Question

(Montgomery & Runger, 2007; )

A company operates four machines three shifts each day. From production records, the following data on the number of breakdowns are collected:

Shift	Machines
	A	B	C	D
1	41	20	12	16
2	31	11	9	14
3	15	17	16	10

Test the hypothesis (using α = 0.05) using the Chi-Square Test that breakdowns are independent of the shift. Find the p-value for this test.

Answer 1

(i)

H0: Null Hypothesis: Breakdown are independent of the shift

HA: Alternative Hypothesis: Breakdown are dependent on shift

Observed Frequencies:

Shift                  Machines

                        A                B             C               D           Total

1                      41             20            12               16              89

2                      31           11               9                 14             65

3                      15            17             16                10              58

Total                  87           48              37                 40             212

Assuming H0, The Expected Frequencies are claculated as follows:

Shift               Machine

                         A               B              C                D                Total

1               87X89/212=36.52     20.15          15.53         16.79            89

2                    26.67                 14.72            11.34          12.26          65

3                    23.80                  13.13              10.12        10.94          58

Total                  87                    48                   37            40              212

O                    E                       (O - E)²?e

41                  36.52                       0.55

20                  20.15                       0.00

12                  15.53                      0.80

16                   16.79                      0.04

31                   26.67                     0.70

11                     14.72                   0.94

9                     11.34                   0.48

14                    12.26                  0.25

15                    23.80                  3.25

17                     13.13                  1.14

16                   10.12                     3.41

10                     10.94                     0.08

--------------------------------------------------------------

                             =     11.6491

= 0.05

ndf = (4 - 1) X (3 - 1) = 6

From Table, critical value of = 12.5916.

Since calculated value of is less than critical value of , Fail to reject H0.

Conclusion:

Breakdowns are indepencent of shift.

(ii)

By Technology, p-value = 0.0703.

Since p-value = 0.0703 is greater than , Fail to reject H0.

Conclusion:

Breakdowns are independent of the shift.

p-value for this test = 0.0703.