## Solutions to Problem Set 5:

Irish Educational Transitions

Statlib is an excellent source of statistical data and programs. Their Irish Educational Transitions dataset has data for 500 Irish school children aged 11 in 1967. The outcome of interest is educational attainment, which for purposes of this analysis I have recoded in `educg`

in three categories: junior, senior and 3rd level.

The predictors of interest are gender, a measure of father's occupational prestige which I have recoded into quartiles in `prestigeg`

, and the score in a reasoning test, which I also recoded into quartiles in `reasong`

. The file `irished.dta`

in the datasets section includes these recodes and drops primary school leavers and a few cases with missing data, for an effective sample size of 435.

. use http://data.princeton.edu/wws509/datasets/irished, clear (Irish Educational Transitions) . desc Contains data from http://data.princeton.edu/wws509/datasets/irished.dta obs: 435 Irish Educational Transitions vars: 9 21 Nov 2016 14:54 size: 15,660 ───────────────────────────────────────────────────────────────────────────────────────────────────────── storage display value variable name type format label variable label ───────────────────────────────────────────────────────────────────────────────────────────────────────── gender float %9.0g gender reason float %9.0g Score in Drumcondra verbal reasoning test educ float %9.0g Educational level attained cert float %9.0g cert Leaving certificate taken prestige float %9.0g Prestige score for father's occupation school float %23.0g school Type of school educg float %9.0g educg RECODE of educ (Educational level attained) prestigeg float %9.0g prestigeg RECODE of prestige (Prestige score for father's occupation) reasong float %9.0g reasong RECODE of reason (Score in Drumcondra verbal reasoning test) ───────────────────────────────────────────────────────────────────────────────────────────────────────── Sorted by:

> library(foreign) > irish <- read.dta("http://data.princeton.edu/wws509/datasets/irished.dta") > head(irish) gender reason educ cert prestige school educg prestigeg reasong 1 male 113 3 no 28 secondary junior Q1 Q4 2 male 110 9 yes 69 secondary senior Q4 Q3 3 male 121 5 no 57 secondary junior Q4 Q4 4 male 82 4 no 18 vocational junior Q1 Q1 5 male 85 4 no 28 vocational junior Q1 Q1 6 male 98 2 no 43 vocational junior Q3 Q2

### [1] Multinomial Logits

(a) Fit a multinomial logit model explaining educational attaintment in terms of gender, parental occupational prestige, and scores in the reasoning test, using the recoded variables.

. mlogit educg i.gender i.prestigeg i.reasong, nolog Multinomial logistic regression Number of obs = 435 LR chi2(14) = 135.45 Prob > chi2 = 0.0000 Log likelihood = -365.44159 Pseudo R2 = 0.1563 ─────────────┬──────────────────────────────────────────────────────────────── educg │ Coef. Std. Err. z P>|z| [95% Conf. Interval] ─────────────┼──────────────────────────────────────────────────────────────── junior │ (base outcome) ─────────────┼──────────────────────────────────────────────────────────────── senior │ gender │ female │ .3051277 .2304158 1.32 0.185 -.146479 .7567344 │ prestigeg │ Q2 │ .8704873 .3023455 2.88 0.004 .2779009 1.463074 Q3 │ 1.189704 .3376027 3.52 0.000 .5280148 1.851393 Q4 │ 1.340194 .3288149 4.08 0.000 .6957287 1.984659 │ reasong │ Q2 │ -.0830386 .2990186 -0.28 0.781 -.6691042 .503027 Q3 │ 1.035168 .3063952 3.38 0.001 .4346448 1.635692 Q4 │ 1.627113 .3488457 4.66 0.000 .9433884 2.310838 │ _cons │ -1.65099 .3203233 -5.15 0.000 -2.278812 -1.023167 ─────────────┼──────────────────────────────────────────────────────────────── 3rd_level │ gender │ female │ .1615339 .3477705 0.46 0.642 -.5200838 .8431517 │ prestigeg │ Q2 │ 1.5331 .5534948 2.77 0.006 .4482699 2.61793 Q3 │ 1.682486 .5878532 2.86 0.004 .5303152 2.834657 Q4 │ 2.226992 .5433381 4.10 0.000 1.162069 3.291915 │ reasong │ Q2 │ 2.110641 1.078963 1.96 0.050 -.0040878 4.225369 Q3 │ 3.233068 1.064462 3.04 0.002 1.146761 5.319375 Q4 │ 4.963772 1.053335 4.71 0.000 2.899273 7.02827 │ _cons │ -5.793072 1.116641 -5.19 0.000 -7.981649 -3.604496 ─────────────┴──────────────────────────────────────────────────────────────── . scalar ml = e(ll)

