Question

Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both an unabraded condition and an abraded condition. Use the paired t test to test H0: μD = 0 versus Ha: μD > 0 at significance level 0.01. (Use μD = μU − μA.) Fabric 1 2 3 4 5 6 7 8 U 36.1 55.0 51.4 38.6 43.2 48.8 25.6 49.5 A 28.5 20.0 46.0 34.5 36.5 52.5 26.5 46.5 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. Reject H0. The data suggests a significant mean difference in breaking load for the two fabric load conditions. Reject H0. The data does not suggest a significant mean difference in breaking load for the two fabric load conditions. Fail to reject H0. The data does not suggest a significant mean difference in breaking load for the two fabric load conditions. Fail to reject H0. The data suggests a significant mean difference in breaking load for the two fabric load conditions.

Answer 1

Number	Before	After	Difference	di-dbar
	36.1	28.5	7.6	0.2025
	55	20	35	775.6225
	51.4	46	5.4	3.0625
	38.6	34.5	4.1	9.3025
	43.2	36.5	6.7	0.2025
	48.8	52.5	-3.7	117.7225
	25.6	26.5	-0.9	64.8025
	49.5	46.5	3	17.2225
Total	348.2	291	57.2	988.14

Test Statistic :-





t = 1.70

Test Criteria :-
Reject null hypothesis if


Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 1.70 ) = 0.0663
Reject null hypothesis if P value < level of significance
P - value = 0.0663 > 0.01, hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

Fail to reject H0. The data does not suggest a significant mean difference in breaking load for the two fabric load conditions.