A pet food company conducted an experiment to compare five different cat foods. A sample of 30 cats from the population

Kidney	Shrimp	Chicken Liver	Salmon	Beef
2.38	2.25	2.28	1.79	1.75
2.63	2.69	2.24	1.85	1.53
2.32	2.25	2.42	1.97	1.18
2.48	2.45	2.69	2.04	1.65
2.59	2.35	2.24	2.26	1.68
2.62	2.37	2.57	1.58	1.31

Kidney

Shrimp

Chicken Liver

Salmon

Beef

2.38

2.25

2.28

1.79

1.75

2.63

2.69

2.24

1.85

1.53

2.32

2.25

2.42

1.97

1.18

2.48

2.45

2.69

2.04

1.65

2.59

2.35

2.24

2.26

1.68

2.62

2.37

2.57

1.58

1.31

tutor

tutor Added an answer on May 17,2019

First, note that this is a test of mean differences for more than 2 groups, which means the one-way ANOVA is appropriate for the analysis of this data. To conduct this, one must ensure that the assumptions that are needed in order to conduct data analysis using the one-way ANOVA. The assumptions of randomness, independence, and equality of variances must be tested in case all procedures need to be presented.

Here, I will use the Excel data analysis tool and present the results here.

First, you need to activate the data tool in Excel.

How to activate the Data analysis tool in Excel. : Go to file > options > under the Excel options select Add-Ins within the "manage" drop-down options select Excel-Add-ins and Click GO Check the analysis toolpak and click OK.

Analyzing one-way ANOVA data using Excel.

Click on Data and click the Data analysis toolpak and the data analysis tool will pop-up, select the ANOVA single: factor and select the input range and check the "Labels in first row" option.

Then click okay, I will present the output for the above data here.

Anova: Single Factor

SUMMARY
Groups	Count	Sum	Average	Variance
Kidney	6	15.02	2.503	0.017
Shrimp	6	14.36	2.393	0.027
Chicken Liver	6	14.44	2.407	0.036
Salmon	6	11.49	1.915	0.054
Beef	6	9.1	1.517	0.051


ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	4.237	4	1.059	28.634	0.000	2.7587
Within Groups	0.925	25	0.037

Total	5.162	29

The output for the above data is an F - statistic and the p-value that needs to be observed. Because the p -value is quite smalle, we reject the null hypothesis and conclude that the differences in the mean are statistically significant.

Hence, the post-hoc analysis can be conducted.

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