Question

The answers to the practice quiz are below I need to understand why the answers are correct.

Consider the One-Way ANOVA table (some values are intentionally left blank) for the pizza delivery times of 20 randomly assigned deliveries at four pizza franchises(A, B, C, D) (5 observations per franchise or Group).

H₀:MeanA=MeanB=MeanC=MeanD         H₁: At least one mean is different.

Please answer questions 1, 2, 3, and 4, based on the following table.

ANOVA
Source of Variation	SS	df	MS	F
Between Groups	420.75	-----	------	-----
Within Groups	-----------	-----	------
Total	461.57

1) What is the sum of squares within groups(franchises) value (SSW) or error sum of squares?

A) 28.34

B) 34.56

C) 38.75

D) 40.82

2) What are the ' Between groups' and 'Total' degrees of freedom?

A) 3, 19

B) 4, 19

C) 3, 20

D) 4, 20

3) What are the 'Within groups' degrees of freedom?

A) 15

B) 16

C) 17

D) 18

4) What is the Tukey-Kramer critical value at Alpha = 0.05?

A) 3.65

B) 4.05

C) 4.85

D) 5.15

Consider the following 2-way ANOVA table, wherein some of the values are left intentionally left blank. Row Factor (Location) has five levels, and Column factor (Variety) has four levels. Answer the following questions(5, 6 and 7) based on the 2- Way ANOVA table.

ANOVA
Source of Variation	SS	df	MS	FStat
Row factor(Location)	824.2	4	---	---
Column factor(Variety)	764.95	3	---	---
Interaction (Location X Variety)	43.80	--	---	---
Error	-------	--	---
Total	2010.00	99

5) How many treatment combinations are there in the experiment?

A) 12

B) 15

C) 20

D) 100

6) Based on the above ANOVA table, is Row factor(Location) significant alpha 0.05?

A) Yes, because F_Stat = 43.72> F_Crit = 2.5

B No, because F_Stat = 11.51 < F_Crit = 4.53

C) Yes, because F_Stat = 17.57 > F_Crit = 1.53

D) No, because F_Stat = 6.57 < F_Crit = 8.53

7) Based on the above ANOVA table, is Interaction between location and variety factors significant at Alpha = 0.05?

A) Yes, because F_Stat = 4.57 > F_Crit = 2.53

B) No, because F_Stat = 0.77 < F_Crit = 1.89

C) Yes, because F_Stat = 4.57 > F_Crit = 3.53

D) No, because F_Stat = 0.134 < F_Crit = 2.81

1. Answer D

2. Answer A

3. Answer B

4. Answer B

5. Answer C

6. Answer A

7. Answer B

2-Way ANOVA( Questions 5,6, and 7)

Row factor(Location)	824.2	4	206.05	43.72
Column factor(Variety)	764.95	3	254.9833	54.10
Interaction (Location X Variety)	43.8	12	3.65	0.77
Error	377.05	80	4.713125
Total	2010	99

Answer 1

One way ANOVA table:

Given SSB= 420.75

SST= 461.57

SSW= SST-SSB= 40.82

DFB=k-1=3

DFW= N-k= 20-4= 16

DFT= N-1=19

MSB=SSB/DFB= 140.25

MSW= SSW/DFW= 2.55125

F= MSB/MSW= 54.97305

1) Sum of squares within-: SSW= SST-SSB= 461.57- 420.75= 40.82

2) N= 20 and K=4 groups

The degreeof freedom between groups= k-1= 3

The degree of freedom for Total= N-1= 20-1 =19

3) The degree of freedom for Within groups= N-k= 20-4= 16

4) Tukey-kramer:

Using q table Critical value: