A log-linear model was fitted to data concerning Nazi membership (No versus Yes)
ID: 3312849 • Letter: A
Question
A log-linear model was fitted to data concerning Nazi membership (No versus Yes) by teachers in Germany before the end of World War II. Teachers were classified by religion (None, Protestant, and Catholic) and by residence (Rural versus Urban) Here are the results from using glm:
> levels(nazi.freq$religion)
[1] "None" "Protestant" "Catholic"
> levels(nazi.freq$membership)
[1] "No" "Yes"
> levels(nazi.freq$residence)
[1] "Rural" "Urban"
> glm.nazi <- glm(Freq~religion*membership*residence,data=nazi.freq,
+ family=poisson(link="log"))
> print(summary(glm.nazi),signif.stars=FALSE)
Call:
glm(formula = Freq ~ religion * membership * residence, family = poisson(link = "log"),
data = nazi.freq)
Deviance Residuals:
[1] 0 0 0 0 0 0 0 0 0 0 0 0
Coefficients:
Estimate Std. Error z value
(Intercept) 6.83518 0.03279 208.445
religionProtestant 0.29891 0.04327 6.907
religionCatholic 0.35775 0.04275 8.369
membershipYes -1.77259 0.08605 -20.600
residenceUrban 0.74143 0.03984 18.608
religionProtestant:membershipYes 1.46613 0.09636 15.215
religionCatholic:membershipYes 0.68668 0.10190 6.739
religionProtestant:residenceUrban 0.13716 0.05212 2.632
religionCatholic:residenceUrban -0.33496 0.05329 -6.285
membershipYes:residenceUrban -0.67412 0.11762 -5.731
religionProtestant:membershipYes:residenceUrban -0.07841 0.13039 -0.601
religionCatholic:membershipYes:residenceUrban -0.01632 0.14238 -0.115
Pr(>|z|)
(Intercept) < 2e-16
religionProtestant 4.94e-12
religionCatholic < 2e-16
membershipYes < 2e-16
residenceUrban < 2e-16
religionProtestant:membershipYes < 2e-16
religionCatholic:membershipYes 1.60e-11
religionProtestant:residenceUrban 0.00849
religionCatholic:residenceUrban 3.28e-10
membershipYes:residenceUrban 9.96e-09
religionProtestant:membershipYes:residenceUrban 0.54760
religionCatholic:membershipYes:residenceUrban 0.90873
(Dispersion parameter for poisson family taken to be 1)
Null deviance: 7.2204e+03 on 11 degrees of freedom
Residual deviance: -1.5254e-13 on 0 degrees of freedom
AIC: 126.31
Number of Fisher Scoring iterations: 2
What was the odds ratio for membership in the Nazi Party for an urban Protestant teacher versus an urban teacher with no religious affiliation in Germany based on the observed data? Report the result to two decimal places. If the answer cannot be found using this model, report the result as 1.00.
Explanation / Answer
The odds ratio for membership in the Nazi Party for an urban Protestant teacher versus an urban teacher with no religious.
This is logistic regression application. The equation of Logistic regression
ln[p/(1-p)] = a + bX + e
where p is the probability of event.
Odds ratio = p/(1-p)
With the given data result, the base class for the refrence for other levels is -
religion = 'None' ; membership = 'No' ; residence = 'Rural'
odds ratio for an urban Protestant teacher:
religionProtestant:membershipYes:residenceUrban -0.07841
As the estimate is negative, odds ration will be 1/exp(coefficient of estimate)
odds ratio = 1/exp(-0.07841)
= 1.08
odds ratio for urban teacher with no religious:
As the estimate is negative, odds ration will be 1/exp(coefficient of estimate)
odds ratio = 1/exp(-0.67412)
= 1.96
The analysis has been done on the basis of provided code and result in question. For the better analysis, we need full data set.
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