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.

Question

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  +   0.0003  +

                0.0317  +   0.0000   +   ***         = 0.2746

DF = ??, p-value = ???

 a) Carefully define the null and alternative hypotheses of the X ² test of the above table. 

b) Find the missing values ‘*’, ‘**’, ‘***’ ??, ??? 

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

Answer 1

a)Null

P_Accounting(A) = P_Sales(A) = P_Marketing(A)

P_Accounting(B) = P_Sales(B) = P_Marketing(B)

P_Accounting(C) = P_Sales(C) = P_Marketing(C) - the final statement is actually redundant, but it is usually included.

Alternate

P_Accounting(A) ≠ P_Sales(A) or P_Accounting(A) ≠ P_Marketing(A) or

P_Accounting(B) ≠ P_Sales(B) or P_Accounting(B) ≠ P_Marketing(B)

(the other conditions, P_Sales(A) ≠ P_Marketing(A) or P_Sales(B) ≠ P_Marketing(B) or P_Accounting(C) ≠ P_Sales(C) or P_Accounting(C) ≠ P_Marketing(C) or P_Sales(C) ≠ P_Marketing(C) are redundant, but they are usually added.

b) * (12.960)   +  (7.830)   +    (6.210)  = 27 Also, 12 + 8 + 7 = 27

** 29 * 38 / 100 = 11.02

*** (8-8.74)^2/8.74 = .0627

Alternatively, you can get this value by calculating .2746 - the sum of the other 8 values

DF (3-1)(3-1) = 2*2 = 4

p = chisq.dist.rt(.2746,4) = .9995

c) Nrmally, when a p-value is greater than alpha, say, 0.05, we fail to reject lack of independence.

These results are too good, though; this would imply that people are allocating grades based upon having a certain percentage in each bin for each job, so testing independence of grading by title is inappropriate.