Question

In: Statistics and Probability

Conduct an appropriate Analysis of Variance for both sets of data below. Use an alpha level...

Conduct an appropriate Analysis of Variance for both sets of data below. Use an alpha level of 0.05 for all hypothesis tests and confidence intervals.

Please state the following:

The problem statement

Your approach and assumptions

The ANOVA output (consider using Excel's Data Analysis/Analysis ToolPak)

The null and alternate hypothesis statements

The result of the test and its interpretation

Post hoc analysis if appropriate.

District A	District B	District C	District D
73	85	97	61
63	59	86	67
89	84	76	84
75	70	78	67
70	80	76	69

&

	Agent A	Agent B	Agent C
Home 1	210	218	226
Home 2	192	190	198
Home 3	183	187	185
Home 4	227	223	237
Home 5	242	240	237

Expert Solution

1)

District A	District B	District C	District D
73	85	97	61
63	59	86	67
89	84	76	84
75	70	78	67
70	80	76	69

H0:all means are equal

Ha:not all means are equal

excel ouptut-

Anova: Single Factor

SUMMARY
Groups	Count	Sum	Average	Variance
District A	5	370	74	91
District B	5	378	75.6	121.3
District C	5	413	82.6	81.8
District D	5	348	69.6	73.8


ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	437.35	3	145.7833333	1.585032165	0.232069928	3.238871517
Within Groups	1471.6	16	91.975

Total	1908.95	19

since p-value=0.232>alpha=0.05

null hypothesis is rejected

so, conclusion is not all means are equal

---------------

using tukey Kramer methhod to find which means are different

one way anova

	Sample	Sample		Absolute	Std. Error	Critical
Group	Mean	Size	Comparison	Difference	of Difference	Range	Results
1: District A	74	5	Group 1 to Group 2	1.6	4.429597423	18.073	Means are not different
2: District B	75.6	5	Group 1 to Group 3	8.6	4.429597423	18.073	Means are not different
3: District C	82.6	5	Group 1 to Group 4	4.4	4.429597423	18.073	Means are not different
4: District D	69.6	5	Group 2 to Group 3	7	4.429597423	18.073	Means are not different
			Group 2 to Group 4	6	4.429597423	18.073	Means are not different
Other Data			Group 3 to Group 4	13	4.429597423	18.073	Means are not different
Level of significance	0.05
Numerator d.f.	4
Denominator d.f.	15
MSW	98.10667
Q Statistic	4.08

--------------------------------------------------------------------------------------------------------------

2)

using Excel's Data Analysis/Analysis ToolPak for two way anova-without replciation

Anova: Two-Factor Without Replication

SUMMARY	Count	Sum	Average	Variance
Home 1	3	654	218	64
Home 2	3	580	193.3333333	17.33333333
Home 3	3	555	185	4
Home 4	3	687	229	52
Home 5	3	719	239.6666667	6.333333333

Agent A	5	1054	210.8	590.7
Agent B	5	1058	211.6	512.3
Agent C	5	1083	216.6	566.3


ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
home	6488.666667	4	1622.166667	68.8330976	3.10615E-06	3.837853355
agent	98.8	2	49.4	2.096181047	0.185356807	4.458970108
Error	188.5333333	8	23.56666667

Total	6776	14

since p-value for agent<alpha(0.05),

so,rejct H0,

we conclude not all mean value for agents are equal

-----------

p-value in case of home =0.185>alpha(0.05),fail to reject H0

so, conlcude that all means are equal in case of Home

orchestra answered 2 years ago

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