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A pet food company conducted an experiment to compare five different cat foods. A sample of 30 cats from the population at a local animal shelter was selected. Six cats were randomly assigned to each of the five products. Each was then presented with 3 ounces of the selected food in a dish at feeding time. The researchers defined the variable to be measured as the number of ounces of food that the cat consumed within a 10-minute time interval. The results for this experiment are summarized in the accompanying table.

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

A. At the 0.05 level of significance, is there evidence of a difference in the mean amount of food eaten among the various products?

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            
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    
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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