this one has an answer in the back of the book, but I cannot understand it really.
For the Major League Baseball season from 1950 through 2008, there were fifty-nine nine-innings games in which one of the teams did not manage to get a hit. The data in the table give the number of no-hitters per season over this period. Assume that the data follow a Poisson distribution,
a) Estimate $\lambda$ and compare the observed and expected frequencies.
$$P_X(k;\lambda) = e^{-\lambda} \frac{\lambda^k}{k!} \; \quad k=0,1,2,...$$
\begin{array}{|c|c|} \hline \text{Number of No-Hitters}& \text{Frequency} \\ \hline 0& 6\\ \hline 1& 19\\ \hline 2& 12\\ \hline 3& 13\\ \hline 4& 9\\ \hline \end{array}
Now I worked out the maximum likelihood to be $$\lambda = \frac{\sum_\limits{i=1}^{n}k_i}{n}$$
I believe that is correct, but what I do not understand comparing the observed and expected frequencies, the book's is their answer as below:
First it has the expected value worked out based on the observed values. $\lambda = [0(6) + 1(19) + 2(12) + 3(13) + 4(9)]/59 = 2.00$
This is probably very simple, but I'm not seeing it, how did he get the third column of the table below?
\begin{array}{|c|c|c|} \hline \text{Number of No-Hitters}& \text{Observed Freq.}& \text{Expected Freq.} \\ \hline 0& 6 & 8.0\\ \hline 1& 19 & 16.0\\ \hline 2& 12 & 16.0\\ \hline 3& 13 & 10.6\\ \hline 4& 9 & 8.4\\ \hline \end{array}
You then plug in the estimated value of $\lambda$ into the PMF to get the prediction for the probabilities, then multiply that by $n$ to get the prediction for the frequencies. So for example $59 \cdot e^{-2} 2^0/0!$ is about $8$, which is the predicted frequency of $0$. Presumably all the individual frequencies shouldn't be too far apart from the data; if they are, then your speculation that the underlying distribution was Poisson was probably wrong.