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 hoursSample 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)=2.010Let the null and alternate hypotheses be H0: u = 6500 against the claim Ha: u ≠ 6500Applying Z test Z= x`- u/ s/√nz= 6700-6500/500/√50Z= 200/70.7113z= 2.82=2.82Applying t testt= x`- u /s/√n t= 6700-6500/600/√50t= 2.82a. 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∝=1.645For 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±116.3366583.336 ,6816.336d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range....
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