Answer:
- 179.981
Step-by-step explanation:
The hypothesis :
H0 : μL - μP ≥ 25
H0 : μL - μP < 25
The sample mean difference ;Xd
d = L - P
Xd = Σd/n
d = -198,-336,-70,-18,-122,-9,-50,-5,-163,-86
Xd = - 1057 / 10
Xd = - 105.7
Using calculator ;
Standard deviation of difference, Sd = 103.845
The test statistic :
T = Xd ÷ (Sd/√n)
T = -105.7 ÷ (103.845/√10)
T = - 3.219
Decision region :
Reject H0 ; If Pvalue < α
The Pvalue : df = n - 1 ; 10 - 1 = 9
Pvalue(-3.219, 9) ; two-tailed = 0.00525
Hence, reject H0
B.) The confidence interval for difference in mean :
Xd ± Tcritical[Sd/√n]
Tcritical at 95%, df = 9
Tcritical = 2.262
C.I = -105.7 ± 2.262[103.845/√10]
C.I = -105.7 ± 74.281076
Lower boundary: - 105.7 - 74.281076 = - 179.9810