How do I find the expected values for this type of Chi squared question? 


I have been doing exam papers and preparing for my exam, but I always can't do these type of questions in statistics, after I get to know how to do one, I'll be able to do these, which are common on my papers. We have to use chi squared distribution, but I am unable to find the expected values; I don't know because they do not tell the expectation.
 A: What you have here can be either construed as a chi-square test of homogeneity or a chi-square test of independence (the calculations are the same in any case; given the set up, it would usually be thought of as one of homogeneity, but the question is framed as one of independence).
A chi-square test of independence is used to test association in contingency tables like these.
Let $p_{i,j}$ be the population proportion in cell $(i,j)$ and let $p_{i,.}$ and $p_{.,j}$ be the proportions in each margin, respectively. 
Let $O_{i,j}$ be the observed count in cell $(i,j)$ and let $O_{i,.}$ and $O_{.,j}$ be the counts in each margin (the row and column sums), respectively. 
If the two categorical variables are independent, $p_{i,j}=p_{i,.} \cdot p_{.,j}$.
How to compute expected values
There are a total of $N$ observations in the table.
Therefore, if we knew $p_{i,.}$ and $p_{.,j}$ we'd estimate the expected number in each cell as $Np_{i,.}p_{.,j}$. However, we must estimate those marginal proportions from the data, $N\hat p_{i,.}\hat p_{.,j}$ where $\hat p_{i,.} = O_{i,.}/N$ and $\hat p_{.,j} =O_{.,j}/N$.
Hence $E_{i,j}=N\hat p_{i,.}\hat p_{.,j}= N\cdot O_{i,.}/N\cdot O_{.,j}/N = O_{i,.}\cdot O_{.,j}/N$ under the assumption of independence.
So for example, the expected value in the upper left corner is the column sum for the first column (120) times the row sum for the first row (139) divided by the overall total (250), giving 66.72.
See $\text{here}$ for the same information presented in a slightly different way, and see Nick Stauner's answer $\text{here}$ for additional explanation.
