Question

The manager of a computer software company wishes to study the number of hours senior executives by type of industry spend at their desktop computers. The manager selected a sample of five executives from each of three industries. At the .05 significance level, can she conclude there is a difference in the mean number of hours spent per week by industry?
Banking Retail Insurance
12 8 10
10 8 8
10 6 6
12 8 8
10 10 10

Answer 1

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

10                    10                          10

The results of excel are:            

Anova: Single Factor      

     

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.

ANOVA      

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

Within Groups       24           12           2                                                             

Total                     46.93      14                                                                

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.