In the book Essentials of Marketing Research, William R. Dillon, Thomas J. Madde
ID: 2934353 • Letter: I
Question
In the book Essentials of Marketing Research, William R. Dillon, Thomas J. Madden, and Neil A. Firtle discuss the relationship between delivery time and computer- assisted ordering. A sample of 40 firms shows that 16 use computer-assisted ordering, while 24 do not. Furthermore, past data are used to categorize each firm's delivery times as below the industry average, equal to the industry average, or above the industry average. The results obtained are given in the table below A Contingency Table Relating Delivery Time and Computer-Assisted Ordering Delivery Time Equal to Industry Average 12 4 16 Computer Assisted Ordering Below Industry Average Above Industry Average Row Total 24 16 40 0 es 10 2 10 Column total (a) Test the hypothesis that delivery time performance is independent of whether computer-assisted ordering is used. What do you conclude by setting -05? Reject # Ho: independence (b) Verify that a chi-square test is appropriate the average of the E,j is greater The test is valid because the number of cells exceeds than and the smallest E is greater than (c) Is there a difference between delivery-time performance between firms using computer-assisted ordering and those not using computer-assisted ordering? No YesExplanation / Answer
Back-up Theory
Suppose we have a contingency table consisting of r rows and c columns, i.e., (r x c) cells, each row representing a particular level of one attribute and each column representing a particular level of another attribute. Let oij be the observed frequency in the cell at ith row-jth column and eij be the corresponding expected frequency.
H0: The two attributes are independent. vs H1: H0 is false.
Test Statistic
2 = (i = 1 to r, j = 1 to c){(oij - eij)2/eij}. [This can also be simplified as 2 = (i = 1 to r, j = 1 to c)(o2ij/eij) – n, where n = total frequency = (i = 1 to r, j = 1 to c)(oij) which is also equal to (i = 1 to r, j = 1 to c)(eij).]
Under H0, 2 is distributed as Chi-square distribution with degrees of freedom = (r - 1)(c - 1).
Under H0, eij = (oi. . o.j)/n where oi. = (j = 1 to c)(oij) = sum of all observed frequencies in the ith row and o.j = (i = 1 to r)(oij) = sum of all observed frequencies in the jth column.
Decision Criterion
Reject H0 if 2cal > 2(r - 1)(c - 1), . i.e. reject H0 if 2cal > 2tab (i.e., calculated value of 2 is greater than the upper % point of Chi-square distribution with degrees of freedom = (r - 1)(c - 1), which can be read off from standard table).
Now, to work out the solution,
In the given problem, the two attributes are:
r = 2, c = 3. Hence, degrees of freedom = 1 x 2 = 2.
From the calculations shown below,
Since 2cal (8.929) > 2tab(5.99), H0 is rejected implying that the two attributes in the given problem are not independent.
Calculations
CONTINGENCY
(2 x 3)
r =
2
c =
3
Oij
DF =
2
1
2
3
Oi.
1
4
12
8
24
2
10
4
2
16
O.j
14
16
10
40
Eij
1
2
3
Total
1
8.4
9.6
6
24
2
5.6
6.4
4
16
Total
14
16
10
40
Chi-square
1
2
3
Total
1
2.304762
0.6
0.67
3.571429
2
3.457143
0.9
1
5.357143
Total
5.761905
1.5
1.67
0
8.928571
The answers to the specific question parts given:
Part (a): Reject H0 of independence.
Part (b): All conditions of chi-square test are valid.
Part (c): yes [since hypothesis of independence is rejected, delivery time does depend on the type of ordering.]
DONE
CONTINGENCY
(2 x 3)
r =
2
c =
3
Oij
DF =
2
1
2
3
Oi.
1
4
12
8
24
2
10
4
2
16
O.j
14
16
10
40
Eij
1
2
3
Total
1
8.4
9.6
6
24
2
5.6
6.4
4
16
Total
14
16
10
40
Chi-square
1
2
3
Total
1
2.304762
0.6
0.67
3.571429
2
3.457143
0.9
1
5.357143
Total
5.761905
1.5
1.67
0
8.928571
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