Question

listed below are the lead concentrations (in μg/g) measured in different ayurveda medicines. ayurveda is a traditio medical system commonly used in india. the lead concentrations listed here are from medicines manufactured in th united states. assume that a simple random sample has been selected. use a 0.10 significance level to test the cl that the mean lead concentration for all such medicines is less than 14.0 μg/g. 3.02 6.46 6.02 5.52 20.52 7.49 11.96 20.52 11.51 17.53 identify the null and alternative hypotheses. 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.)

Explanation:

Step1: Calculate the sample mean $\bar{x}$

The formula for the sample mean is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
Here, $n = 10$ and $x_{i}$ are the data points: $3.02,6.46,6.02,5.52,20.52,7.49,11.96,20.52,11.51,17.53$.
$\sum_{i=1}^{10}x_{i}=3.02 + 6.46+6.02+5.52+20.52+7.49+11.96+20.52+11.51+17.53=110.55$
$\bar{x}=\frac{110.55}{10}=11.055$

Step2: Calculate the sample standard deviation $s$

The formula for the sample standard deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$.
First, calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$:
$(3.02 - 11.055)^{2}=(-8.035)^{2}=64.561225$
$(6.46-11.055)^{2}=(-4.595)^{2}=21.114025$
$(6.02 - 11.055)^{2}=(-5.035)^{2}=25.351225$
$(5.52-11.055)^{2}=(-5.535)^{2}=30.636225$
$(20.52-11.055)^{2}=(9.465)^{2}=89.586225$
$(7.49-11.055)^{2}=(-3.565)^{2}=12.709225$
$(11.96-11.055)^{2}=(0.905)^{2}=0.819025$
$(20.52-11.055)^{2}=(9.465)^{2}=89.586225$
$(11.51-11.055)^{2}=(0.455)^{2}=0.207025$
$(17.53-11.055)^{2}=(6.475)^{2}=41.925625$
$\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=64.561225+21.114025+25.351225+30.636225+89.586225+12.709225+0.819025+89.586225+0.207025+41.925625 = 376.505$
$s=\sqrt{\frac{376.505}{9}}\approx6.46$

Step3: Calculate the test statistic $t$

The formula for the $t$-test statistic in a one - sample $t$-test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$.
Here, $\mu = 14.0$, $\bar{x}=11.055$, $s = 6.46$, $n = 10$
$t=\frac{11.055-14.0}{6.46/\sqrt{10}}=\frac{-2.945}{2.043}\approx - 1.44$

Answer:

$-1.44$