SOLUTION: Listed below are the lead concentrations in ug/g measured in different traditional medicines. Use a 0.01 significance level to test the claim that the mean lead concentration f
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Question 1168842: Listed below are the lead concentrations in ug/g measured in different traditional medicines. Use a 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 16 ug/g. Assume that the sample is a simple random sample.
5 125 19.5 22.5 4.5 5.5 10.5 13 8.5 22
Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses?
nned to find test statistic
and p-value
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's perform the hypothesis test step-by-step.
**1. Define the Hypotheses**
* **Null Hypothesis (H₀):** The mean lead concentration is greater than or equal to 16 ug/g.
* H₀: µ ≥ 16
* **Alternative Hypothesis (H₁):** The mean lead concentration is less than 16 ug/g.
* H₁: µ < 16 (This is a left-tailed test)
**2. Set the Significance Level**
* α = 0.01
**3. Calculate the Sample Statistics**
* Data: 5, 125, 19.5, 22.5, 4.5, 5.5, 10.5, 13, 8.5, 22
* Sample size (n) = 10
* **Sample Mean (x̄):**
* Sum of the data: 5 + 125 + 19.5 + 22.5 + 4.5 + 5.5 + 10.5 + 13 + 8.5 + 22 = 236
* x̄ = 236 / 10 = 23.6
* **Sample Standard Deviation (s):**
* Using a calculator or statistical software, we find s ≈ 36.88
**4. Calculate the Test Statistic (t)**
* Since the population standard deviation is unknown and the sample size is small (n < 30), we use a t-test.
* Formula: t = (x̄ - µ) / (s / √n)
* Calculation:
* t = (23.6 - 16) / (36.88 / √10)
* t = 7.6 / (36.88 / 3.162278)
* t = 7.6 / 11.662
* t ≈ 0.6517
**5. Calculate the Degrees of Freedom (df)**
* df = n - 1 = 10 - 1 = 9
**6. Calculate the P-value**
* Using a t-distribution table or statistical software, we find the p-value for a left-tailed t-test with t ≈ 0.6517 and df = 9.
* Because the t value is positive, and we are looking for a left tailed test, the p-value will be larger than 0.5.
* Using a t-table or calculator, the p-value is approximately 0.738
**7. Make a Decision**
* Compare the p-value (0.738) to the significance level (0.01).
* Since 0.738 > 0.01, we fail to reject the null hypothesis.
**8. Draw a Conclusion**
* There is not enough evidence to reject the claim that the mean lead concentration is greater than or equal to 16 ug/g at the 0.01 significance level.
* Therefore, we cannot support the claim that the mean lead concentration is less than 16 ug/g.
**Summary**
* **Test Statistic (t):** ≈ 0.6517
* **P-value:** ≈ 0.738
* **Decision:** Fail to reject the null hypothesis.
* **Conclusion:** The data does not provide sufficient evidence to conclude that the mean lead concentration is less than 16 ug/g.
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