A company, studying the relation between job satisfaction and length of service of employees, classified employees into three le
ngth-of-service groups (less than 5 years, 5-10 years, more than 10 years). Suppose µ1 = 65, µ2 = 80, µ3 = 95, and σ= 3, and that ANOVA model Yij=μi+ϵij is applicable.
a. Draw a representation of this model
b. Find E(MSTR) and E(MSE) if 25 employees from each group are selected at random for intensive interviewing about job satisfaction.
c. Is E(MSTR) substantially large than E(MSE) here?
a. Draw a representation of this model
b. Find E(MSTR) and E(MSE) if 25 employees from each group are selected at random for intensive interviewing about job satisfaction.
c. Is E(MSTR) substantially large than E(MSE) here?
1 answer:
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.
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Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