91Ó°ÊÓ

We eliminate home decorating type for our solution. Then we calculate:

Press STAT. 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. We will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the $p$-value of the test.

The calculator displays:

$F=2.1226578$

$p=0.1624495$

Factor

$df=2$

$SS=4093.3333$

$MS=2046.6667$

Error

$df=12$

$SS=11570.4$

$MS=31.05157$

We use a solution sheet to conduct the hypothesis tests, and we have:

a) The null hypothesis that three mean commuting mileages are the same is:

$H_0:{\mathrm{μ}}_n={\mathrm{μ}}_h={\mathrm{μ}}_c$

b) The alternate hypothesis is that at least any two of the magazines have different mean lengths.

c) The degree of freedom in the numerator - $df\left(num\right)$is $2$,

and the degree of freedom in the denominator - $\mathrm{df}\left(\mathrm{denom}\right)$is $12$.

d) We use the $F$ distribution.

e) The value of the test statistic (F-value) is $2.123$.

f) The $P$-value for the test is $0.1624$.

g). The graph of the distribution is

h)

i. Level of significance $\mathrm{Î\pm }$ is $0.05$.

ii. Decision: We do not reject the null hypothesis

iii. Reason for decision: $P$-value is $0.1624$ which is greater than the $0.05$ level of significance.

iv. Conclusion: There is no sufficient evidence to conclude that the mean lengths of the magazines are different.

We accept the null hypothesis.