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Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C.

To get the results on the calculator:
Press STAT, then Press 1 EDIT. Put the data into the lists $L_1,L_2,L_3$
Press STAT, and arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter$L_1,L_2,L_3$. Press ENTER. The values in the foregoing ANOVA table arequickly produced by the calculator, including the test statistic and the $p$-value of the test.
The calculator shows :

$F=0.66853555$

$p=0.530548$

FACTOR

$df=2$

$SS=23.212$

$MS=11.606$

ERROR

$df=12$

$SS=208.324$

$\begin{array}{r}MS=17.36\end{array}$

The solution sheet to lead the hypothesis tests, and then:
The null hypothesis that three mean commuting mileages are the identical is:
$H_0:{\mathrm{μ}}_l={\mathrm{μ}}_t={\mathrm{μ}}_j$
The alternate hypothesis is that at least two of the means are dissimilar.
The degree of freedom in the numerator - $df\left(num\right)$is$2$,
and the degree of freedom in the denominator - $\mathrm{df}\left(denom\right)$is $12$

Use the$F$ distribution for the test of three different means.
The value of the test statistic (F-value) is $0.668$
The $P$-value for the test is $0.5305$

The graph of the distribution is,

Level of significance $\mathrm{Î\pm }$is $0.10$
Decision: Do not reject the null hypothesis
Reason for decision: $P$-value is $0.53$which is greater than the$0.10$ level of significance.
Conclusion: There is insufficient evidence to conclude that the means are different.