Answer:
a. At the 0.05 level of significance, there is evidence that the mean life is different from 6,500 hours.
b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .
The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.
c. CI [6583.336 ,6816.336]
d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.
Step-by-step explanation:
Population mean = u= 6500 hours.
Population standard deviation = σ=500 hours.
Sample size =n= 50
Sample mean =x`= 6,700 hours
Sample standard deviation=s= 600 hours.
Critical values, where P(Z > Z) =∝ and P(t >) =∝
Z(0.10)=1.282
Z(0.05)=1.645
Z(0.025)=1.960
t(0.01)(49)= 1.299
t(0.05)= 1.677
t(0.025,49)=2.010
Let the null and alternate hypotheses be
H0: u = 6500 against the claim Ha: u ≠ 6500
Applying Z test
Z= x`- u/ s/√n
z= 6700-6500/500/√50
Z= 200/70.7113
z= 2.82=2.82
Applying t test
t= x`- u /s/√n
t= 6700-6500/600/√50
t= 2.82
a. At the 0.05 level of significance, there is evidence that the mean life is different from 6,500 hours.
Yes we reject H0 for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645
For t test we reject H0 as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.
b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .
The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.
c. The 95 % confidence interval of the population mean life is estimated by
x` ± z∝/2 (σ/√n )
6700± 1.645 (500/√50)
6700±116.336
6583.336 ,6816.336
d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.
