Six samples of each of four types of cereal grain grown in a certain region were

Question

Six samples of each of four types of cereal grain grown in a certain region were analyzed to determine thiamin content, resulting in the following data (microgram of thiamin/gram of grain). Wheat: 5.2, 4.5, 6.0, 6.1, 6.7, 5.8 Barley: 6.5, 8.0, 6.1, 7.5, 5.9, 5.6 Maize: 5.8, 4.7, 6.4, 4.9, 6.0, 5.2 Oats: 8.3, 6.1, 7.8, 7.0, 5.5, 7.2 Conduct an ANOVA test at a significance level of 0.05 to determine if at least two of the grains differ with respect to population average thiamin content. (Include the null hypothesis, alternative hypothesis, decision and conclusion along with the analysis)

Explanation / Answer

Null hypothesis = H0 = All the 4 crops have equal thiamin content

Alternative hypothesis = HA = Atleast 2 crops among these 4 have different thiamin content

I am using R software to solve this problem. At first we need to load the data into R environment as below:

Wheat <- c(5.2,4.5,6.0,6.1,6.7,5.8)
Barley <- c(6.5,8.0,6.1,7.5,5.9,5.6)
Maize <- c(5.8,4.7,6.4,4.9,6.0,5.2)
Oats <- c(8.3,6.1,7.8,7.0,5.5,7.2)
Crops <- c(rep("Wheat",6),rep("Barley",6),rep("Maize",6),rep("Oats",6))
ThiaminContent <- c(Wheat,Barley,Maize,Oats)
Data <- data.frame(Crops,ThiaminContent)

Now we can conduct an anova test using the aov() function in R as below:

fit <- aov(ThiaminContent ~ Crops, data = Data)

Print the summary of the fit using the summary() function as below:

summary(fit)

Df Sum Sq Mean Sq F value Pr(>F)
Crops 3 8.983 2.9944 3.957 0.0229 *
Residuals 20 15.137 0.7568   
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Here we can see that the F value is 3.957 and the p value is 0.0229. As this p value is less than 0.05, we can reject the null hypothesis and conclude that atleast 2 crops among these 4 have different thiamin content.

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