The Null Hypothesis: There will be no difference in the number of votes for each candidate. 

The first step is to enter the observed number of votes for each candidate.  

Then look up in a Table of Chi-Square values to determine what value is the cutoff for rejecting the null hypothesis. The degrees of freedom for three candidates is the number of candidates minus one. For this example, the degrees of freedom is 2. For an alpha of .05, degrees of freedom of 2, and a two-tailed test, the Chi-Square value will need to be greater than 5.992. 

The second step is to total the number of observed frequencies and divide it by the number of candidates. The result is the expected frequency for each candidate. This is the expectation by chance. In this example, 197 + 120 + 214 = 531. Then, 531 / 3 = 177. 

 The third step is to subtract the expected frequencies from the observed frequencies. The results are difference scores. 

The fourth step is to square each difference score. 

The fifth step is to divide each squared difference score by the expected frequency. The results are weighted squared difference scores. 

The sixth step is to sum the weighted squared difference scores. The result is the Chi-Square value. 

Our result now needs to be evaluated based on the Chi-Square value we determined in the first step, from the table to be our cut off for rejecting the null hypothesis. 
 

When we looked up in a Table of Chi-Square values to determine what value is the cutoff for rejecting the null hypothesis, using the degrees of freedom of 2, and an alpha of .05, the Chi-Square value needs to be greater than 5.992. 
Our obtained Chi-Square value of 28.35 is greater than 5.992; so we reject the null hypothesis. 

REMINDER: 

The Null Hypothesis: There will be no difference in the number of votes for each candidate. 

Interpretation: The number of votes received by each candidate was not a factor of chance.