> # make sure outcome is a factor or a matrix! > irish$educg <- as.factor(irish$educg) > library(nnet) > ml <- multinom(educg ~ gender + prestigeg + reasong , data=irish) # weights: 27 (16 variable) initial value 477.896346 iter 10 value 369.535450 iter 20 value 365.464546 final value 365.441586 converged > summary(ml) Call: multinom(formula = educg ~ gender + prestigeg + reasong, data = irish) Coefficients: (Intercept) genderfemale prestigegQ2 prestigegQ3 prestigegQ4 senior -1.650999 0.3051297 0.8704957 1.189714 1.340206 3rd level -5.792979 0.1615402 1.5331076 1.682500 2.227006 reasongQ2 reasongQ3 reasongQ4 senior -0.08303942 1.035163 1.627145 3rd level 2.11053104 3.232968 4.963707 Std. Errors: (Intercept) genderfemale prestigegQ2 prestigegQ3 prestigegQ4 senior 0.3203241 0.2304163 0.3023462 0.3376034 0.3288158 3rd level 1.1165939 0.3477700 0.5534933 0.5878517 0.5433370 reasongQ2 reasongQ3 reasongQ4 senior 0.2990188 0.3063954 0.3488479 3rd level 1.0789145 1.0644124 1.0532858 Residual Deviance: 730.8832 AIC: 762.8832

(b) Interpret the coefficient for females in both equations.

. di exp(_b[senior:2.gender]), exp(_b[3rd_level:2.gender]) 1.3567983 1.1753123

> exp(coef(ml)[, "genderfemale"]) -1 senior 3rd level 0.3568010 0.1753197

The relative probability of senior rather than junior level is 36% higher, and the relative probability of 3rd rather than junior level is 18% higher, for females than for males in the same quartile of parental occupational prestige and reasoning scores.

(c) Compute the average marginal effect of gender on the probability of achieving 3rd level, using the continuous approximation "by hand".

We follow the calculations in the R and Stata logs. First we predict probabilities, then calculate the marginal effect, and finally average. I use equation numbers for brevity. I also calculate marginal effects for all 3 categories but only the last one was required.

. predict p1 p2 p3, pr . gen sumpb = p2 * _b[2:2.gender] + p3 * _b[3:2.gender] . gen me1 = p1 * ( - sumpb) . gen me2 = p2 * (_b[2:2.gender] - sumpb) . gen me3 = p3 * (_b[3:2.gender] - sumpb) . sum me1 me2 me3 Variable │ Obs Mean Std. Dev. Min Max ─────────────┼───────────────────────────────────────────────────────── me1 │ 435 -.0547583 .0164814 -.0751537 -.0197663 me2 │ 435 .0581615 .0107572 .0377484 .0749559 me3 │ 435 -.0034032 .0079186 -.022972 .0049748

> probs <- predict(ml, type = "probs") > b <- c(0, coef(ml)[,"genderfemale"]) > pb <- probs[, 2] * b[2] + probs[, 3] * b[3] > me <- matrix(0, nrow(probs), ncol(probs)) > for(j in 1:ncol(probs)) { + me[, j] <- probs[, j] * (b[j] - pb) + } > apply(me, 2, mean) [1] -0.054758571 0.058161404 -0.003402833

We find that the average probability of achieving third level is practically the same for females and comparable males, only 0.34 percentage points lower. (The other two marginal effects tell us they have a lower probability of junior level compensated almost exactly by a higher probability of senior level, namely -5.48 and 5.82 for a net difference of 0.34.)

(d) Reestimate the average marginal effect of gender on the probability of achieving 3rd level using the exact discrete calculation, also "by hand".

