Answer:
Below see details
Explanation:
A) It is attached. Please see the picture
B) First to calculate the overall mean,
μ=65∗25/75+80∗25/75+95∗25/75
μ=65∗25/75+80∗25/75+95∗25/75 = 80
Next to calculate E(MSTR) = σ2+(1/r−1) ∑ni(μi−μ)^2 = 5634
And E(MSE) = σ^2= 9
C) Yes, it is substantially large than E(MSE) in this case.
D) If we sampled 25 employees from each group, we are likely to get a F statistics to indicate differences of job satisfactions among three types of length of service of employees.
Answer:
beam with a span length of 10 ft, a width of 8 in, and an effective depth of 20 in. Normal weight concrete is used for the beam. This beam carries a total factored load of 9.4 kips. The beam is reinforced with tensile steel, which continues uninterrupted into the support. The concrete has a strength of 4000 psi, and the yield strength of the steel is 60,000 psi. Using No. 3 bars and 60,000 psi steel for stirrups, do the followings: