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Listed below are the lead concentrations in $\mu \mathrm{g} / \mathrm{g}$ measured in different traditional medicines. Use a 0.10 significance level to test the claim that the mean lead concentration for all such medicines is less than $15 \mu \mathrm{g} / \mathrm{g}$. Assume that the sample is a simple random sample.
$\begin{array}{llllllllll}8.5 & 6 & 17 & 13 & 14.5 & 17.5 & 3 & 17 & 6 & 4.5\end{array}$
Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses?
A.
\[
\begin{array}{l}
H_{0}: \mu> 15 \mu \mathrm{g} / \mathrm{g} \\
H_{1}: \mu< 15 \mu \mathrm{g} / \mathrm{g}
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: \mu=15 \mu \mathrm{g} / \mathrm{g} \\
H_{1}: \mu> 15 \mu \mathrm{g} / \mathrm{g}
\end{array}
\]
c.
\[
\begin{array}{l}
H_{0}: \mu=15 \mu \mathrm{g} / \mathrm{g} \\
H_{1}: \mu \neq 15 \mu \mathrm{g} / \mathrm{g}
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \mu=15 \mu \mathrm{g} / \mathrm{g} \\
H_{1}: \mu< 15 \mu \mathrm{g} / \mathrm{g}
\end{array}
\]
Determine the test statistic.
(Round to two decimal places as needed.)

Step 1 ：The null and alternative hypotheses for a hypothesis test are typically a statement of no effect or status quo, and what we are testing against the null hypothesis. In this case, we are testing the claim that the mean lead concentration for all such medicines is less than 15 μg/g. So, the null hypothesis would be that the mean lead concentration is equal to 15 μg/g, and the alternative hypothesis would be that the mean lead concentration is less than 15 μg/g. This corresponds to option D.

Step 2 ：The test statistic for a hypothesis test for a mean is calculated as \((\text{sample mean} - \text{hypothesized mean}) / (\text{standard deviation} / \sqrt{\text{sample size}})\). We can calculate this using the given data.

Step 3 ：Given data = [ 8.5 6. 17. 13. 14.5 17.5 3. 17. 6. 4.5], the sample mean is 10.7, the standard deviation is 5.696977756280567, and the sample size is 10. The hypothesized mean (mu) is 15.

Step 4 ：Substituting these values into the formula, we get the test statistic as -2.3868434318061547.

Step 5 ：Rounding to two decimal places, the test statistic is -2.39.

Step 6 ：So, the null and alternative hypotheses are: \[H_{0}: \mu=15 \mu \mathrm{g} / \mathrm{g} \] \[H_{1}: \mu<15 \mu \mathrm{g} / \mathrm{g} \] and the test statistic is \(\boxed{-2.39}\).