To do this calculation we need to set gender to female and then to male, predict the probabilities each time, and then compute discrete differences and average

. gen keep = gender . forvalues g = 1/2 { 2. quietly replace gender = `g' 3. predict p`g'1 p`g'2 p`g'3, pr 4. } . forvalues j=1/3 { 2. gen de`j' = p2`j' - p1`j' 3. } . sum de1 de2 de3 Variable │ Obs Mean Std. Dev. Min Max ─────────────┼───────────────────────────────────────────────────────── de1 │ 435 -.0544882 .0163955 -.074616 -.0214939 de2 │ 435 .0579666 .0105146 .0418357 .0742811 de3 │ 435 -.0034784 .0080581 -.0213806 .0043799 . quietly replace gender = keep

> library(dplyr) > males <- mutate(irish, gender = factor(1, levels=1:2, labels=c("male","female"))) > pmale <- predict(ml, type = "probs", newdata = males) > females <- mutate(irish, gender = factor(2, levels=1:2, labels=c("male","female"))) > pfemale <- predict(ml, type = "probs", newdata = females) > mpfemale <- apply(pfemale, 2, mean) > mpmale <- apply(pmale, 2, mean) > data.frame(rbind(female = mpfemale, male = mpmale, diff = mpfemale - mpmale)) junior senior X3rd.level female 0.44118814 0.42318310 0.135628766 male 0.49567669 0.36521657 0.139106745 diff -0.05448855 0.05796653 -0.003477979

The results are *very* similar to the continuous calculation. Girls have almost the same probability of achieving third level as comparable boys, on average 0.35 percentage points lower. (The other two discrete differences show a lower probability of junior level of 5.45 percentage points compensated by a higher probability of senior level of 5.80 points.)

(e) How would you go about testing the goodness of fit of this model considering that we have individual data? No need to do anything, just explain what you would do.

The deviance does not have a chi-squared distribution with individual data, so we need to create groups. Fortunately all predictors are categorical variables, with 32 possible combinations of gender and quartiles of occupational prestige and resoning score. We could group the data into these 32 groups and then the deviance distribution should be well approximated by a chi-squared distribution as the groups would average almost 14 observations each.

### [2] Sequential Logits

(a) Fit a sequential logit model using the continuation ratio method, where you first focus on the probability of going beyond the junior level, and then look at the conditional probability of achieving 3rd form among those going beyond the junior level.

. gen beyond = educg > 1 . logit beyond i.gender i.prestigeg i.reasong, nolog Logistic regression Number of obs = 435 LR chi2(7) = 100.36 Prob > chi2 = 0.0000 Log likelihood = -250.36968 Pseudo R2 = 0.1670 ─────────────┬──────────────────────────────────────────────────────────────── beyond │ Coef. Std. Err. z P>|z| [95% Conf. Interval] ─────────────┼──────────────────────────────────────────────────────────────── gender │ female │ .2836095 .2231757 1.27 0.204 -.1538069 .7210259 │ prestigeg │ Q2 │ .9807345 .2927419 3.35 0.001 .406971 1.554498 Q3 │ 1.268714 .3279591 3.87 0.000 .6259258 1.911502 Q4 │ 1.503489 .3171323 4.74 0.000 .8819216 2.125057 │ reasong │ Q2 │ .1002454 .2892327 0.35 0.729 -.4666402 .667131 Q3 │ 1.221205 .3002627 4.07 0.000 .6327005 1.809709 Q4 │ 2.141073 .3344413 6.40 0.000 1.48558 2.796565 │ _cons │ -1.70499 .3157896 -5.40 0.000 -2.323927 -1.086054 ─────────────┴──────────────────────────────────────────────────────────────── . scalar sl = e(ll) . estimates store beyond . di exp(_b[2.gender]) - 1 .32791427 . gen third = educg == 3 . logit third i.gender i.prestigeg i.reasong if beyond, nolog Logistic regression Number of obs = 232 LR chi2(7) = 34.81 Prob > chi2 = 0.0000 Log likelihood = -115.21048 Pseudo R2 = 0.1312 ─────────────┬──────────────────────────────────────────────────────────────── third │ Coef. Std. Err. z P>|z| [95% Conf. Interval] ─────────────┼──────────────────────────────────────────────────────────────── gender │ female │ -.0541974 .33231 -0.16 0.870 -.705513 .5971182 │ prestigeg │ Q2 │ .6368128 .5673881 1.12 0.262 -.4752474 1.748873 Q3 │ .4121018 .5852127 0.70 0.481 -.734894 1.559098 Q4 │ .8775737 .535457 1.64 0.101 -.1719027 1.92705 │ reasong │ Q2 │ 2.108388 1.090134 1.93 0.053 -.0282357 4.245011 Q3 │ 2.180404 1.066049 2.05 0.041 .0909867 4.269821 Q4 │ 3.327179 1.044294 3.19 0.001 1.2804 5.373958 │ _cons │ -4.133982 1.126399 -3.67 0.000 -6.341682 -1.926281 ─────────────┴──────────────────────────────────────────────────────────────── . scalar sl = sl + e(ll) . estimates store third . di exp(_b[2.gender]) - 1 -.05275488

