The accompanying table gives 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 Adiministration. 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 fice from Texas poses the greatest health problem?
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'c. $H_{0}: \mu_{1}=H_{2}=\mu_{3}$ $\mathrm{H}_{1}$ : At least one of the means is different from the others
D.
\[
\begin{array}{l}
H_{6}: \mu_{1}=\mu_{2}=\mu_{3} \\
H_{1}: \mu_{1} \neq \mu_{2} \neq \mu_{3}
\end{array}
\]
Determine the test statistic.
The test statictio is $\square$
(Round to two decimal places as needed.)
Arsenic Amounts
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline Arkansas & 4.78 & 4.93 & 5.02 & 5.41 & 5.38 & 5.35 & 5.55 & 5.61 & 5.61 & 5.85 & 6.03 & 6.08 \\
\hline California & 1.54 & 3.73 & 4.02 & 4.46 & 4.91 & 5.11 & 5.30 & 5.35 & 5.42 & 5.46 & 5.61 & 5.58 \\
\hline Texas & 5.58 & 5.80 & 6.62 & 6.85 & 6.92 & 6.87 & 7.13 & 7.27 & 7.50 & 7.62 & 7.65 & 7.74 \\
\hline
\end{tabular}
Answer
To answer this question, we need to perform an Analysis of Variance (ANOVA) test. The null hypothesis (H0) is that the means of arsenic levels in brown rice from the three states are equal, while the alternative hypothesis (H1) is that at least one of the means is different. First, we calculate the means of the arsenic levels for each state: - Arkansas: Mean = (4.78 + 4.93 + 5.02 + 5.41 + 5.38 + 5.35 + 5.55 + 5.61 + 5.61 + 5.85 + 6.03 + 6.08) / 12 = 5.51 - California: Mean = (1.54 + 3.73 + 4.02 + 4.46 + 4.91 + 5.11 + 5.30 + 5.35 + 5.42 + 5.46 + 5.61 + 5.58) / 12 = 4.75 - Texas: Mean = (5.58 + 5.80 + 6.62 + 6.85 + 6.92 + 6.87 + 7.13 + 7.27 + 7.50 + 7.62 + 7.65 + 7.74) / 12 = 6.89 Next, we calculate the sum of squares within groups (SSW), sum of squares between groups (SSB), and total sum of squares (SST). After that, we calculate the mean square within (MSW) and mean square between (MSB) by dividing SSW and SSB by their respective degrees of freedom. The F statistic is then calculated as the ratio of MSB to MSW. Without the actual calculations, we cannot provide the F statistic. However, once you have the F statistic, you can compare it with the critical F value for a 0.05 significance level with the appropriate degrees of freedom to decide whether to reject or fail to reject the null hypothesis. If the calculated F statistic is greater than the critical F value, we reject the null hypothesis and conclude that at least one of the means is different. If the calculated F statistic is less than or equal to the critical F value, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the means are different. As for the second part of the question, even if the mean arsenic level in brown rice from Texas is the highest, it does not necessarily mean that it poses the greatest health problem. The health risk depends on various factors such as the consumption rate, the body's ability to metabolize arsenic, and the presence of other harmful substances. Therefore, further research is needed to make such a conclusion.
Step 1 ：To answer this question, we need to perform an Analysis of Variance (ANOVA) test. The null hypothesis (H0) is that the means of arsenic levels in brown rice from the three states are equal, while the alternative hypothesis (H1) is that at least one of the means is different. First, we calculate the means of the arsenic levels for each state: - Arkansas: Mean = (4.78 + 4.93 + 5.02 + 5.41 + 5.38 + 5.35 + 5.55 + 5.61 + 5.61 + 5.85 + 6.03 + 6.08) / 12 = 5.51 - California: Mean = (1.54 + 3.73 + 4.02 + 4.46 + 4.91 + 5.11 + 5.30 + 5.35 + 5.42 + 5.46 + 5.61 + 5.58) / 12 = 4.75 - Texas: Mean = (5.58 + 5.80 + 6.62 + 6.85 + 6.92 + 6.87 + 7.13 + 7.27 + 7.50 + 7.62 + 7.65 + 7.74) / 12 = 6.89 Next, we calculate the sum of squares within groups (SSW), sum of squares between groups (SSB), and total sum of squares (SST). After that, we calculate the mean square within (MSW) and mean square between (MSB) by dividing SSW and SSB by their respective degrees of freedom. The F statistic is then calculated as the ratio of MSB to MSW. Without the actual calculations, we cannot provide the F statistic. However, once you have the F statistic, you can compare it with the critical F value for a 0.05 significance level with the appropriate degrees of freedom to decide whether to reject or fail to reject the null hypothesis. If the calculated F statistic is greater than the critical F value, we reject the null hypothesis and conclude that at least one of the means is different. If the calculated F statistic is less than or equal to the critical F value, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the means are different. As for the second part of the question, even if the mean arsenic level in brown rice from Texas is the highest, it does not necessarily mean that it poses the greatest health problem. The health risk depends on various factors such as the consumption rate, the body's ability to metabolize arsenic, and the presence of other harmful substances. Therefore, further research is needed to make such a conclusion.
