In: Statistics and Probability
On Saturday morning, Jennifer Bentford received a call at her home from the production supervisor at Bentford Electronics Plant1. The supervisor indicated that she and the supervisors from Plants 2, 3, and 4 had agreed that something must be done to improve company morale and thereby increase the production output of their plants. Jennifer, president of Bentford Electronics, agreed to set up a Monday morning meeting with the supervisors to see if they could arrive at a plan for accomplishing these objectives.
By Monday, each supervisor had compiled a list of several ideas, including a four-day work week and interplant competitions of various kinds. A second meeting was set for Wednesday to discuss the issue further.
Following the Wednesday afternoon meeting, Jennifer and her plant supervisors agreed to implement a weekly contest called the NBE Game of the Week winner and would receive 10 points. The second-place plant would receive 7 points, and the third- and fourth-place plants would receive 3 points and 1 point, respectively. The contest would last 26 weeks. At the end of that period, a $200,000 bonus would be divided among the employees in the four plants proportional to the total points accumulated by each plant.
The announcement of the contest created a lot of excitement and enthusiasm at the four plants. No one complained about the rules because the four plants were designed and staffed to produce equally.
At the close of the contest, Jennifer called the supervisors into a meeting, at which time she asked for data to determine whether the contest had significantly improved productivity. She indicated that she had to know this before she could authorize a second contest. The supervisors, expecting this request, had put together the following data:
Units Produced (4 Plants Combined) |
Before-Contest Frequency |
During-Contest Frequency |
0 – 2,500 |
11 |
0 |
2,501 – 8,000 |
23 |
20 |
8,001 – 15,000 |
56 |
83 |
15,001 – 20,000 |
15 |
52 |
105 days |
155 days |
Jennifer examined the data and indicated that the contest looked to be a success, but she wanted to base her decision to continue the contest on more than just an observation of the data. “Surely there must be some way to statistically test the worth of this contest,” Jennifer stated. “I have to see the results before I will authorize the second contest.”
Jennifer examined the data and indicated that the contest looked to be a success, but she wanted to base her decision to continue the contest on more than just an observation of the data. “Surely there must be some way to statistically test the worth of this contest,” Jennifer stated. “I have to see the results before I will authorize the second contest.”
Just need an explanation on how to solve this.I think it's a chi-square contingency. If possible, use stat crunch.
Expected value = (Row Total*Column Total)/Overall Total
The calculations are:
Col 1 | Col 2 | Total | ||
Row 1 | Observed | 11 | 0 | 11 |
Expected | 4.44 | 6.56 | 11.00 | |
O - E | 6.56 | -6.56 | 0.00 | |
(O - E)² / E | 9.68 | 6.56 | 16.24 | |
Row 2 | Observed | 23 | 20 | 43 |
Expected | 17.37 | 25.63 | 43.00 | |
O - E | 5.63 | -5.63 | 0.00 | |
(O - E)² / E | 1.83 | 1.24 | 3.07 | |
Row 3 | Observed | 56 | 83 | 139 |
Expected | 56.13 | 82.87 | 139.00 | |
O - E | -0.13 | 0.13 | 0.00 | |
(O - E)² / E | 0.00 | 0.00 | 0.00 | |
Row 4 | Observed | 15 | 52 | 67 |
Expected | 27.06 | 39.94 | 67.00 | |
O - E | -12.06 | 12.06 | 0.00 | |
(O - E)² / E | 5.37 | 3.64 | 9.01 | |
Total | Observed | 105 | 155 | 260 |
Expected | 105.00 | 155.00 | 260.00 | |
O - E | 0.00 | 0.00 | 0.00 | |
(O - E)² / E | 16.88 | 11.44 | 28.32 | |
28.32 | chi-square | |||
3 | df | |||
3.11E-06 | p-value |