Question

Assume that the differences are normally distributed. Complete parts ​(a) through ​(d) below. Observation 1 2 3 4 5 6 7 8 Upper X Subscript i 48.3 48.3 49.7 44.0 51.9 53.4 45.6 48.8 Upper Y Subscript i 48.9 49.5 51.4 49.1 52.2 54.7 49.7 48.9 ​(b) Compute d overbar and s Subscript d. d overbarequals negative 1.800 ​(Round to three decimal places as​ needed.) s Subscript dequals nothing ​(Round to three decimal places as​ needed.) c) Test if mu Subscript dless than0 at the alphaequals0.05 level of significance. What are the correct null and alternative​ hypotheses? ​P-value=

Answer 1

x	y	d=x-y
48.3	48.9	-0.6
48.3	49.5	-1.2
49.7	51.4	-1.7
44	49.1	-5.1
51.9	52.2	-0.3
53.4	54.7	-1.3
45.6	49.7	-4.1
48.8	48.9	-0.1
	Total	-14.4

b) = - 1.800

= 1.827567

c) The null and alternative hypothesis is

Level of significance = 0.05

Test statistic is

n = 8

Degrees of freedom = n - 1 = 8 - 1 = 7

P-value = P(T < - 2.79) = 0.0135

P-value < 0.05 we reject null hypothesis.

Conclusion: Yes, mu Subscript d less than 0 at the alphaequals0.05 level of significance.