> irish <- mutate(irish, beyond = as.numeric(educg) > 1) > sl1 <- glm(beyond ~ gender + prestigeg + reasong, data=irish, family=binomial) > summary(sl1) Call: glm(formula = beyond ~ gender + prestigeg + reasong, family = binomial, data = irish) Deviance Residuals: Min 1Q Median 3Q Max -2.1568 -0.9986 0.5066 0.9328 1.9349 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.7050 0.3158 -5.399 6.70e-08 *** genderfemale 0.2836 0.2232 1.271 0.203803 prestigegQ2 0.9807 0.2927 3.350 0.000808 *** prestigegQ3 1.2687 0.3280 3.869 0.000110 *** prestigegQ4 1.5035 0.3171 4.741 2.13e-06 *** reasongQ2 0.1002 0.2892 0.347 0.728899 reasongQ3 1.2212 0.3003 4.067 4.76e-05 *** reasongQ4 2.1411 0.3344 6.402 1.53e-10 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 601.10 on 434 degrees of freedom Residual deviance: 500.74 on 427 degrees of freedom AIC: 516.74 Number of Fisher Scoring iterations: 4 > irish <- mutate(irish, third = as.numeric(educg) > 2) > sl2 <- glm(third ~ gender + prestigeg + reasong, data=filter(irish, beyond), family=binomial) > summary(sl2) Call: glm(formula = third ~ gender + prestigeg + reasong, family = binomial, data = filter(irish, beyond)) Deviance Residuals: Min 1Q Median 3Q Max -1.208 -0.766 -0.588 1.148 2.587 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -4.1340 1.1264 -3.670 0.000242 *** genderfemale -0.0542 0.3323 -0.163 0.870445 prestigegQ2 0.6368 0.5674 1.122 0.261710 prestigegQ3 0.4121 0.5852 0.704 0.481313 prestigegQ4 0.8776 0.5355 1.639 0.101229 reasongQ2 2.1084 1.0901 1.934 0.053104 . reasongQ3 2.1804 1.0660 2.045 0.040822 * reasongQ4 3.3272 1.0443 3.186 0.001442 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 265.23 on 231 degrees of freedom Residual deviance: 230.42 on 224 degrees of freedom AIC: 246.42 Number of Fisher Scoring iterations: 6

(b) Interpret the coefficients for females in both equations in terms of odds or conditional odds. Is this result broadly consistent with the results of part 1?

> b <- c(coef(sl1)["genderfemale"], coef(sl2)["genderfemale"]) > exp(b) - 1 genderfemale genderfemale 0.32791427 -0.05275488

The odds of going beyond the junior level are 32.8% higher for females than males in the same quartile of parental occupational prestige and reasoning score.

Among those who go beyond the junior level, the (conditional) odds of achieving third level are 5.3% lower for females than males in the same category of occupational prestige and reasoning score.

The results are generally consistent with the previous model in estimating a higher probability of going beyond the junior level, but then a lower probability of going beyond the senior level conditional on having gotten there. It is not immediately obvious whether these two effects cancel each other in terms of the probability of achieving third level, but marginal effects can clear than up.

(c) Predict the average probability of reaching 3rd level if everyone was male and then if everyone was female, and compare your results with the corresponding answer from part 1.

