Question

A consumer research organization tests three brands of tires to see how many miles they can be
driven before they should be replaced. One tyre of each brand is tested in each of five types of
cars. The results (in thousands of miles) are as follows: (10)
Type of car Brand A Brand B Brand C
I
6 9 4
II
3 2 7
III
2 3 6
IV
8 8 5
V 9 1 8
Compute the ANOVA and interpret your result.

Answer 1

The life time duration (thousands of miles)    of the 5 types of tires on 3 brands of cars.

Total number of observations:  N = k n = 15
number of groups = k = 3
number of rows = cases in each group = n = 5
degrees of freedom inside group = DF_g = n - 1 = 4

type  car    A           B            C
tire    ====================
I              6           9             4
II             3            2            7
III            2            3            6
IV           8            8            5
V            9            1            8
========================
mean    5.6          4.6          6.0           μ₁ = 5.6 ,    μ₂= 4.6      μ_3 = 6.0

overall grand mean = mean of means = μ = 16.2/3 = 5.4

ANOVA (analysis of variance) is to test whether the mean of three or more than three groups are equal or not.  Null hypothesis is that all the brands of cars use the tires to the same life time.  It is like multiple  t tests.  We calculate the F test statistic.

variance between groups = n * (μ₁ - μ)²  + n (μ₂ - μ)²  + n ( μ_3 - μ)²
Hence sum of squares SST  = 5 * [ 0.2² + 0.8² + 0.6² ]  = 5.2
MST mean squares variability between groups:  SST /( k-1) = 5.2/2 = 2.6

variance within groups
SSE = sum of square due to Error
     Car A:  (6-5.6)²+(3-5.6)²+(2-5.6)²+(8-5.6)²+(9-5.6)² =  37.2
     Car B = (9-4.6)²+(2-4.6)²+(3-4.6)²+(8-4.6)²+(1-4.6)² =  48.4
     Car C=  ( 4-6.0)² + (7 -6.0)² +(6 -6.0)² +(5 -6.0)² +(8 -6.0)² =10

     SSE within = 37.2 + 48.4 + 10 = 95.6
    degrees of freedom within groups = N-k = 15 - 3 = 12  = (n-1)k = 4 * 3
    MSE = mean square error = SSE / df = 95.6 / 12 = 7.97

F for  F- Test =  SSE between groups / SSE within groups = 2.6 / 7.97 = 0.326

Upper critical values for F distribution at 1%, 5% and 10% significance levels at degrees of freedom (2, 12)  are      6.927    , 3.885,  2.807    respectively.

Since the F test statistic  0.326 << the above critical values,  there is strong probability that the groups have all the same mean.  We dont reject the null hypothesis.

In this case the p value is  0.72  or 72%  >> 10% or α value of 0.10.  So we dont reject the null hypothesis.

Interpretation is that  the mean life duration of each tire does not depend much on the type of car it is fitted on.  All tires are can be used with all brands of cars giving on an average the same life time duration.