Step 1 :Given the lead concentrations in traditional medicines, we are asked to test the claim that the mean lead concentration for all such medicines is less than $13 \mu g / g$ at a 0.10 significance level.
Step 2 :The null and alternative hypotheses for this test are: \n $H_{0}: \mu=13 \mu g / g$ \n $H_{1}: \mu<13 \mu g / g$
Step 3 :We calculate the sample mean, standard deviation, and size from the given data.
Step 4 :The sample mean is $9.95 \mu g / g$, the sample standard deviation is $7.16 \mu g / g$, and the sample size is 11.
Step 5 :We then calculate the test statistic using the formula: \n $t = \frac{\bar{x} - \mu_{0}}{s / \sqrt{n}}$
Step 6 :Substituting the values, we get the test statistic as $-1.41$.
Step 7 :We also calculate the degrees of freedom as $n - 1 = 10$.
Step 8 :Using the test statistic and degrees of freedom, we calculate the P-value for a one-tailed test.
Step 9 :The P-value is $0.094$.
Step 10 :Final Answer: \n The null and alternative hypotheses are: \n $H_{0}: \mu=13 \mu g / g$ \n $H_{1}: \mu<13 \mu g / g$ \n The test statistic is \(\boxed{-1.41}\). \n The P-value is \(\boxed{0.094}\).