Workers at a private business was given a grade (A, B, C) based on their yearly performance. The data was recorded along with the department they belong. The results obtained using MINITAB are shown below: Expected counts are printed below observed counts                              A                       B                   C               Total                          12                     8                     7 Accounting   (12.960)        (7.830)           (6.210)          *                            17                  10                    8   Sales           (16.800)        (10.150)        (8.050)          35                          19                  11                     8                 Marketing     (18.240)           (**)            (8.740)         38  Total                  48                  29              23              100   Chi-sq = 0.0711 +         0.0037 + 0.1005 +                 0.0024 +       0.0022 + 0.003  +                 0.0317 +       0.0000 +     ***       = 0.2746 Degrees of Freedom = ??,    p-value = ???   What is the conclusion of this test? Give reasons for your answer

Workers at a private business was given a grade (A, B, C) based on their yearly performance. The data was recorded along with the department they belong. The results obtained using MINITAB are shown below:

Expected counts are printed below observed counts 

 

                          A                       B                   C               Total

                         12                     8                     7

Accounting   (12.960)        (7.830)           (6.210)          *

                           17                  10                    8

  Sales           (16.800)        (10.150)        (8.050)          35

                         19                  11                     8                

Marketing     (18.240)           (**)            (8.740)         38

 Total                  48                  29              23              100

 

Chi-sq = 0.0711 +         0.0037 + 0.1005 +

                0.0024 +       0.0022 + 0.003  +

                0.0317 +       0.0000 +     ***       = 0.2746

Degrees of Freedom = ??,    p-value = ???

 

What is the conclusion of this test? Give reasons for your answer 

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