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A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed $24$individuals and asked them their daily one-way commuting mileage.

Consider the null hypothesis that there is no significance difference between three mean commuting mileages. To test the mean commuting mileage are same at $5\%$level of significance, use Ti-83 calculator. For this, click on STAT $→$press $1→$EDIT, then put the above given weights of different groups data into the list $L1,L2,andL3$. The screenshot is given as below.

Now Again press STAT $→$arrow over the TESTS localid="1649235797890" $→$arrow down to ANOVA localid="1649239023592" $→$Press var then select1, press var then select 2 , and press var then select 3 to fill the values of localid="1649235791952" $L1,L2,andL3$. The screen shot is given as below:

Press ENTER. The screenshot of the obtained output of is given as below:

Use the above results to solve the below given subparts.

The null hypothesis that three mean commuting mileages are same is given as below:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2={\mathrm{μ}}_3$

The alternate hypothesis will be at least any two of the means are different.

The degree of freedom in the numerator is $\mathbf{2}$, and the degree of freedom in the denominator is $21$.

The $F$distribution is used for the test of three different means.

The value of the test statistic is $9.09$.

The $P-$value for the test is $\mathbf{0}\mathbf{.}\mathbf{001}$.

The graph of the distribution is given as below:

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

Decision: Null hypothesis will be rejected.

Reason for decision: $P$-value is $0.001$which is less than the $0.05$level of significance.

Three mean commuting mileages are not same.