Question

12.26 Constructing an ANOVA table Refer to Exercise 12.5 (page 655). Using the table of group means and standard deviations, construct an ANOVA table similar to that on page 662. Based on the F statistic and degrees of freedom, compute the P-value. What do you conclude?

This is the question that is referred to...

Joystick types. Example 12.1 (page 645) describes a study designed to compare different joystick types on the time it takes to complete a navigation mission. In Exercise 12.3 (page 653), you described the ANOVA model for this study. The three joystick types are designated 1, 2, and 3. The following table summarizes the time (seconds) data.

Joystick	x̅	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Answer 1

Here we will perform an ANOVA analysis first with these 3 differet type of Joysticks catalyst.

Here hypothesis are

H₀: All Joysticks are having same mean time it takes to complete a navigation mission

H_a : Alt least one Joystick is having different meantime it takes to complete a navigation mission

Now we will find mean and variance of the given Catalysts

Joystick	x̅	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Here Degree of freedom Between groups = 3-1 = 2

Degree of freedom within groups = (20-1) + (20-1) + (20-1) = 57

First we will calculate Grand mean x_GM = [20 * 279 + 20 * 245 + 20 * 258] / 60= 260.6667

SS(Between Groups) =

= 20 * (279 - 260.6667)² + 20 * (245 - 260.6667)² + 20 * (258 - 260.6667)²

= 11773.33

SS(Within Groups) =

SS(W) = (20-1) * 78² + (20-1) * 68² + (20-1) * 80² = 325052

Now MSB = SSB/dF =11773.33/2 = 5886.67

MSW = SSW/dF = 325052/57 = 5702.67

Here F = 5886.67/5702.67 = 1.0323

Pr(F > 1.0323) = FDIST(1.0323; 2; 57) = 0.3628

F_critical= F_0.05,2,57 = 3.1588

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	11773.3	2	5886.67	1.03226531	0.3628	3.1588
Within Groups	325052	57	5702.67
Total	120.238	59