This is a chi-squared test for the independence of two categorical variables,
given data in a $3 \times 3$ contingency table.
Under the assumption that 'Religious Preference' and 'Party Affiliation' categories are independent,
one estimates the expected cell counts by using the multiplication rule
for independence, and the row and column totals. Here is how to find $E_{11}$ in the cell for 'Catholic and CDU-CSU'.
$$\hat P(\text{Catholic}) = 549/1739,\;
\hat P(\text{CDU-CSU}) = 756/1739,\,\text{ thus }\\
\hat P(\text{Catholic}\cap\text{CDU-CSU}) = \frac{549}{1739}\times\frac{756}{1739}.$$
Then the expected count for this cell is the grand sample size $1739$ times the
estimated probability, $E_{11} = \hat E(\text{Cath.}\cap\text{CDU-CSU})= 1739\hat P(\text{Cath.}\cap\text{CDU-CSU})$ $=
549(756)/1739 = 238.6682.$
While observed counts are integers, expected counts need not be integers;
do not round expected counts.
The general rule for an expected cell count is "its row total times its
column total divided by the grand total."
Then the contribution to chi-squared for cell $(1,1)$ is
$$\frac{(X_{11} - E_{11})^2}{E_{11}} = \frac{(308 - 238.6682)^2}{238.6682} = 20.141.$$
The chi-squared statistic for this test is the sum of all $3 \times 3 = 9$
contributions to the chi-squared statistic:
$$Q = \sum_{i=1}^3 \sum_{j=1}^3 \frac{(X_{ij} - E_{ij})^2}{E_{ij}}.$$
Provided that all $E_{ij} > 5,$ the statistic $Q \stackrel{aprx}{\sim}
\mathsf{Chisq}(\nu = 4).$ (Some authors would say it is OK for a few of the $E_{ij}$ to be as small as 3.)
In general $\nu = (r-1)(c-1),$ where the
table has $r$ rows and $c$ columns. A rigorous argument for this
computation of degrees of freedom $\nu$ is difficult to give at an
elementary level.
It may help to notice, in the current example, that if only the four observed counts $X_{11}, X_{12}, X_{21}, X_{22}$ were given along
with the row and column totals, then it would be possible to reconstruct
all of 9 of the observed counts. That is to say, given the structure of
row and column totals used to find the $E_{ij},$ only four
of the $X_{ij}$ provide additional information.
The Minitab 17 printout for this problem is as follows:
Chi-Square Test for Association: Relig, Worksheet columns
Rows: Relig Columns: Worksheet columns
CDU SPD NotVoted All
Cath 308 162 79 549
238.7 206.8 103.5
20.141 9.699 5.820
Prot 284 273 80 637
276.9 239.9 120.1
0.181 4.559 13.415
None 164 220 169 553
240.4 208.3 104.3
24.284 0.658 40.129
All 756 655 328 1739
Cell Contents: Count
Expected count
Contribution to Chi-square
Pearson Chi-Square = 118.885, DF = 4, P-Value = 0.000
First, notice that the expected count and contribution to chi-squared shown
for cell $(1,1)$ in the Minitab table agree with our earlier computations.
[In printouts, Minitab rounds numbers to make displays that fit nicely
on a page; internally about 12-place accuracy is maintained throughout
the computation.] Also notice that the smallest of the $E_{ij} \approx 104,$
very safely larger than 5.
Second notice that $Q$ is quite large. If the categories were independent
we would expect this statistic to be about 4 (the mean of a chi-squared
distribution is its degrees of freedom). The value that cuts 5% from the
upper tail of $\mathsf{Chisq}(4)$ is 9.4877. This is the critical value
for a test at the 5% level.
Sometimes, it is helpful to look at some of the larger contributions
to chi-squared. For example, in cell $(1,1),$ 308 Catholics voted for CDU-CSU,
whereas, under the null hypothesis of independence, one would have expected
only about 239 such votes.
However, the P-value for $Q = 118.885$ is smaller than 0.0005 (printed as 0.000). So we could also reject at the 1% level
and at the 0.1% level. (Maybe for "x%" in your question, you should
say something such as: "Reject at any significance level down to 0.0005.") Whatever phrase you choose, the data are very strongly inconsistent with the hypothesis
that voting behavior and religious preference are independent.
