Questions: 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. 2.98, 6.54, 5.99, 5.46, 20.47, 7.46, 11.98, 20.53, 11.47, 17.54 Identify the null and alternative hypotheses. H₀: μ=14 H₁: μ<14 Identify the test statistic. -1.45 Identify the P-value. 0.091 State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. the null hypothesis. There is not sufficient evidence at the 0.01 significance level to support the claim that the mean lead concentration for all Ayurveda medicines manufactured in the United States is less than 14.0 μg/g.

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. 2.98, 6.54, 5.99, 5.46, 20.47, 7.46, 11.98, 20.53, 11.47, 17.54 Identify the null and alternative hypotheses. H₀: μ=14 H₁: μ<14 Identify the test statistic. -1.45 Identify the P-value. 0.091 State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. the null hypothesis. There is not sufficient evidence at the 0.01 significance level to support the claim that the mean lead concentration for all Ayurveda medicines manufactured in the United States is less than 14.0 μg/g.
Transcript text: Listed below are the lead concentrations (in $\mu \mathrm{g} / \mathrm{g}$ ) 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 $14.0 \mu \mathrm{~g} / \mathrm{g}$. \[ \begin{array}{llllllllll} 2.98 & 6.54 & 5.99 & 5.46 & 20.47 & 7.46 & 11.98 & 20.53 & 11.47 & 17.54 \end{array} \] Identify the null and alternative hypotheses. \[ \begin{array}{l} \mathrm{H}_{0}: \mu=14 \\ \mathrm{H}_{1}: \mu=14 \end{array} \] Identify the test statistic. \[ -1.45 \] Identify the P -value. \[ 0.091 \] State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. $\square$ the null hypothesis. There $\square$ sufficient evidence at the 0.01 significance level to $\square$ the claim that the mean lead concentration for all Ayurveda medicines manufactured in the United States is less than $14.0 \mu \mathrm{~g} / \mathrm{g}$.
failed

Solution

failed
failed

Solution Steps

Step 1: Calculate Sample Mean and Standard Deviation

The lead concentrations measured in different Ayurveda medicines are: \[ \{2.98, 6.54, 5.99, 5.46, 20.47, 7.46, 11.98, 20.53, 11.47, 17.54\} \] The sample mean \( \bar{x} \) is calculated as: \[ \bar{x} = \frac{\sum x_i}{n} = \frac{2.98 + 6.54 + 5.99 + 5.46 + 20.47 + 7.46 + 11.98 + 20.53 + 11.47 + 17.54}{10} = 11.042 \] The sample standard deviation \( s \) is calculated as: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \approx 6.4719 \]

Step 2: Calculate Standard Error

The standard error \( SE \) is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{6.4719}{\sqrt{10}} \approx 2.0466 \]

Step 3: Calculate Test Statistic

The test statistic \( t \) for the left-tailed test is calculated as: \[ t = \frac{\bar{x} - \mu_0}{SE} = \frac{11.042 - 14.0}{2.0466} \approx -1.4453 \]

Step 4: Calculate P-value

For a left-tailed test, the P-value is calculated as: \[ P = T(t) \approx 0.0911 \]

Step 5: Conclusion about the Null Hypothesis

The null hypothesis \( H_0: \mu = 14 \) is tested against the alternative hypothesis \( H_1: \mu < 14 \). Given that the significance level \( \alpha = 0.01 \) and the P-value \( P \approx 0.0911 \):

  • Since \( P > \alpha \), we fail to reject the null hypothesis.

Final Answer

Fail to reject the null hypothesis. There is not sufficient evidence at the \( 0.01 \) significance level to support the claim that the mean lead concentration for all Ayurveda medicines manufactured in the United States is less than \( 14.0 \, \mu g/g \).

\(\boxed{\text{Fail to reject } H_0}\)

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
For a standard normal distribution, find: P(z>2.27) Express the probability as a decimal rounded to 4 decimal places.
failedAnswered
The data represent the age of world leaders on their day of inauguration. Find the five-number summary, and construct a boxplot for the data. Comment on the shape of the distribution. 57, 65, 44, 69, 65, 67, 52, 54, 53, 43, 48, 62, 67, 61, 63 The five-number summary is Choose the correct boxplot of the data below: A B. C. Choose the correct description of the shape of the distribution. A. The distribution is skewed to the left. B. The distribution is skewed to the right. C. The distribution is roughly symmetric. D. The shape of the distribution cannot be determined from the boxplot.
failedAnswered
All of the following are true about job interviews, except: Interview questions should target how candidates previously performed in a similar job position Interview structure should inform job candidates about job expectations and the work culture Interview performance is a poor predictor of on-the-job performance and length of employment Interview questions should be unstructured to allow for conversation flow between the interviewer and the job candidate
failedAnswered
m=(y2-y1)/(x2-x1) so (6-2)/(-4-2)=-4/6
failedAnswered
Find the prime factorization of 245.