..oo T-Mobile LTE 6:12 AM Done AT&T; Expert O&A; the biling department of a nati
ID: 3296910 • Letter: #
Question
..oo T-Mobile LTE 6:12 AM Done AT&T; Expert O&A; the biling department of a national cable service company is conducting a survey to see how astomers pay their monthly bh the able company accepts ments of se by mail, by credit card, or electronic funds transfer (EFT) from a bank account. The cable company randomly sampied 200 of Its customers to determine if there is a relationship between a customer's age and the payment method used. The sample results are shown below Payment method Age of customers 50 25 25 Under 30 30-49 S0and over Mail Credit card EFT 20 10 20 15Explanation / Answer
Solution
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: customer’s age and method of payment.
r = 3, c = 3. Hence, degrees of freedom = 2 x 2 = 4.
From the calculations shown below, 2cal = 12
Using Excel Function for chi-square, 2tab, 0.05 = 9.49
Since 2cal > 2tab, H0 is rejected implying that the two attributes in the given problem are not not independent.
Conclusion
There is evidence to suggest that method of payment is dependent on the age of the customer.
DONE
Calculations
CONTINGENCY
(3 x 3)
r =
3
c =
3
Oij
DF =
4
1
2
3
Oi.
1
20
50
30
100
2
10
25
15
50
3
20
25
5
50
O.j
50
100
50
200
Eij
1
2
3
Total
1
25
50
25
100
2
12.5
25
12.5
50
3
12.5
25
12.5
50
Total
50
100
50
200
Chi-square
1
2
3
Total
1
1
0
1
2
2
0.5
0
0.5
1
3
4.5
0
4.5
9
Total
6
0
6
12
CONTINGENCY
(3 x 3)
r =
3
c =
3
Oij
DF =
4
1
2
3
Oi.
1
20
50
30
100
2
10
25
15
50
3
20
25
5
50
O.j
50
100
50
200
Eij
1
2
3
Total
1
25
50
25
100
2
12.5
25
12.5
50
3
12.5
25
12.5
50
Total
50
100
50
200
Chi-square
1
2
3
Total
1
1
0
1
2
2
0.5
0
0.5
1
3
4.5
0
4.5
9
Total
6
0
6
12
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