Part a)
Matched Pair
Part b)
Test Statistic :-
S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √(58.5 / 10-1) = 2.5495
d̅ = Σ di/n = -15 / 10 = -1.5
t = d̅ / ( S(d) / √(n) )
t = -1.5 / ( 2.5495 / √(10) )
t = -1.8605
part c)
Critical value t(α) = t(0.05) = -1.833
Part d)
Test Criteria :-
Reject null hypothesis if t < - t(α)
Critical value t(α) = t(0.05) = -1.833
t < - t(α) = -1.8605 < -1.833
Result :- Reject null hypothesis
Yes, there is sufficient evidene to support the claim
2)
Confidence Interval :-
d̅ ± t(α/2, n-1) Sd / √(n)
t(α/2) = t(0.05 /2) = 1.8331
-1.5 ± t(0.05/2) * 2.5495/√(10)
Lower Limit = -1.5 - t(0.05/2) 2.5495/√(10)
Lower Limit = -3.3241
Upper Limit = -1.5 + t(0.05/2}) 2.5495/√(10)
Upper Limit = 0.3241
95% Confidence interval is ( -3.3241 , 0.3241
)
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