ISS PREVIOUS YEAR 2016 PAPER-2 Q.NO.3 SOLUTION

ISS PREVIOUS YEAR 2016 PAPER-2 Q.NO.3 SOLUTION SET-A

Question 3:

A certain task can be accomplished in a factory by four different workers on five different types of machines. A sample study in context of a two-way design without repeated value is being made with two fold objectives of examining whether the four workers differ with respect to mean productivity and whether it is same for five machines. The researcher involved in this study reports gathered data as under :

Sum of squares for variance between machines = 35.2

Sum of squares for variance between workers = 53.8

Sum of squares for total variance = 174.2

The F-statistic between machines is equal to :

(a) 0.95(b) 1.24(c) 1.48(d) 1.75

Solution:

This is a two-way ANOVA without repetition problem.

Here, number of workers = 4number of machines = 5

We need to find the F-statistic for machines.

Step 1: Write the given sums of squares

Sum of squares due to machines, SS(machines) = 35.2

Sum of squares due to workers, SS(workers) = 53.8

Total sum of squares, SS(total) = 174.2

So, error sum of squares is:

SS(error) = SS(total) - SS(machines) - SS(workers)

SS(error) = 174.2 - 35.2 - 53.8

SS(error) = 85.2

Step 2: Degrees of freedom

For machines: df(machines) = c - 1 = 5 - 1 = 4

For workers: df(workers) = r - 1 = 4 - 1 = 3

For error: df(error) = (r - 1)(c - 1) = (4 - 1)(5 - 1) = 3 × 4 = 12

Step 3: Mean sum of squares

Mean square for machines:

MS(machines) = SS(machines) / df(machines)

MS(machines) = 35.2 / 4 = 8.8

Mean square for error:

MS(error) = SS(error) / df(error)

MS(error) = 85.2 / 12 = 7.1

Step 4: Calculate F-statistic for machines

F = MS(machines) / MS(error)

F = 8.8 / 7.1

F = 1.2394

So,

F ≈ 1.24

Final Answer:

The F-statistic between machines is

1.24

Correct option: (b)

ISS PREVIOUS YEAR 2016 PAPER-2 Q.NO.3 SOLUTION

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