Q:

# Listed below are the lead concentrations in mu​g/g measured in different traditional medicines. Use a 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 17 mu​g/g. 13 21 3.5 18.5 21.5 8.5 15.5 19 9 5.5

Accepted Solution

A:
Answer:$$t=\frac{13.5-17}{\frac{6.57}{\sqrt{10}}}=-1.68$$  $$p_v =P(t_{(9)}<-1.68)=0.064$$  If we compare the p value and the significance level given $$\alpha=0.01$$ we see that $$p_v>\alpha$$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis. We can say that at 1% of significance the mean lead concentration for all is significantly less than 17 mu​g/g.Step-by-step explanation:Data given and notation  The mean and sample deviation can be calculated from the following formulas:$$\bar X =\frac{\sum_{i=1}^n x_i}{n}$$$$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}$$$$\bar X=13.5$$ represent the sample mean  $$s=6.57$$ represent the sample standard deviation  $$n=10$$ sample size  $$\mu_o =17$$ represent the value that we want to test  $$\alpha=0.01$$ represent the significance level for the hypothesis test.  t would represent the statistic (variable of interest)  $$p_v$$ represent the p value for the test (variable of interest)  State the null and alternative hypotheses.  We need to conduct a hypothesis in order to check if the population mean is less than 17, the system of hypothesis are :  Null hypothesis:$$\mu \geq 17$$  Alternative hypothesis:$$\mu < 17$$  Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  $$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$$ (1)  t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  Calculate the statistic  We can replace in formula (1) the info given like this:  $$t=\frac{13.5-17}{\frac{6.57}{\sqrt{10}}}=-1.68$$  P-value  We need to calculate the degrees of freedom first given by: $$df=n-1=10-1=9$$ Since is a one-side left tailed test the p value would given by:  $$p_v =P(t_{(9)}<-1.68)=0.064$$  Conclusion  If we compare the p value and the significance level given $$\alpha=0.01$$ we see that $$p_v>\alpha$$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis. We can say that at 1% of significance the mean lead concentration for all is significantly less than 17 mu​g/g.