Solutions: https://www.paulamoraga.com/course-aramco/99-problems-5glm-solutions.html

Whyte, et al 1987 (Dobson, 1990) reported the number of deaths due to AIDS in Australia per 3 month period from January 1983 to June 1986. The data are explanatory variable \(X\): time (measured in multiple of 3 month after January 1983), and response variable \(Y\): number of deaths in Australia due to AIDS.

```
<- 1:14
x <- c(0, 1, 2, 3, 1, 4, 9, 18, 23, 31, 20, 25, 37, 45)
y cbind(x, y)
```

```
## x y
## [1,] 1 0
## [2,] 2 1
## [3,] 3 2
## [4,] 4 3
## [5,] 5 1
## [6,] 6 4
## [7,] 7 9
## [8,] 8 18
## [9,] 9 23
## [10,] 10 31
## [11,] 11 20
## [12,] 12 25
## [13,] 13 37
## [14,] 14 45
```

- Fit a Poisson regression model using time as explanatory variable
- Interpret the coefficient estimate of time
- Test if time has an effect on the number of death due to AIDS

The following R output reports the result of an experiment for pesticide which attempted to kill beetles. Beetles were exposed to gaseous carbon disulphide at various concentrations (in mg/L) for five hours and the number of beetles killed were noted.

```
:
Callglm(formula = cbind(Killed, Survived) ~ Dose, family = binomial)
> summary(modl)
:
CoefficientsPr(>|z|)
Estimate Std. Error z value -14.82300 1.28959 -11.49 <2e-16 ***
(Intercept) 0.24942 0.02139 11.66 <2e-16 ***
Dose : 284.2024 on 7 degrees of freedom
Null deviance: 7.3849 on 6 degrees of freedom
Residual deviance> vcov(modl)
(Intercept) Dose1.66302995 -0.0274363845
(Intercept) -0.02743638 0.0004573762
Dose
```

- Write down the fitted model.
- Perform a hypothesis test to determine the significance of Dose. You need to state the hypothesis, state the calculated test statistic, state the reference distribution under the null hypothesis, set a decision rule and state the conclusion.
- Estimate the odds ratio as well as its significance when Dose increases 4 units. Interpret the odds ratio.
- Provide the 95% confidence interval for the probability of killing when the value of Dose equals 62.