91Ó°ÊÓ

Assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim. 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 (based on data from "Lead, Mercury, and Arsenic in US and Indian Manufactured Ayurvedic Medicines Sold via the Internet," by Saper et al., Journal of the American Medical Association, Vol. 300, No. 8). Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than \(14 \mu \mathrm{g} / \mathrm{g}\).\(\begin{array}{cccccccccc}3.0 & 6.5 & 6.0 & 5.5 & 20.5 & 7.5 & 12.0 & 20.5 & 11.5 & 17.5\end{array}\)

Step by step solution

01

Identify the null and alternative hypotheses

To test the claim that the mean lead concentration is less than 14 \(\frac{\mu g}{g}\), set up the hypotheses as follows:Null Hypothesis (\(H_0\)): \( \mu \geq 14 \)Alternative Hypothesis (\(H_1\)): \( \mu < 14 \)

02

Calculate the sample mean and standard deviation

First, calculate the sample mean (\( \bar{x} \)) and the sample standard deviation (\( s \)) for the given data:Data: 3.0, 6.5, 6.0, 5.5, 20.5, 7.5, 12.0, 20.5, 11.5, 17.5Sample Mean: \(\bar{x} = \frac{\sum x_i}{n} = \frac{111.5}{10} = 11.15 \)Sample Standard Deviation: \(s = 6.032\)

03

Calculate the test statistic

Use the t-test for the mean since the population standard deviation is not known. The test statistic is calculated as:\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} = \frac{11.15 - 14}{\frac{6.032}{\sqrt{10}}} = -1.48 \]

04

Determine the critical value or P-value

Using a significance level of 0.05 and a t-distribution with \(n-1=10-1=9\) degrees of freedom, find the critical value for a one-tailed test. The critical value (\(t_{0.05, 9}\)) is approximately \(-1.833 \).Alternatively, the P-value for the test statistic \(-1.48\) can be found using statistical tables or software, which gives a P-value of approximately 0.087.

05

Compare the test statistic and make a conclusion

Since the test statistic \(-1.48\) is greater than the critical value \(-1.833\) and the P-value (0.087) is greater than the significance level (0.05), we do not reject the null hypothesis. Thus, there is not enough evidence to support the claim that the mean lead concentration for all such medicines is less than \(14 \mu \mathrm{g}/\mathrm{g}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

null hypothesis

In hypothesis testing, the null hypothesis (\(H_0\)) is a statement that assumes no effect or no difference in the context of the question being tested. For example, in the context of the given exercise about lead concentration in Ayurveda medicines, the null hypothesis is that the mean lead concentration is greater than or equal to 14 \(\mu \mathrm{g}/\mathrm{g}\). This serves as the default or starting assumption and is what we're trying to find evidence against. It's important to note that the null hypothesis is not necessarily what you believe to be true, but rather a setup for statistical testing. Think of it as the 'status quo' that you aim to challenge with your data findings.

alternative hypothesis

The alternative hypothesis (\(H_1\)) is the statement that you want to test for. It represents a new effect or difference that you are trying to detect. In our example, this hypothesis states that the mean lead concentration is less than 14 \(\mu \mathrm{g}/\mathrm{g}\). This contradicts the null hypothesis. If data provides enough evidence against the null hypothesis, it supports the alternative hypothesis. Essentially, it's what you want to prove.

t-test

When you're dealing with sample data and don't know the population standard deviation, the t-test is a good choice to determine if there's a significant difference between the sample mean and the hypothesized population mean. In this case, since we're examining whether the mean lead concentration is below a certain level, we use the t-test formula:\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]where \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. The t-test helps you understand if observed data significantly deviates from the hypothesized mean when standard deviation is unknown.

P-value

The P-value quantifies how likely it is to observe data at least as extreme as what we obtained, assuming the null hypothesis is true. In this context, a P-value tells you how probable it is to get a sample mean of 11.15 when the true mean is 14 \(\mu \mathrm{g}/\mathrm{g}\). For our exercise, the P-value is approximately 0.087. This means there's an 8.7% probability of observing a sample mean of 11.15 or less if the true mean were 14. Lower P-values (typically < 0.05) indicate stronger evidence against the null hypothesis.

significance level

The significance level, denoted by \(\alpha\), is the criterion used to decide whether to reject the null hypothesis. It represents the probability of making a Type I error, which occurs when you reject a true null hypothesis. In our example, the significance level is set at 0.05. This means there's a 5% risk of concluding that the mean lead concentration is less than 14 \(\mu \mathrm{g}/\mathrm{g}\) when it is actually equal to or greater than 14. You compare the P-value to your significance level to make your decision. If the P-value is less than 0.05, you reject the null hypothesis; otherwise, you do not reject it.