Question

Lead in Medicine Listed below are the lead concentrations (in \({\rm{\mu g > g}}\)) 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 \({\rm{\mu g/g}}\).

3.0 6.5 6.0 5.5 20.5 7.5 12.0 20.5 11.5 17.5

Short answer

There is not enough evidence to support the claim that the mean lead concentration for all such medicines is less than 14\(\mu g/g\).

Step-by-step solution

Given information

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.

Check the requirements

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

Thus, the sample size (n) of lead concentrations measured is 10.

Describe the hypothesis

Null hypothesis\(\left( {{H_0}} \right)\)is a statement of claim thatthe mean lead concentration is 14\({\rm{\mu g/g}}\).

Alternate hypothesis\(\left( {{H_1}} \right)\)is a statement of claim that the mean lead concentration is less than 14\({\rm{\mu g/g}}\).

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

Mathematically, it can be expressed as

\(\begin{array}{l}{H_0}:\mu = 14{\rm{ }}\\{H_1}:\mu < 14\end{array}\)

The hypothesis is left-tailed.

Calculate the test statistic

Formula for test statistics is given by,

\(t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\)

Where ,\(\bar x\)is sample mean and s is standard deviation of sample.

The mean is computed as,

\(\begin{array}{c}\bar x = \frac{{\sum {{x_i}} }}{n}\\ = \frac{{3.0 + 6.5 + 6.0 + ... + 17.5}}{{10}}\\ = 11.05\end{array}\)

The standard deviation is computed as,

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

By substituting the values, test statistics is given by,

\(\begin{array}{c}t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ = \frac{{11.05 - 14}}{{\frac{{6.461}}{{\sqrt {10} }}}}\\ = - 1.444\end{array}\)

Calculate the critical value

For the significance level of 0.05,

The degree of freedom is,

\(\begin{array}{c}{\rm{Degree of freedom}} = n - 1\\ = 10 - 1\\{\rm{ }} = 9\end{array}\)

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\({{\rm{t}}_{{\rm{0}}{\rm{.05}}}}\)but the given test is left-tailed therefore use -1.833 as a critical value.

Thus, the critical value\({{\rm{t}}_{0.05}}\)is -1.833.

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

Compare test statistics and critical value.

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.

Conclusion

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