In: Statistics and Probability
The following sample data reflect shipments received by a large firm from three different vendors and the quality of those shipments. (You may find it useful to reference the appropriate table: chi-square table or F table)
Vendor | Defective | Acceptable | ||
1 | 28 | 126 | ||
2 | 12 | 78 | ||
3 | 33 | 246 | ||
a. Select the competing hypotheses to determine whether quality is associated with the source of the shipments.
H0: Quality and source of shipment (vendor) are independent.; HA: Quality and source of shipment (vendor) are dependent.
H0: Quality and source of shipment (vendor) are dependent.; HA: Quality and source of shipment (vendor) are independent.
b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)
FIND THE TEST STATISTIC.
a)
H0: Quality and source of shipment (vendor) are independent.; HA: Quality and source of shipment (vendor) are dependent.
b)
Observed Frequencies | |||
Shipment Quality | |||
Vendor | Defective | Acceptable | Total |
1 | 28 | 126 | 154 |
2 | 12 | 78 | 90 |
3 | 33 | 246 | 279 |
Total | 73 | 450 | 523 |
Expected Frequencies | |||
Shipment Quality | |||
Vendor | Defective | Acceptable | Total |
1 | 21.49522 | 132.5048 | 154 |
2 | 12.56214 | 77.43786 | 90 |
3 | 38.94264 | 240.0574 | 279 |
Total | 73 | 450 | 523 |
fo-fe | |
6.50478 | -6.50478 |
-0.56214 | 0.562141 |
-5.94264 | 5.942639 |
(fo-fe)^2/fe | |
1.968445 | 0.319326 |
0.025155 | 0.004081 |
0.906845 | 0.14711 |
Chi square = sum(Oi-Ei)^2/Ei
Chi square = 3.371
df = (r-1)*(c-1) = (3-1)*(2-1) = 2*1 = 2
critical value, =CHISQ.INV.RT(0.05,2) = 5.99
Since chisq < critical value, i fail to reject Ho and conclude that Quality and source of shipment (vendor) are independent.