df2

getwd()

[1] "/Users/jaclynbazsika/Documents/Documents/MSDS650/w2"

df2 = read.table("fastfood-2.txt", header=TRUE)

r = c(t(as.matrix(df2)))
r 

 [1] 31 27 24 31 28 31 45 29 46 21 18 48 42 36 46 32 17 40

t(df2)

      [,1] [,2] [,3] [,4] [,5] [,6]
Item1   31   31   45   21   42   32
Item2   27   28   29   18   36   17
Item3   24   31   46   48   46   40

f = c("Item1", "Item2", "Item3") 
k = 3 
n = 6   

tm = gl(k, 6, n*k, factor(f)) 
tm 

 [1] Item1 Item1 Item1 Item1 Item1 Item1 Item2 Item2 Item2 Item2 Item2 Item2 Item3 Item3 Item3 Item3 Item3 Item3
Levels: Item1 Item2 Item3

blk = gl(n, k, k*n)            
blk 

 [1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6
Levels: 1 2 3 4 5 6

av = aov(r ~ tm + blk)

summary(av) 

            Df Sum Sq Mean Sq F value Pr(>F)  
tm           2  538.8  269.39   4.959 0.0319 *
blk          5  559.8  111.96   2.061 0.1547  
Residuals   10  543.2   54.32                 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

With the P-value >0.05(blk), it can be concluded that the null hypothesis can not be rejected. However, for tm, the p-value is < 0.05. This would render the data statistically insignifcant; therefore, the null hypothesis is rejected. For tm, the mean sales volume of new menu items are all equal. For blk, the mean sales volume is not equal.