listed below are the lead concentrations (in μg/g) measured in different ayurveda medicines. ayurveda is a…
listed below are the lead concentrations (in μg/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. assume that a simple random sample has been selected. use a 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 μg/g. 2.95 6.50 5.96 5.46 20.51 7.51 11.96 20.52 11.48 17.50 h0: μ = 14.0 h1: μ < 14.0 (type integers or decimals. do not round.) identify the test statistic. (round to two decimal places as needed.)
Answer
Explanation:
Step1: Calculate sample mean $\bar{x}$
$\bar{x}=\frac{2.95 + 6.50+5.96+5.46+20.51+7.51+11.96+20.52+11.48+17.50}{10}=\frac{109.35}{10} = 10.935$
Step2: Calculate sample standard - deviation $s$
First, calculate the sum of squared differences from the mean. [ \begin{align*} \sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}&=(2.95 - 10.935)^{2}+(6.50 - 10.935)^{2}+(5.96 - 10.935)^{2}+(5.46 - 10.935)^{2}+(20.51 - 10.935)^{2}+(7.51 - 10.935)^{2}+(11.96 - 10.935)^{2}+(20.52 - 10.935)^{2}+(11.48 - 10.935)^{2}+(17.50 - 10.935)^{2}\ &=(-7.985)^{2}+(-4.435)^{2}+(-4.975)^{2}+(-5.475)^{2}+(9.575)^{2}+(-3.425)^{2}+(1.035)^{2}+(9.585)^{2}+(0.545)^{2}+(6.565)^{2}\ &=63.76+19.67+24.75+30.00+91.68+11.73+1.07+91.88+0.29+43.10\ &=377.93 \end{align*} ] $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}=\sqrt{\frac{377.93}{9}}\approx6.46$
Step3: Calculate the t - test statistic
The formula for the one - sample t - test statistic is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\mu = 14.0$, $\bar{x}=10.935$, $s\approx6.46$, and $n = 10$. $t=\frac{10.935 - 14.0}{6.46/\sqrt{10}}=\frac{-3.065}{2.04}\approx - 1.50$
Answer:
$-1.50$