df_{without Early}-df_{proportional}\) = 21566-21565=1

\(\chi^2_{.05}(1)\) = 3.8414588 Since 3.8414588 > 0.08, choose model without ‘Early’.

4. Model

  • Model : \[logit{P(Y_{Talk}\le j)}=\alpha_j+\beta_1X_{Care}+\beta_2X_{Parent}\]

That is, \(logit{P(Y_{Talk}\le j)}=\alpha_j-0.35023X_{Care}-0.80328X_{Parent},\) \(where \alpha_{Never}= -4.22657, \alpha_{Year}= -3.11453, \alpha_{Month}= -1.02188, \alpha_{Week}= 1.13272 .\) Since \(\beta_1<0\), when there is someone to take care of the child(i.e., \(X_{Care}=1\)), \(p(having more time to talk)\) increases. Since \(\beta_2<0\), when the mother of the child responded to the questionnaire(i.e., \(X_{Parent}=1\)), \(p(having more time to talk)\) increases.

5. Discussion

The result implies that the exsistence of child care which is indicated by Care is related to child-parent interaction, and implies that gender of the respondent of the questionnaire which is indicated by Parent is related to child-parent interaction. It is suggested that further study should be conducted for discovering the reasons that why only 0.1440714% of the respondents responded that there is no one to take care of their child and why only 0.1813801% of the respondents was father.

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