. forvalues g = 1/2 { 2. quietly replace gender = `g' 3. quietly estimates restore beyond 4. predict p`g'b, pr 5. quietly estimates restore third 6. predict p`g'tb, pr 7. gen p`g't = p`g'b * p`g'tb 8. } . quietly replace gender = keep . gen des = p2t - p1t . sum p2t p1t des Variable │ Obs Mean Std. Dev. Min Max ─────────────┼───────────────────────────────────────────────────────── p2t │ 435 .1399573 .1444416 .0029064 .4548946 p1t │ 435 .1348909 .1441322 .0024251 .452615 des │ 435 .0050663 .0036268 .0004813 .0124282

> pb_male <- predict(sl1, type = "response", newdata = males) > pt.b_male <- predict(sl2, type = "response", newdata = males) > pt_male <- pb_male * pt.b_male > pb_female <- predict(sl1, type = "response", newdata = females) > pt.b_female <- predict(sl2, type = "response", newdata = females) > pt_female <- pb_female * pt.b_female > c(female = mean(pt_female), male = mean(pt_male), diff = mean(pt_female) - mean(pt_male)) female male diff 0.139957258 0.134890925 0.005066332

We find a difference of half a percentage point, not unlike the 0.34 we found in the previous model, indicating essentially no gender differences in the overall probability of reaching third level.

(d) Compare the sequential and multinomial logit models in terms of parsimony, goodness of fit, and how well they represent gender differences in educational attaintment.

. di ml, sl -365.44159 -365.58016

> c(logLik(ml), logLik(sl1) + logLik(sl2)) [1] -365.4416 -365.5802

The models have very similar log-likelihoods, -365.4 for the multinomial logit and -365.6 for the sequential logit, with a tiny difference in favor of multinomial logits, and exactly the same number of parameters, so there really isn't much to choose here. From what we have seen they also lead to similar predictions by gender, so from this point of view we could declare it a tie.

Taking a closer look, the generally small gender differences are confined to the contrast between junior and senior levels, which happens to be one of the two contrasts in the multinomial logit model, so that would be a point in its favor. On the other hand the sequential logit model shows that girls have a higher probability of going beyond junior but then a lower conditional probability of continuing to third, which I think shows more clearly what's going on. This would tip the balance towards sequential logits.

(e) Comment of the coefficients that correspond to children in the top quartile of the reasoning test. Do we need marginal effects to determine if they have a higher probability of reaching 3rd level than comparable children with lower scores?

In case it was not clear this question refers just to the sequential logit model. In this model children in the top quartile of reasoning scores, compared with children in lower quartiles with the same gender and parental occupational prestige, have a higher probability of going beyond junior level and *also* a higher conditional probability of continuing to third level given that they went beyond junior, so obviously the overall probability is also higher. Marginal effects are clearly not needed to settle this issue.

(If you think in terms of the multinomial logit model the answer is not so clear. Kids in the upper quartile of reasoning have higher relative probabilities of senior compared to junior and generally much higher relative probabilities or third compared to junior, so it looks like the overall probability of third level should be higher, but I would compute marginal effects to make sure. In fact they do.)

### [3] Ordered Logits

(a) Fit an ordered logit model to the same data using the same predictors.

. ologit educg i.gender i.prestigeg i.reasong, nolog Ordered logistic regression Number of obs = 435 LR chi2(7) = 125.58 Prob > chi2 = 0.0000 Log likelihood = -370.37744 Pseudo R2 = 0.1450 ─────────────┬──────────────────────────────────────────────────────────────── educg │ Coef. Std. Err. z P>|z| [95% Conf. Interval] ─────────────┼──────────────────────────────────────────────────────────────── gender │ female │ .1974119 .1987321 0.99 0.321 -.192096 .5869197 │ prestigeg │ Q2 │ .9903856 .2794842 3.54 0.000 .4426066 1.538165 Q3 │ 1.165486 .3021775 3.86 0.000 .5732289 1.757743 Q4 │ 1.47537 .2896877 5.09 0.000 .907592 2.043147 │ reasong │ Q2 │ .2097262 .2829288 0.74 0.459 -.344804 .7642565 Q3 │ 1.238899 .2832053 4.37 0.000 .6838272 1.793971 Q4 │ 2.354729 .2989482 7.88 0.000 1.768801 2.940656 ─────────────┼──────────────────────────────────────────────────────────────── /cut1 │ 1.693513 .3016261 1.102337 2.28469 /cut2 │ 4.170666 .3600605 3.464961 4.876372 ─────────────┴──────────────────────────────────────────────────────────────── . scalar ol = e(ll) . scalar odf = e(df_m)

