Question

64. 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. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of $10\%$, test the hypothesis that the three formulas produce the same mean weight gain

Linda's rats	Tuan's rats	Javier's rats
43.5	47.0	51.2
39.4	40.5	40.9
41.3	38.9	37.9
46.0	46.3	45.0
38.2	44.2	48.6

Short answer

There is insufficient evidence to conclude that the means are different. Hence, get the null hypothesis.

Step-by-step solution

Given information

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.

Using Calculator

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

$df=12$

$SS=208.324$

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

Hypothesis test

The solution sheet to lead the hypothesis tests, and then:
The null hypothesis that three mean commuting mileages are the identical is:
$H_0:{\mu }_l={\mu }_t={\mu }_j$
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 $0.5305$

Graphical representation of the distribution

The graph of the distribution is,

Decision, Reason and explanation 

Level of significance $\alpha$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.