# problem-11.6

problem-11.6 Before using either function, the data is put into the form of a numeric variable to record the values and a factor to record the laboratory.

    > lab1 = c(4.13, 4.07, 4.04, 4.07, 4.05)
    > lab2 = c(3.86, 3.85, 4.08, 4.11, 4.08)
    > lab3 = c(4.00, 4.02, 4.01, 4.01, 4.04)
    > lab4 = c(3.88, 3.89, 3.91, 3.96, 3.92)
    > chems = stack(data.frame(lab1,lab2,lab3,lab4))
    > boxplot(values ~ ind, data=chems) # var.equal unlikely
    > oneway.test(values ~ ind, data=chems, var.equal=FALSE)

            One-way analysis of means (not assuming equal variances)

    data:  values and ind
    F = 18.71, num df = 3.000, denom df = 7.914, p-value = 0.0005927

The null hypothesis is that each lab has the same mean. The small p-value suggests that the data is not consistent with this assumption.