In: Statistics and Probability
A wholesaler has recently developed a computerized sales
invoicing system. Prior to implementing this system, a manual
system was used. The distribution of the number of errors per
invoice for the manual system is as follows:
Errors per Invoice | 0 | 1 | 2 | 3 | More Than 3 |
Percentage of Invoices | 85% | 9% | 4% | 1% | 1% |
After implementation of the computerized system, a random sample of
500 invoices gives the following error distribution:
Errors per Invoice | 0 | 1 | 2 | 3 | More Than 3 |
Number of Invoices | 472 | 11 | 9 | 5 | 3 |
pi | Ei | fi | (f – E) ^ 2/E | ||
0.85 | 425 | 472 | 5.1976 | ||
0.09 | 45 | 11 | 25.6889 | ||
0.04 | 20 | 9 | 6.0500 | ||
0.01 | 5 | 5 | .0000 | ||
0.01 | 5 | 3 | .8000 | ||
Chi-Square | 37.73650 | p-value 0.0000001096 | |||
(a) Show that it is appropriate to carry out a
chi-square test using these data.
Each Ei ≥
(b) Use the Excel output shown above to determine whether the error percentages for the computerized system differ from those for the manual system at the .05 level of significance. What do you conclude?
H0: Conclude systems .
Solution:
Given: The distribution of the number of errors per invoice for the manual system is as follows:
Errors per Invoice | 0 | 1 | 2 | 3 | More Than 3 |
Percentage of Invoices | 85% | 9% | 4% | 1% | 1% |
After implementation of the computerized system, a random sample of 500 invoices gives the following error distribution:
Errors per Invoice | 0 | 1 | 2 | 3 | More Than 3 |
Number of Invoices | 472 | 11 | 9 | 5 | 3 |
We are given following output for Chi-square test of goodness of fit:
pi | Ei | fi | (f – E) ^ 2/E |
0.85 | 425 | 472 | 5.1976 |
0.09 | 45 | 11 | 25.6889 |
0.04 | 20 | 9 | 6.05 |
0.01 | 5 | 5 | 0 |
0.01 | 5 | 3 | 0.8 |
Chi-Square | 37.7365 |
and p-value = 0.0000001096
Part a) Show that it is appropriate to carry out a chi-square test using these data.
Yes it is appropriate to carry out a chi-square test using these data, since each
Part b) Use the Excel output shown above to determine whether the error percentages for the computerized system differ from those for the manual system at the .05 level of significance.
What do you conclude?
H0: the error percentages for the computerized system does not differ from those for the manual system
Vs
H1: the error percentages for the computerized system differ from those for the manual system
level of significance = 0.05 and p-value = 0.0000001096.
Decision rule: Reject H0, if p-value < 0.05 level of significance, otherwise we fail to reject H0.
Since p-value = 0.0000001096< 0.05 level of significance, we reject H0.
Thus we conclude that: the error percentages for the computerized system differ from those for the manual system.