(12 2 marks for a, b, c and 3 wattimena, in coal marks for el \"ceilings\" e) th
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(12 2 marks for a, b, c and 3 wattimena, in coal marks for el "ceilings" e) the of pillars holding up the underground stable (1) or not mines The was binary, each pillar classed either stable (O). 12 The variables are: 1-yes & o no Stability of the pillar, stable coded ratio of the pillar, wh.ratio Strength/Stress ratio of the pillar, ss.ratio Using R we examine two multiple logistic regression models ratio ss ratio m(Stal both) Coefficients Error z value Pr Estimate Std 1.375 3.5682 4.9064 (Intercept) 0.4120 0.820 0.6943 0.5696 wh. ratio 0.0163 2.403 2.3641 5.6802 Ss. ratio Null deviance 40.168 on 28 degrees of freedom Residual deviance: 15.533 on 26 degrees of freed 21.533 [1] (15.533,26) [1] 14 anova (model.both, test "chi sg") Terms added sequentially first to last) Chi) Resid. 40.168 Pr 28 NULL 39.906 27 wh. ratio 1 0.2616 15.533 7.934e-07 26 ss ratio 1 24.3738 -glm stable ss. ratio, family binomial) summary (model Coefficients Estimate Std. Error z value Pr(> Izl) 2.711 2.691 0.00713 (Intercept) -7.293 2.520 0.01174 ss ratio 5.318 Null 40.168 on 28 degrees of freedom Residual deviance: 16.282 on 27 degrees of freedom AIC: 20.282 40,168,28) [1] 0.06388317 1-pchis (16.272,27) [1] 0.947563 anova (model ss, test "Chisq") to last) Terms added sequentially (first Chi) Df Deviance Resid. Df Resid. Dev Pr( 28 40.168 27 16.282 1.022e-06 NULL ss ratio 1 23.886 Page 2Explanation / Answer
a. Residual Deviance is a measure of goodness of fit of a model. Higher numbers always indicates bad fit. The residual deviance shows how well the response variable is predicted by a model that includes independent variables.
Residual deviance of model.both is 15.533 with p-value 0.9467914
Residual deviance of model.ss is 16.282 with p-value 0.947563
Comparing residual deviance, the best model is model.both, but if we compare the p-values, it is almost same for both the models. So both models are almost equal with respect to residual deviance but model.ss has only one independent variable (simple), so the best model id model.ss.
b. The regression equation is ln(1-p/p) = exp(-(-7.293+5.318ss.ratio))
If ss.ratio is increased by 0.5, then exp(-(-7.293+5.318(ss.ratio+0.5)) = exp(-(-7.293+5.318(ss.ratio))-5.318*0.5)
= exp(-(-7.293+5.318(ss.ratio)) * exp(-2.659) = 0.0700182 * exp(-(-7.293+5.318(ss.ratio))
So, exp(-(-7.293+5.318ss.ratio)) >> exp(-(-7.293+5.318(ss.ratio+0.5))
So, it will increase the value of probability p, and have more chance to predict 1 (yes)
c. AIC of model.both is 21.533
AIC of model.ss is 20.282
As, the AIC of model.ss is lower, we choose model.ss
d. ss.ratio = 2
The logistic regression is p = 1/1+exp(-(-7.293+5.318ss.ratio)
p = 1/1+exp(-(-7.293+5.318*2) = 0.9658
So, stable = 1 (yes)
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