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I fitted a ordinal logistic regression model of the form

mod <- polr(myResponse ~ Academic_YearCode + GenderCode, Hess = TRUE, data = dat) 

myResponse has 3 groups 1 (no improvement), 2 (some improvement), 3 (best improvement), Academic_YearCode has either 0 or 1 and GenderCode is either 0(Male) or 1 (female). This data was collected over many schools. This is the model summary:

Coefficients:
                  Value Std. Error t value
Academic_YearCode  0.3875    0.06872   5.638
GenderCode         0.8876    0.07047  12.596

Intercepts:
     Value    Std. Error t value 
1|2  -1.9745   0.0697   -28.3085
2|3  -0.3758   0.0547    -6.8655

I derived for each year, the prob. of a MALE or FEMALE child in either of the no improvement, some improvement and best improvement as follows to see whether the improvement overall has improved from year 0 to year 1 and whether this improvement is higher for male than female.

case1 <- data.frame(GenderCode = 0, Academic_YearCode = 0)
predict(mod, case1, type = 'probs')
    1         2         3 
0.1219099 0.2852266 0.5928635 

case2 <- data.frame(GenderCode = 0, Academic_YearCode = 1)
predict(mod, case2, type = 'probs')
    1          2          3 
0.08612118 0.23181089 0.68206793 

case3 <- data.frame(GenderCode = 1, Academic_YearCode = 0)
predict(mod, case3, type = 'probs')
  1          2          3 
0.05406043 0.16632442 0.77961515 

case4 <- data.frame(GenderCode = 1, Academic_YearCode = 1)
predict(mod, case4, type = 'probs')
    1          2          3 
0.03734299 0.12364425 0.83901276 

Following this tutorial, I realised I need to do a mixed effects logistic regression: How to use ordinal logistic regression with random effects?. So I did:

library(ordinal)
modOrdinal <- clmm(myResponse ~ Academic_YearCode + GenderCode + 
                  (1|School_updated), data = dat, Hess=T, nAGQ = 17)

Coefficients:
              Estimate Std. Error z value Pr(>|z|)    
Academic_YearCode  0.41132    0.07030   5.851 4.89e-09 ***
GenderCode         0.96070    0.07262  13.229  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Threshold coefficients:
    Estimate Std. Error z value
1|2  -1.8615     0.2099  -8.868
2|3  -0.1724     0.2058  -0.838

1) How do I interpret these coefficients?

2) How do I work the probabilities for the individual response groups for both MALE and FEMALE child for each year by hand?

Thank you

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