> library(MASS) > ol <- polr(educg ~ gender + prestigeg + reasong, data = irish) > summary(ol) Call: polr(formula = educg ~ gender + prestigeg + reasong, data = irish) Coefficients: Value Std. Error t value genderfemale 0.1974 0.1987 0.9933 prestigegQ2 0.9904 0.2795 3.5436 prestigegQ3 1.1655 0.3022 3.8570 prestigegQ4 1.4754 0.2897 5.0929 reasongQ2 0.2097 0.2829 0.7413 reasongQ3 1.2389 0.2832 4.3745 reasongQ4 2.3547 0.2989 7.8767 Intercepts: Value Std. Error t value junior|senior 1.6935 0.3016 5.6146 senior|3rd level 4.1707 0.3601 11.5832 Residual Deviance: 740.7549 AIC: 758.7549

(b) Interpret the estimate of the first cutpoint in terms of odds or probabilities, keeping in mind the reference cell used in the model.

. di exp(_b[/cut1]), invlogit(_b[/cut1]) 5.4385543 .84468563

> cut1 <- ol$zeta[1] > c(exp(cut1), plogis(cut1)) junior|senior junior|senior 5.4385069 0.8446845

For males in the lower quartile of parental occupational prestige and reasoning score, the odds of being in junior level are 5.44 to 1, which gives a probability of 84.5%,

(c) Interpret the coefficient of females in terms of (i) a latent variable, and (ii) the odds of progressing past junior and senior levels.

. di _b[educg:2.gender]/(_pi/sqrt(3)) .10883886 . di exp(_b[educg:2.gender]) 1.2182457

> b <- coef(ol)["genderfemale"] > c(b/(pi/sqrt(3)), exp(b)) genderfemale genderfemale 0.1088369 1.2182413

In terms of a latent scale of educational achievement, girls are on average about one-tenth of a standard deviation higher than boys in the same category of parental occupational prestige and reasoning score. These girls also have 21.8% higher odds of going beyond junior level, as well as 21.8% higher odds of going beyond senior level, than boys in the same category of occupational prestige and reasoning score. (I think this is clearer than saying they have 21.8% higher odds or achieving third level rather than junior or senior, as well as 21.8% higher odds of achieving senior or third level rather than junior.)

(d) Predict the probability of reaching 3rd level if everyone was male and then if everyone was female, and compare your result with the corresponding answer from part 2.

. forvalues g = 1/2 { 2. quietly replace gender = `g' 3. predict o`g'1 o`g'2 o`g'3, pr 4. } . quietly replace gender = keep . gen ode = o23 - o13 . sum o23 o13 ode Variable │ Obs Mean Std. Dev. Min Max ─────────────┼───────────────────────────────────────────────────────── o23 │ 435 .1505351 .1414167 .0184647 .464272 o13 │ 435 .1302775 .1265154 .0152071 .4156715 ode │ 435 .0202576 .015174 .0032576 .0486004

> pol_male <- predict(ol, type = "probs", newdata = males) > pol_female <- predict(ol, type = "probs", newdata = females) > mo_male <- apply(pol_male, 2, mean) > mo_female <- apply(pol_female, 2, mean) > data.frame(rbind(female = mo_female, male = mo_male, diff = mo_female - mo_male)) junior senior X3rd.level female 0.44597704 0.4034877 0.15053524 male 0.48405429 0.3856677 0.13027799

We see that on average girls have a 15% probability of reaching third level while comparable boys have an average probability of 13%, a difference of two percentage points. This is larger than we had estimated using multinomial or sequential logit models.

(e) How well does this model stack up against the previous two? Make sure you consider parsimony, goodness of fit, and how well the model reflects gender differences in educational attaintment.

. di -2 * ml + 2 * 16 762.88317 . di -2 * ol + 2 * 9 758.75489

> c(ml = AIC(ml), ol = AIC(ol)) ml ol 762.8832 758.7549

The model is clearly more parsimonious than the previous ones, using only 9 parameters instead of 16, a savings of 7. The log-likelihood of -370.4 is 4.9 points lower than the multinomial logit model, so the fit is a bit worse. Using Akaike's criterion with a deviance penalty of two points per parameter we obtain a small advantage of 4.1 points for the ordered logit model.

In terms of representing gender differences, however, the ordered logit model is not as good as the previous ones, because it assumes a uniform shift in academic achievement. This should result in girls having higher progressions from junior to senior as well as senior to third, whereas the evidence indicates that they have a somewhat higher probability of progressing to senior level but then a somewhat lower probability of continuing to third level, so in the end we see no gender differences in the probability of achiving third level.