Question

Assume that the differences are normally distributed. Complete parts (a) through (d) below. 1 4 5 7 8 Observation X 2 49.1 3 44.6 6 50.8 46.0 49.8 48.8 46.7 50.0 Y 49.4 47.6 46.8 54.6 50.4 50.4 46.8 52.7 (a) Determine d =X-Y for each pair of data. 1 3 4 5 6 7 8 Observation 2 d; (Type integers or decimals.) (b) Computed and sd- (Round to three decimal places as needed.) Sda (Round to three decimal places as needed.) (c) Test if Md <0 at the a=0.05 level of significance. What are the correct null and alternative hypotheses? OB. Ho Hd = 0 H:Hd < 0 O A. Ho Hd > 0 H:Hd <0 OC. Ho Hd < 0 H:Hg > 0 OD. Ho: 400 H:Hd = 0 P-value = (Round to three decimal places as needed.) Choose the correct conclusion below. O A. Do not reject the null hypothesis. There is sufficient evidence that 10 <0 at the a = 0.05 level of significance. O B. Do not reject the null hypothesis. There is insufficient evidence that Hd < at the c = 0.05 level of significance. O C. Reject the null hypothesis. There is insufficient evidence that Hd < at the a=0.05 level of significance. OD. Reject the null hypothesis. There is sufficient evidence that Hd <0 at the a = 0.05 level of significance. (d) Compute a 95% confidence interval about the population mean difference Hd

Answer 1

Part a)

Number	Before	After	Difference
	46	49.4	-3.4
	49.1	47.6	1.5
	44.6	46.8	-2.2
	49.8	54.6	-4.8
	48.8	50.4	-1.6
	50.8	50.4	0.4
	46.7	46.8	-0.1
	50	52.7	-2.7
Total	385.8	398.7	-12.9

Part b)

d̅ = Σ di/n = -12.9 / 8 = -1.612

S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √(30.909 / 8-1) = 2.101

Part c)

To Test :-

H0 :- µd = 0

H1 :- µd < 0

Test Statistic :-
t = d̅ / ( S(d) / √(n) )
t = -1.6125 / ( 2.1013 / √(8) )
t = -2.1705

P - value = P ( t > 2.1705 ) = 0.0333

Reject null hypothesis if P value < α level of significance
P - value = 0.0333 < 0.05, hence we reject null hypothesis
Conclusion :- Reject null hypothesis

Option D is correct

Part d)

Confidence Interval :-
d̅ ± t(α/2, n-1) Sd / √(n)
t(α/2) = t(0.05 /2) = 1.895
-1.6125 ± t(0.05/2) * 2.1013/√(8)

Lower Limit = -3.369
Upper Limit = 0.144

95% Confidence interval is ( -3.369 , 0.144 )