In: Statistics and Probability
A local company is concerned about the number of days missed by
its employees due to illness. A random sample of 10 employees is
selected. The number of days absent in one year is listed below. An
incentive program is offered in an attempt to decrease the number
of days absent. The number of days absent in one year after the
incentive program is listed below. Test the claim that the
incentive program cuts down on the number of days missed by
employees. Use΅= 0.05. Assume that the distribution is normally
distributed.
Employee | A | B | C | D | E | F | G | H | I | J |
Days Absent Before Incentive |
3 | 8 | 7 | 2 | 9 | 4 | 2 | 0 | 7 | 5 |
Days absent after incentive | 1 | 7 | 7 | 0 | 8 | 2 | 0 | 1 | 5 | 5 |
A) Hypothises :
B) Critical value: d, sd, tcritical,
C) Test statistic, an explanation on the decision to reject or fail to reject Ho
D) The conclusion about the decision.
Let
is the population mean number of days absent before the
incentive
Let
the population mean number of days absent after the incentive.
Hypothesis:
H0:
H1:
(right tailed test)
b)
degrees of freedom = n-1 = 10 -1 = 9
alpha = 0.05
critical value = t0.05,9 = 1.833
Employee | Before | After | Before - After |
A | 3 | 1 | 2 |
B | 8 | 7 | 1 |
C | 7 | 7 | 0 |
D | 2 | 0 | 2 |
E | 9 | 8 | 1 |
F | 4 | 2 | 2 |
G | 2 | 0 | 2 |
H | 0 | 1 | -1 |
I | 7 | 5 | 2 |
J | 5 | 5 | 0 |
Sum | 11 | ||
Average ![]() |
1.1 | ||
std dev Sd | 1.10 |
C) test statistic t =
= 1.1/ (1.1/sqrt(10)) = 3.161
Since test statistic t > critical value hence We Reject H0.
D) Since test is rejected there is a sufficient evidence to conclude that the the incentive program cuts down on the number of days missed by employees.