Question

Four brands of lithium ion batteries are to be compared by testing each brand in five Cars. Twenty Cars are randomly selected and divided randomly into four groups of five Cars each. Then each group of cars uses a different brand of battery. The lifetimes of the batteries, to the nearest hour, are as follows. Brand A Brand B Brand C Brand D 1500 1345 1500 1450 1540 1560 1400 1500 1280 1230 1230 1450 1345 1453 1450 1460 1300 1234 1235 1510 

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c:c} A & B & C & D\\ \hline 1500 & 1345 & 1500 & 1450 \\ \hline 1540 & 1560 &1400 & 1500\\ \hline 1280 & 1230 & 1230 & 1450 \\ \hline 1345 & 1453 & 1450 & 1460\\ \hline 1300 & 1234 & 1235 & 1510\\ \hline \end{array}$

Means in groups

$\mu_1=\frac{1500+1540+1280+1345+1300}{5}=1393$

$\mu_2=\frac{1345+1560+1230+1453+1234}{5}=1364.4$

$\mu_3=\frac{1500+1400+1230+1450+1235}{5}=1363$

$\mu_4=\frac{1450+1500+1450+1460+1510}{5}=1474$

we see that D is a best battery.

But this observation must be confirmed by a statistical test.

null hypothesis

$H_0:\mu_1=\mu_2=\mu_3=\mu_4$

For we calculate:

1) General average

$\mu=\frac{\mu_1+\mu_2+\mu_3+\mu_4}{4}=\frac{1393+1364.4+1363+1474}{4}=1398.6$

2) Total sum of squared deviations

$SS=\sum_{j=1..4}\sum_{i=1..5}(A_{i,j}-\mu)^2=\\ (1500-1398.6)^2+(1540-1398.6)^2+(1280-1398.6)^2+(1345-1398.6)^2+(1300-1398.6)^2+\\ (1345-1398.6)^2+(1560-1398.6)^2+(1230-1398.6)^2+(1453-1398.6)^2+(1234-1398.6)^2+\\ (1500-1398.6)^2+(1400-1398.6)^2+(1230-1398.6)^2+(1450-1398.6)^2+(1235-1398.6)^2+\\ (1450-1398.6)^2+(1500-1398.6)^2+(1450-1398.6)^2+(1460-1398.6)^2+(1510-1398.6)^2=\\ =244200.8 ​$

3) Sums of squared deviations in groups

$SS_{ing}=\sum_{j=1..4}\sum_{i=1..5}(A_{i,j}-\mu_j)^2=\\ (1500-1393)^2+(1540-1393)^2+(1280-1393)^2+(1345-1393)^2+(1300-1393)^2+\\ (1345-1364.4)^2+(1560-1364.4)^2+(1230-1364.4)^2+(1453-1364.4)^2+(1234-1364.4)^2+\\ (1500-1363)^2+(1400-1363)^2+(1230-1363)^2+(1450-1363)^2+(1235-1363)^2+\\ (1450-1474)^2+(1500-1474)^2+(1450-1474)^2+(1460-1474)^2+(1510-1474)^2=\\ =203422.2$

3) Sums of squared deviations between groups

$SS_{bg}=\sum_{j=1..4}n_j\cdot (\mu_i-\mu)^2=\\ 5\cdot \left( (1393-1398.6)^2 + (1364-1398.6)^2 +(1363-1398.6)^2 +(1474-1398.6)^2\right)=40767.6$

4) Variance between groups

$D_{bg}=\frac{SS_{bg}}{m-1}=\frac{40767.6}{4-1}=13589.2$

5) Variance in groups

$D_{ing}=\frac{SS_{ing}}{N-m}=\frac{203422.2}{20-4}=12714.585$

6) The value of the Fisher criterion

$F=\frac{D_{bg}}{D_{ing}}=\frac{13589.2 }{12714.585 }=1.069$

7) p-value of Fisher criterion

1-pF(1.069,4,16)=0.41>0.05

Conclusions: There are no sufficient grounds for rejecting the null hypothesis at the reliability level of 0.95 therefore all brands are approximately equivalent.