Question

A company operates three machines during three shifts each day. From production records, the data in the table below were collected. At the .05 level of significance test to determine if the number of breakdowns is independent of the shift.

		Machine
Shift		A	B	C
1	Observed	46	11	13
	Expected	40.27933	14.07821	15.64246
2	Observed	37	10	11
	Expected	33.3743	11.6648	12.96089
3	Observed	20	15	16
	Expected	29.34637	10.25698	11.39665

• A. Yes, you can reject the claim that the number of breakdowns is independent of the shift because the p-value = 0.0423
• B. Yes, you can reject the claim that the number of breakdowns is independent of the shift because the p-value = 0.9577
• C. No, you cannot reject the claim that the number of breakdowns is independent of the shift because the p-value = 0.0423
• D. No, you cannot reject the claim that the number of breakdowns is independent of the shift because the p-value = 0.9577

Answer 1

Solution:

Given:

Level of significance = 0.05

		Machine
Shift		A	B	C
1	Observed	46	11	13
	Expected	40.27933	14.07821	15.64246
2	Observed	37	10	11
	Expected	33.3743	11.6648	12.96089
3	Observed	20	15	16
	Expected	29.34637	10.25698	11.39665

We have to test if   the number of breakdowns is independent of the shift or not.

Hypothesis :

H0:  the number of breakdowns is independent of the shift

Vs

H1:  the number of breakdowns is NOT independent of the shift.

Test statistic:

Formula for Chi square independence

Where

O_ij = Observed frequencies for i^th row and j^th column.

E_ij = Expected frequencies for i^th row and j^th column.

Thus we need to make following table:

Oij	Eij	Oij2/Eij
46	40.27933	52.5331
11	14.07821	8.5948
13	15.64246	10.8039
37	33.3743	41.0196
10	11.6648	8.5728
11	12.96089	9.3358
20	29.34637	13.6303
15	10.25698	21.9363
16	11.39665	22.4627
N = 179

Thus

P-value:

Use following Excel command:

=CHISQ.DIST.RT( X² , df )

where

df = ( R - 1) X (C - 1)

R = Number of Rows = 3

C = Number of Columns = 3

thus

df = ( 3-1) X ( 3-1)

df = 2 X 2

df = 4

thus

=CHISQ.DIST.RT(9.8894,4)

=0.0423

Thus P-value = 0.0423

Decision Rule:
Reject H0, if P-value < 0.05 level of significance, otherwise we fail to reject H0

Since P-value = 0.0423 < 0.05 level of significance, we reject H0.

That is we reject claim: the number of breakdowns is independent of the shift

Thus correct answer is:

A. Yes, you can reject the claim that the number of breakdowns is independent of the shift because the p-value = 0.0423