Question

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 0.10 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 μg/g.
2.98 6.54 5.96 5.50 20.50 7.52 12.04 20.51 11.52 17.52
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: $2.98,6.54,5.96,5.50,20.50,7.52,12.04,20.51,11.52,17.52$.
$\sum_{i=1}^{10}x_{i}=2.98 + 6.54+5.96+5.50+20.50+7.52+12.04+20.51+11.52+17.52=110.19$
$\bar{x}=\frac{110.19}{10}=11.019$

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 $x_{i}$:
For $x_1 = 2.98$: $(2.98 - 11.019)^{2}=(-8.039)^{2}=64.625521$
For $x_2 = 6.54$: $(6.54 - 11.019)^{2}=(-4.479)^{2}=19.962441$
For $x_3 = 5.96$: $(5.96 - 11.019)^{2}=(-5.059)^{2}=25.6$
For $x_4 = 5.50$: $(5.50 - 11.019)^{2}=(-5.519)^{2}=30.45$
For $x_5 = 20.50$: $(20.50 - 11.019)^{2}=(9.481)^{2}=90$
For $x_6 = 7.52$: $(7.52 - 11.019)^{2}=(-3.499)^{2}=12.24$
For $x_7 = 12.04$: $(12.04 - 11.019)^{2}=(1.021)^{2}=1.042441$
For $x_8 = 20.51$: $(20.51 - 11.019)^{2}=(9.491)^{2}=90.08$
For $x_9 = 11.52$: $(11.52 - 11.019)^{2}=(0.501)^{2}=0.251$
For $x_{10}=17.52$: $(17.52 - 11.019)^{2}=(6.501)^{2}=42.26$
$\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=64.625521+19.962441 + 25.6+30.45+90+12.24+1.042441+90.08+0.251+42.26=376.511445$
$s=\sqrt{\frac{376.511445}{9}}\approx\sqrt{41.834605}\approx6.47$

Step3: Calculate the test statistic $t$

The formula for the test statistic in a one - sample $t$ - test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$.
Here, $\mu = 14.0$, $\bar{x}=11.019$, $s = 6.47$, $n = 10$
$t=\frac{11.019-14.0}{6.47/\sqrt{10}}=\frac{- 2.981}{2.045}\approx - 1.46$

Answer:

$-1.46$