Typical methods of recruiting new cashiers and clerks are newspaper advertisemen
ID: 3177931 • Letter: T
Question
Typical methods of recruiting new cashiers and clerks are newspaper advertisements, notices in the store, and employee referrals. A recent study of new employees at one grocery chain had supervisors classify new recruits as "successful" and "unsuccessful". A summary of the number of "successful" and "unsuccessful" recruits by source is given below. Note that all 542 employees were screened through interviews and subsequently hired. The "successful/unsuccessful" assessment was made after a 3-month probationary period. a) Is there evidence to suggest that some methods of recruitment are better than others? Use the classical approach at a significance level of 5%. b) Is there evidence to suggest there is a difference in the proportion of "successful" and "unsuccessful" recruits hired through "In store notice". Use the p-value approach at a significance level of 1%.Explanation / Answer
Solution
Back-up Theory
We have a contingency table consisting of r = 3 rows and c = 2 columns, i.e., (r x c) cells, each row representing a particular method of recruitment and each column representing the success (i.e., successful or unsuccessful) of the recruited employees.
Let oij be the observed frequency in the cell at ith row-jth column and eij be the corresponding expected frequency.
To test if some method(s) is/are better than other method(s) of recruitment, we form:
H0: Success (i.e., successful or unsuccessful) of the recruited employees is independent of the method of recruitment. 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 or can be obtained using Excel Function.
Now, to work out the solution,
Part (a)
In the given problem,
r = 3, c = 2. Hence, degrees of freedom = 2 x 1 = 2. = 5%
From the calculations shown below,
2cal = 2.774, 2tab = 5.99,
Since 2cal < 2tab, H0 is accepted.
=> There is not enough evidence to suggest that any one method is better than the others.
Calculations
Oij (given)
Oi1
O12
Oi.
O1i
54
36
90
O2j
112
90
202
O3j
126
124
250
O.i
292
250
542
Eij
Ei1
Ei2
Ei.
E1j
48.48708
41.51292
90
E2j
108.8266
93.17343
202
E3j
134.6863
115.3137
250
E.i
292
250
542
chi-square
1
2
Total
1
0.626811
0.732115
1.358926
2
0.092539
0.108085
0.200624
3
0.56021
0.654325
1.214535
Total
1.279559
1.494525
2.774085
DONE
Part (b)
Let P1 and P2 be the respective proportion of ‘successful’ employees and ‘unsuccessful’ employees for ‘in-store notice’ method. Then we want to test if P1 = P2 or are they different.
H0: P1 = P2 = ½ Vs HA: P1 P2
Under H0, expected number ‘successful’ employees and ‘unsuccessful’ employees, both would be 202 x ½ = 101.
Following the very steps as in Part (a). chi-square = 1.476
Under H0, chi-square has 1 degree of freedom.
Using Excel Function, p-value for 1,476 under 1 degree of freedom is 0.224 which is much more than the given level of significance of 0.01. Hence H0 is accepted
=> There is no evidence to suggest that the proportion of ‘successful’ employees and ‘unsuccessful’ employees are different.
DONE
Oij (given)
Oi1
O12
Oi.
O1i
54
36
90
O2j
112
90
202
O3j
126
124
250
O.i
292
250
542
Eij
Ei1
Ei2
Ei.
E1j
48.48708
41.51292
90
E2j
108.8266
93.17343
202
E3j
134.6863
115.3137
250
E.i
292
250
542
chi-square
1
2
Total
1
0.626811
0.732115
1.358926
2
0.092539
0.108085
0.200624
3
0.56021
0.654325
1.214535
Total
1.279559
1.494525
2.774085
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