Answer:
a) H0:  There is no association between level of education and TV station preference (Independence)
H1: There is association between level of education and TV station preference (No independence)
b)  
c) 
d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence 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:
                                   High school   Some College   Bachelor or higher  Total
Public Broadcasting       15                       15                          10                     40
Commercial stations      5                         25                         10                     40   
Total                                20                      40                          20                    80
We need to conduct a chi square test in order to check the following hypothesis:
Part a
H0:  There is no association between level of education and TV station preference (Independence)
H1: There is association between level of education and TV station preference (No independence)
The level os significance assumed for this case is  
The statistic to check the hypothesis is given by:
 
Part b
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:
                                   High school   Some College   Bachelor or higher  Total
Public Broadcasting       10                       20                         10                     40
Commercial stations      10                        10                         20                     40   
Total                                20                      30                          30                    80
Part b
And now we can calculate the statistic:
 
Now we can calculate the degrees of freedom for the statistic given by:
 
Part c
In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is 
Part d
And we can calculate the p value given by:
 
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(33.75,2,TRUE)"
Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.