1.Write the logistic regression equation to model the odds of distress as a function of temperature. Call: glm(formula = as.factor(distress) ~ temp, family = "binomial",     data = ONE)   Deviance Residuals:     Min       1Q   Median       3Q      Max  -1.0611  -0.7613  -0.3783   0.4524   2.2175    Coefficients:             Estimate Std. Error z value Pr(>|z|)  (Intercept)  15.0429     7.3786   2.039   0.0415 * temp         -0.2322     0.1082  -2.145   0.0320 * --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1   (Dispersion parameter for binomial family taken to be 1)       Null deviance: 28.267  on 22  degrees of freedom Residual deviance: 20.315  on 21  degrees of freedom AIC: 24.315   Number of Fisher Scoring iterations: 5 Logit = log (odds) = log{p/(1-p)}= α+β1X1+ β2X2 +…+βiXi

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1.Write the logistic regression equation to model the odds of distress as a function of temperature.

Call:

glm(formula = as.factor(distress) ~ temp, family = "binomial",

    data = ONE)

 

Deviance Residuals:

    Min       1Q   Median       3Q      Max 

-1.0611  -0.7613  -0.3783   0.4524   2.2175 

 

Coefficients:

            Estimate Std. Error z value Pr(>|z|) 

(Intercept)  15.0429     7.3786   2.039   0.0415 *

temp         -0.2322     0.1082  -2.145   0.0320 *

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

 

(Dispersion parameter for binomial family taken to be 1)

 

    Null deviance: 28.267  on 22  degrees of freedom

Residual deviance: 20.315  on 21  degrees of freedom

AIC: 24.315

 

Number of Fisher Scoring iterations: 5

Logit = log (odds) = log{p/(1-p)}= α+β1X1+ β2X2 +…+βiXi

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