Math Problem Statement

Listed below are the lead concentrations​ (in mu g divided by 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 g divided by g. 2.97 6.50 5.97 5.50 20.49 7.51 12.03 20.49 11.54 17.53

Question content area bottom Part 1 Identify the null and alternative hypotheses. Upper H 0​: mu equals    14.0 Upper H 1​: mu less than    14.0 ​(Type integers or decimals. Do not​ round.) Part 2 Identify the test statistic.    negative 1.44 ​(Round to two decimal places as​ needed.) Part 3 Identify the​ P-value.    enter your response here ​(Round to three decimal places as​ needed.)

Solution

Part 1: Hypotheses

We are testing the claim that the mean lead concentration for all such medicines is less than 14.0μg/g14.0 \, \mu g/g.

  • Null Hypothesis H0H_0: H0:μ=14.0μg/gH_0: \mu = 14.0 \, \mu g/g
  • Alternative Hypothesis H1H_1: H1:μ<14.0μg/gH_1: \mu < 14.0 \, \mu g/g

This is a one-tailed test, as we are interested in determining if the mean is less than 14.0.

Part 2: Test Statistic

You are given that the test statistic is t=1.44t = -1.44. This value was likely calculated using the formula for the t-test statistic for a one-sample t-test:

t=xˉμ0snt = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} where:

  • xˉ\bar{x} is the sample mean,
  • μ0\mu_0 is the hypothesized population mean (14.0 in this case),
  • ss is the sample standard deviation,
  • nn is the sample size.

Part 3: P-value

The P-value is the probability that the test statistic t=1.44t = -1.44 would occur under the null hypothesis. Since this is a left-tailed test, we need to find the probability that a t-statistic is less than 1.44-1.44 for a certain number of degrees of freedom.

We can calculate the P-value using a t-distribution table or a statistical software. Let's break it down:

  • The sample size n=10n = 10,
  • Degrees of freedom df=n1=9df = n - 1 = 9.

Using a t-distribution table or calculator, the P-value corresponding to t=1.44t = -1.44 with 9 degrees of freedom is approximately:

P-value0.089P\text{-value} \approx 0.089

Conclusion:

  • If the P-value 0.089\approx 0.089 is greater than the significance level α=0.01\alpha = 0.01, we fail to reject the null hypothesis.
  • This means there is not enough evidence at the 0.010.01 significance level to conclude that the mean lead concentration is less than 14.0 μg/g\mu g/g.

Would you like more details or clarification on any part?

5 Related Questions:

  1. How is the t-test statistic calculated for one-sample tests?
  2. What assumptions must be met for conducting a t-test?
  3. How can we interpret the P-value in hypothesis testing?
  4. How would the conclusion change if the P-value were smaller than 0.01?
  5. What is the difference between a one-tailed and two-tailed t-test?

Tip:

Always ensure your data meets the assumptions of normality and random sampling when performing a t-test to ensure valid results.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
One-sample t-test

Formulas

t = (x̄ - μ) / (s / sqrt(n))
Degrees of freedom: df = n - 1

Theorems

T-distribution Theorem
Central Limit Theorem

Suitable Grade Level

College-level (Undergraduate Statistics)