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Data are given on the amount of arsenic present in the three different states.

Let\({\mu _1},{\mu _2},{\mu _3}\)be the population mean arsenic levels for three states.

The null hypothesis to test the difference in the presence of arsenic in the 3 samples is as follows:

\(\begin{aligned}{l}{H_0}:{\mu _1} = {\mu _2} = {\mu _3}\\{H_1}:\;{\rm{at}}\;{\rm{least}}\;{\rm{one}}\;{\rm{mean}}\;{\rm{is}}\;{\rm{different}}\end{aligned}\)

If the computed F-statistic is greater than the critical value, the null hypothesis is rejected at a 5% level of significance.

If the computed F-statistic is less than the critical value, the null hypothesis fails to reject at a 5% level of significance.

Let n denote the sample sizes.

As the 3 samples are of equal sizes which are 12(n).

Let\({\bar x_i}\)denote the sample means.

The 3 sample means are computed below:

\(\begin{aligned}{c}{{\bar x}_1} = \frac{{4.8 + 4.9 + \ldots + 6.1}}{{12}}\\ = 5.475\\{{\bar x}_2} = \frac{{1.5 + 3.7 + \ldots + 5.6}}{{12}}\\ = 4.71\\{{\bar x}_3} = \frac{{5.6 + 5.8 + \ldots + 7.7}}{{12}}\\ = 6.97\end{aligned}\)

Let\({s^2}_{\bar x}\)denote the variance of the sample means:

The mean of the 3 sample means is equal to:

\(\begin{aligned}{c}\bar \bar x = \frac{{5.475 + 4.71 + 6.97}}{3}\\ = 5.72\end{aligned}\)

The variance of the sample means is computed below:

\(\begin{aligned}{c}s_{\bar x}^2 = \frac{{\sum\limits_{i = 1}^3 {{{({{\bar x}_i} - \bar \bar x)}^2}} }}{{3 - 1}}\\ = \frac{{{{\left( {5.475 - 5.72} \right)}^2} + {{\left( {4.71 - 5.72} \right)}^2} + {{\left( {6.97 - 5.72} \right)}^2}}}{{3 - 1}}\\ = 1.32\end{aligned}\)

The variance between sample means is equal to:

\(\begin{aligned}{c}ns_{\bar x}^2 = 12\left( {1.32} \right)\\ = 15.826\end{aligned}\)

Therefore, the variance between sample means is equal to 15.826.

Now, the 3 sample variances denoted by\({s_i}^2\)are computed below:

\(\begin{aligned}{c}{s_1}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{1i}} - {{\bar x}_1})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {4.8 - 5.475} \right)}^2} + {{\left( {4.9 - 5.475} \right)}^2} + \ldots + {{\left( {4.9 - 6.1} \right)}^2}}}{{12 - 1}}\\ = 0.175\end{aligned}\)

\(\begin{aligned}{c}{s_2}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{2i}} - {{\bar x}_2})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {1.5 - 4.71} \right)}^2} + \ldots + {{\left( {5.6 - 4.71} \right)}^2}}}{{12 - 1}}\\ = 1.416\end{aligned}\)

\(\begin{aligned}{c}{s_3}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{3i}} - {{\bar x}_3})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {5.6 - 6.97} \right)}^2} + \ldots + {{\left( {7.7 - 6.97} \right)}^2}}}{{12 - 1}}\\ = 0.4788\end{aligned}\)

The variance within samples\(\left( {{s_p}^2} \right)\)is equal to:

\(\begin{aligned}{c}{s_p}^2 = \frac{{{s_1}^2 + {s_2}^2 + {s_3}^2}}{3}\\ = \frac{{0.175 + 1.416 + 0.4788}}{3}\\ = 0.69\end{aligned}\)

Therefore, the variance within samples is equal to 0.690.

Now, the F-statistic is computed as shown below:

\(\begin{aligned}{c}F = \frac{{{\rm{variance}}\;{\rm{between}}\;{\rm{samples}}}}{{{\rm{variance}}\;{\rm{within}}\;{\rm{samples}}}}\\ = \frac{{n{s_{\bar x}}^2}}{{{s_p}^2}}\\ = \frac{{15.826}}{{0.69}}\\ = 22.948\end{aligned}\)

Therefore, the value of the F-statistic is equal to 22.948.

Let k be the number of samples.

It is known that k equals 3.

Thus, the degrees of freedom are calculated as shown below:

\(\begin{aligned}{c}df = \left( {k - 1,k\left( {n - 1} \right)} \right)\\ = \left( {3 - 1,3\left( {12 - 1} \right)} \right)\\ = \left( {2,33} \right)\end{aligned}\)

Thus, the critical value of F at a 5% level of significance with (2,33) degrees of freedom from the F-distribution table is equal to 3.285.

It can be observed that:

\(\begin{aligned}{c}F > {F_{crit}}\\\left( {22.948} \right) > \left( {3.285} \right)\end{aligned}\)

Since the value of the F-statistic is greater than the critical value, the null hypothesis is rejected at a 0.05 level of significance.

Therefore, there is enough evidence to conclude that there is a significant difference in the amount of arsenic for the three states.

In other words, the 3 states do not appear to have the same amount of arsenic.

Also, it can be concluded that the presence of a greater amount of arsenic in Texas will be more dangerous to health.