Answer:
1) H0: There is independence between the marital status and the diagnostic of alcoholic
H1: There is association between the marital status and the diagnostic of alcoholic
2) The statistic to check the hypothesis is given by:
 
3)  
4)  
And we can calculate the p value given by:
 
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(19.72,2,TRUE)"
Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed. 
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Assume the following dataset:
                     Diag. Alcoholic   Undiagnosed Alcoholic    Not alcoholic    Total
Married                     21                              37                            58                116
Not Married              59                             63                            42                164
Total                          80                             100                          100              280 
Part 1
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is independence between the marital status and the diagnostic of alcoholic
H1: There is association between the marital status and the diagnostic of alcoholic
The level os significance assumed for this case is  
Part 2
The statistic to check the hypothesis is given by:
 
Part 3
The table given represent the observed values, we just need to calculate the expected values with the following formula  
And the calculations are given by:
 
 
 
 
 
 
And the expected values are given by:
                     Diag. Alcoholic   Undiagnosed Alcoholic    Not alcoholic    Total
Married             33.143                       41.429                        41.429                116
Not Married     46.857                      58.571                        58.571                164
Total                   80                              100                             100                 280 
And now we can calculate the statistic:
 
Part 4
Now we can calculate the degrees of freedom for the statistic given by:
 
And we can calculate the p value given by:
 
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(19.72,2,TRUE)"
Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.