Question

Arsenic in Rice Listed below are amounts of arsenic in samples of brown rice from three different states. The amounts are in micrograms of arsenic and all samples have the same serving size. The data are from the Food and Drug Administration. Use a 0.05 significance level to test the claim that the three samples are from populations with the same mean. Do the amounts of arsenic appear to be different in the different states? Given that the amounts of arsenic in the samples from Texas have the highest mean, can we conclude that brown rice from Texas poses the greatest health problem?

Arkansas	4.8	4.9	5	5.4	5.4	5.4	5.6	5.6	5.6	5.9	6	6.1
California	1.5	3.7	4	4.5	4.9	5.1	5.3	5.4	5.4	5.5	5.6	5.6
Texas	5.6	5.8	6.6	6.9	6.9	6.9	7.1	7.3	7.5	7.6	7.7	7.7

Short answer

There is sufficient evidence to conclude that there is a significant difference in the amount of arsenic in 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 greater amount of arsenic in Texas will be more dangerous to health.

Step-by-step solution

Given information

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

State the hypotheses

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:

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

State the decision rule

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.

Compute the variance between the groups

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.

Compute the variance within the groups

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.

Compute the test statistic

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:

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

State the decision

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.