Lead in Medicine Listed below are the lead concentrations (in ) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States (based on data from “Lead, Mercury, and Arsenic in US and Indian Manufactured Ayurvedic Medicines Sold via the Internet,” by Saper et al., Journal of the American Medical Association,Vol. 300, No. 8). Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 14 $\mu g/g$.

3.0 6.5 6.0 5.5 20.5 7.5 12.0 20.5 11.5 17.5

Sample of lead concentrations measured in different Ayurveda medicines is given (in $\mu g/g$); 3.0 6.5 6.0 5.5 20.5 7.5 12.0 20.5 11.5 17.5.

Level of significance is 0.05.

Assume that the lead concentrations are normally distributed and the samples are randomly selected.

Thus, the sample size (n) oflead concentrations measuredis 10.

Null hypothesis$\left(H_0\right)$is a statement of claim that the mean lead concentration is 14.

Alternate hypothesis$\left(H_1\right)$is a statement of claim that the mean lead concentration is less than 14.

Here, $\mu$is the true population mean lead concentration.

Mathematically, it can be expressed as

$$\begin{gathered} H_0:\mu =14 \\ {H}_1:\mu <14 \end{gathered}$$

The hypothesis is left-tailed.

Formula for test statistics is given by,

$t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Where , $\overline{x}$is sample mean and s is standard deviation of sample.

The mean is computed as,

$$\begin{gathered} \overline{x}=\frac{\sum x_i}{n} \\ =\frac{3.0+6.5+6.0+...+17.5}{10} \\ =11.05 \end{gathered}$$

The standard deviation is computed as,

$$\begin{gathered} s=\sqrt{\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(3.0-11.05\right)}^2+{\left(6.5-11.05\right)}^2+...+{\left(17.5-11.05\right)}^2}{10-1}} \\ =6.461 \end{gathered}$$

By substituting the values, test statistics is given by,

$$\begin{gathered} t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}} \\ =\frac{11.05-14}{\frac{6.461}{\sqrt{10}}} \\ =-1.444 \end{gathered}$$

For the significance level of 0.05,

The degree of freedom is,

$$\begin{gathered} \text{Degree of freedom}=n-1 \\ =10-1 \\ =9 \end{gathered}$$

In the t-distribution table, find the value corresponding to the row value of degree of freedom 9 and column value of area in one tail 0.05 is 1.833 which is critical value ${\text{t}}_{0.05}$but the given test is left-tailed therefore use -1.833 as a critical value.

Thus, the critical value${\text{t}}_{0.05}$is -1.833.

The critical region is$\left(t:t<-1.833\right)$.

The test statistic is -1.444 and the critical value $t_{0.05}$is -1.833.

According to this, we can conclude that the test statistics -1.444 will not fall in the critical region bounded by the critical value -1.833.

Therefore, we failed to reject the null hypothesis.

It is concluded that there is not enough evidence to support the claim that mean lead concentrations measured in different Ayurveda medicines is less than 14 $\mu g/g$.