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Assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.Listed below are the lead concentrations (in μ\mug / 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 μ\mug / g.

3.06.56.05.520.57.512.020.511.517.5\begin{array} { c c c c c c c c c } { 3.0 } & { 6.5 } & { 6.0 } & { 5.5 } & { 20.5 } & { 7.5 } & { 12.0 } & { 20.5 } & { 11.5 } & { 17.5 } \end{array}

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Answered 7 months ago
Answered 7 months ago
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Given:

α=0.05\alpha=0.05

The mean is the sum of all values divided by the number of values:

x=3.0+6.5+6.0+5.5+20.5+7.5+12.0+20.5+11.5+17.510=110.510=11.05\overline{x}=\dfrac{3.0+6.5+6.0+5.5+20.5+7.5+12.0+20.5+11.5+17.5}{10}=\dfrac{110.5}{10}=11.05

nn is the number of values in the data set.

n=10n=10

The variance is the sum of squared deviations from the mean divided by n1n-1. The standard deviation is the square root of the variance:

s=(3.011.05)2+....+(17.511.05)21016.4612s=\sqrt{\dfrac{(3.0-11.05)^2+....+(17.5-11.05)^2}{10-1}}\approx 6.4612

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