Question

uestion #2: 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 α 0.025. (Assume that the distribution is normally distributed but be sure to verify the remaining requirements for the test.) Employee Days absent before 3 8 7 2 9 4 2 07 5 ncentive Davs absent after1 7 7 l0 8 2 0 1 5 5 ncentive A B C D E F G H I J H mean number of days absent before incentive y mean number of days absent after incentive To test, Ho : μd Given, x- sample mean (before incentive)-4.7, y-3.6 Now, d-n-y-1.1 =1.1005 n=10 Test statistic, T-

Please do not handwrite answer unless it is extremely neat. I posted this question before already but could not read the writing.

Answer 1

We are to test, $H_0: \mu_d = 0$ versus $H_1: \mu_d > 0$ .

Now, we are given, $\bar{x} =$ sample mean (before incentive) = 4.7 and $\bar{y} =$ sample mean (after incentive) = 3.6.

Now, $\bar{d} = \bar{x} - \bar{y}$ = difference in sample means = 4.7 - 3.6 = 1.1.
Also, from the sample data, we get, n = sample size = 10 and $s_d =$ sample s.d. of the differences = 1.1005.

Now, the test statistic is $T = \frac{\sqrt{n} \bar{d}}{s_d} \sim t_{n-1} = t_9$ , under $H_0$.

We get T = 3.1608

Now, p-value = $P(t_9 > 3.1608) = 0.0058$ . (we calculate this value using MS Excel.)

Since the p-value is less than $\alpha = 0.025$, we reject $H_0$.

We conclude that there is enough evidence to suggest that the incentive program cuts down on the number of days missed by employees.