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 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 μg/g
2.96 6.45 6.03 5.48 20.46 7.48 12.03 20.50 11.46 17.50
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.
-1.44
(round to two decimal places as needed.)
identify the p - value
0.093
(round to three decimal places as needed)

Explanation:

Step1: State the hypotheses

The claim is that the mean lead concentration for all such medicines is less than \(14.0\ \mu g/g\). The null hypothesis \(H_0\) is the statement of no difference, so \(H_0:\mu = 14.0\). The alternative hypothesis \(H_1\) is the claim we are testing, so \(H_1:\mu < 14.0\).

Step2: Calculate the test statistic

The formula for the \(t\) - test statistic for a one - sample \(t\) - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\). First, we calculate the sample mean \(\bar{x}\) and sample standard deviation \(s\) from the data set \(\{2.96,6.45,6.03,5.48,20.46,7.48,12.03,20.50,11.46,17.50\}\).
\(n = 10\) (number of data points).
\(\bar{x}=\frac{2.96 + 6.45+6.03+5.48+20.46+7.48+12.03+20.50+11.46+17.50}{10}=\frac{110.35}{10}=11.035\)
The sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\)
\(\sum_{i=1}^{n}(x_i-\bar{x})^2=(2.96 - 11.035)^2+(6.45 - 11.035)^2+(6.03 - 11.035)^2+(5.48 - 11.035)^2+(20.46 - 11.035)^2+(7.48 - 11.035)^2+(12.03 - 11.035)^2+(20.50 - 11.035)^2+(11.46 - 11.035)^2+(17.50 - 11.035)^2\)
\(=(-8.075)^2+(-4.585)^2+(-5.005)^2+(-5.555)^2+(9.425)^2+(-3.555)^2+(0.995)^2+(9.465)^2+(0.425)^2+(6.465)^2\)
\(=65.205625 + 21.022225+25.050025+30.858025+88.830625+12.638025+0.990025+89.596225+0.180625+41.796225\)
\(=375.16765\)
\(s=\sqrt{\frac{375.16765}{9}}\approx6.46\)
\(t=\frac{11.035 - 14.0}{6.46/\sqrt{10}}=\frac{- 2.965}{2.04}\approx - 1.44\)

Step3: Calculate the P - value

For a one - tailed \(t\) - test with \(n-1=9\) degrees of freedom and \(t=-1.44\), we can use a \(t\) - distribution table or statistical software. Using a \(t\) - distribution table (or software like R: pt(-1.44,9)), we find the P - value. The P - value is the probability of getting a \(t\) - statistic as extreme or more extreme than the observed value under the null hypothesis. For \(t=-1.44\) and \(df = 9\), the P - value is approximately \(0.093\).

Answer:

• Null hypothesis (\(H_0\)): \(\mu = 14.0\)
• Alternative hypothesis (\(H_1\)): \(\mu < 14.0\)
• Test statistic: \(-1.44\)
• P - value: \(0.093\)