Answer:
Since the calculated value of F= 5.733 falls in the critical region we reject the null hypothesis  and conclude all three means are not equal.
Explanation:
The given data is 
Banking          Retail            Insurance
12                      8                        10
10                     8                          8
10                     6                          6
12                     8                           8
<u>10                    10                          10</u>
 
The results of excel are:            	
<u><em>Anova: Single Factor  </em></u>    
      
SUMMARY      
Groups    Count	Sum	Average         Variance  
Column 1	5	54            10.8              1.2  
Column 2	5	40               8               2  
Column 3	5	42              8.4                 2.8  
          	
1) Let the null and alternate hypotheses be 
H0: u1=u2=u3  i.e all the three means are equal and
Ha: Not all three means are equal
2) The significance level is set at ∝ =0.05
3)The test statistic to use is 
F= sb²/ sw²
which has F distribution with v1= k-1 →3-1=2 and v2= n-k →15-3=12 degrees of freedom
After calculations the following table is obtained.
<u><em>ANOVA  </em></u>    
Source              SS            df         MS            F        P-value        F crit
of Variation
B/w Groups        22.93         2     11.467      5.733    0.01788        3.885
<u>Within Groups       24           12           2                                                       </u>      	
<u>Total                     46.93      14                                                                </u>
4) The critical region is F ≥ F(0.05, 2,12) = 3.885
5) Since the calculated value of F= 5.733 falls in the critical region we reject the null hypothesis  and conclude all three means are not equal.