Question

		Table 2: Number of items Suspended
		Rocks	Kittens	Daggers	Wookies
	Luke	25	15	30	3
	Rey	30	5	45	2
	Anikin	25	25	40	5

Examine the data in Table 2.  It shows the number of different types of objects that each
Jedi trainee can suspend in the air after 12 weeks of training.
Do an ANOVA analysis - Randomized Block Design at alpha = 0.05.
Put the output in the indicated area.
Is there a difference in the 3 kids overall ability to suspend objects in the air (rows)?
Is there a difference in the type of object that is suspend objects in the air (columns)?
Put the p-factor and coclusion in the indicated space.
Based upon the data in Table 2, perform a Chi-square Test Of Independence to determine
if the type of suspended objects is dependent upon the  Jedi trainee doing the suspending.

Answer 1

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
Luke	4	73	18.25	142.25
Rey	4	82	20.5	424.3333
Anikin	4	95	23.75	206.25
Rocks	3	80	26.66667	8.333333
Kittens	3	45	15	100
Daggers	3	115	38.33333	58.33333
Wookies	3	10	3.333333	2.333333
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows	61.16667	2	30.58333	0.662854	0.549421	5.143253
Columns	2041.667	3	680.5556	14.75015	0.003553	4.757063
Error	276.8333	6	46.13889
Total	2379.667	11

p value for rows : 0.5494

p value > 0.05, do not reject Ho

so,  there is not a difference in the 3 kids overall ability to suspend objects in the air (rows)

.................

p value for column: 0.0035

p value<0.05, reject Ho

so, there is a difference in the type of object that is suspend objects in the air (columns)

..........................

Chi-Square Test of independence
	Observed Frequencies
	0
0	Rocks	Kittens	Daggers	Wookies			Total
Luke	25	15	30	3			73
Rey	30	5	45	2			82
Anikin	25	25	40	5			95
Total	80	45	115	10			250
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	Rocks	Kittens	Daggers	Wookies			Total
Luke	80*73/250=23.36	45*73/250=13.14	115*73/250=33.58	10*73/250=2.92			73
Rey	80*82/250=26.24	45*82/250=14.76	115*82/250=37.72	10*82/250=3.28			82
Anikin	80*95/250=30.4	45*95/250=17.1	115*95/250=43.7	10*95/250=3.8			95
Total	80	45	115	10			250
	(fo-fe)^2/fe
Luke	0.115	0.263	0.382	0.002
Rey	0.539	6.454	1.405	0.500
Anikin	0.9592	3.6497	0.3133	0.3789

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   14.961  
      
Level of Significance =   0.05  
Number of Rows =   3  
Number of Columns =   4  
Degrees of Freedom=(#row - 1)(#column -1) = (3- 1 ) * ( 4- 1 ) =   6  
      
p-Value =   0.0206   [Excel function: =CHISQ.DIST.RT(χ²,df) ]
Decision:    p-value < α , Reject Ho  

type of suspended objects is dependent upon the  Jedi trainee doing the suspending.

.........................

Please revert back in case of any doubt.

Please upvote. Thanks in advance.