Question

The following are the weekly losses of work-hours due to accidents in 10 industrial plants before and after a certain safety program were put into operation

Before	45	73	46	124	33	57	83	34	26	17
After	36	60	44	119	35	51	77	29	24	11

a- Is the assumption that the difference in weekly losses of work-hours is normally distributed reasonable?

b- Test whether the safety program is effective.

Use α=0.05. c- Find a 95% confidence interval on the difference in mean weekly losses of work-hours.

Answer 1

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	45	36	9
	73	60	13
	46	44	2
	124	119	5
	33	35	-2
	57	51	6
	83	77	6
	34	29	5
	26	24	2
	17	11	6
Average	53.8	48.6	5.2
St. Dev.	32.058	31.031	4.077
n	10	10	10

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

This corresponds to a right-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = 9

Hence, it is found that the critical value for this right-tailed test is t_c = 1.833 , for α=0.05 and df = 9

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

(4) Decision about the null hypothesis

Since it is observed that t = 4.033 >tc​=1.833, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0015 , and since p = 0.0015 <0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is greater than μ2​, at the 0.05 significance level.

Confidence Interval

The number of degrees of freedom are df = 10 - 1 = 9 , and the significance level is α=0.05.

Based on the provided information, the critical t-value for α=0.05 and df = 9 degrees of freedom is t_c = 2.262

The 95% confidence for the population mean μ is computed using the following expression

Therefore, based on the information provided, the 95 % confidence for the population mean μ is

CI = (5.2 - 2.917, 5.2 + 2.917)

CI =(2.283,8.117)