A machine which is used in a manufacturing process jams from time to time. It is thought that the frequency of jams might change over time as the machine becomes older. Once every three months the number of jams in a day is counted. The results are as follows.  

$$
{\begin{array}{l}{{\mathrm{Observation~}}i}\\ {{\mathrm{Age~of~machine~}}t_{i}\ {\mathrm{(months)}}}\\ {{\mathrm{Number~of~jams~}}y_{i}}\end{array}}\left|{\begin{array}{l l l l l l l l l}{{1}}&{{2}}&{{3}}&{{4}}&{{5}}&{{6}}&{{7}}&{{8}}\\ {{3}}&{{6}}&{{9}}&{{12}}&{{15}}&{{18}}&{{21}}&{{24}}\\ {{10}}&{{13}}&{{24}}&{{17}}&{{20}}&{{22}}&{{20}}&{{23}}\end{array}}\right|
$$  

Our model is as follows. Given the values of two parameters $\alpha,~\beta$ , the number of jams $y_{i}$ on a dat when the machine has age $t_{i}$ months has a Poisson distribution  

$$
y_{i}\sim\operatorname{Poisson}(\lambda_{i})
$$  

where  

$$
\log_{e}(\lambda_{i})=\alpha+\beta t_{i}.
$$  

Assume that the effect of our prior distribution on the posterior distribution is negligible and that large-sample approximations may be used.  

(a) Let the values of $\alpha$ and $\beta$ which maximise the likelihood be $\hat{\alpha}$ and $\hat{\beta}$ . Assuming that the likelihood is differentiable at its maximum, show that these satisfy the following two equations  

$$
\begin{array}{r c l}{{\displaystyle\sum_{i=1}^{8}(\hat{\lambda}_{i}-y_{i})}}&{{=}}&{{0}}\\ {{\displaystyle\sum_{i=1}^{8}t_{i}(\hat{\lambda}_{i}-y_{i})}}&{{=}}&{{0}}\end{array}
$$  

where  

$$
\log_{e}(\hat{\lambda}_{i})=\hat{\alpha}+\hat{\beta}t_{i}
$$  

and show that these equations are satisfied (to a good approximation) by  

$$
\hat{\alpha}=2.552\;\;\;\;\;\mathrm{and}\;\;\;\;\;\hat{\beta}=0.02638.
$$  

(You may use R to help with the calculations, but show your commands).   
You may assume from now on that these values maximise the likelihood.  

(b) Find an approximate symmetric 95% posterior interval for $\alpha+24\beta$ .(c) Find an approximate symmetric 95% posterior interval for $\exp(\alpha+24\beta)$ , the mean jam-rate per day at age 24 months.  

(You may use R to help with the calculations, but show your commands). 