Question 1168838: Consider a drug testing company that provides a test for marijuana usage. Among 311 tested subjects, results from 25 subjects were wrong (either a false positive or a false negative). Use a 0.01 significance level to test the claim that less than 10 percent of the test results are wrong.
Need to find
Hypothesis test
p-value
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Absolutely, let's break down this hypothesis test step-by-step.
**1. Define the Hypotheses**
* **Null Hypothesis (H₀):** The proportion of wrong test results is 10% or more (p ≥ 0.10).
* **Alternative Hypothesis (H₁):** The proportion of wrong test results is less than 10% (p < 0.10).
This is a left-tailed test.
**2. Set the Significance Level**
* α = 0.01
**3. Calculate the Sample Proportion**
* Sample size (n) = 311
* Number of wrong results (x) = 25
* Sample proportion (p̂) = x/n = 25/311 ≈ 0.0804
**4. Calculate the Test Statistic**
We'll use the z-test for proportions:
$$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$
Where:
* p̂ = sample proportion (0.0804)
* p₀ = hypothesized proportion (0.10)
* n = sample size (311)
$$ z = \frac{0.0804 - 0.10}{\sqrt{\frac{0.10(1-0.10)}{311}}} $$
$$ z = \frac{-0.0196}{\sqrt{\frac{0.09}{311}}} $$
$$ z = \frac{-0.0196}{\sqrt{0.000289389}} $$
$$ z = \frac{-0.0196}{0.01701144} $$
$$ z \approx -1.1522 $$
**5. Calculate the P-value**
Since this is a left-tailed test, we need to find the area to the left of z = -1.1522 in the standard normal distribution.
Using a z-table or calculator, we find:
* P(Z < -1.1522) ≈ 0.1246
Therefore, the p-value is approximately 0.1246.
**6. Make a Decision**
* Compare the p-value (0.1246) with the significance level (0.01).
* Since 0.1246 > 0.01, we fail to reject the null hypothesis.
**7. Conclusion**
There is not sufficient evidence at the 0.01 significance level to support the claim that less than 10 percent of the test results are wrong.
**Summary**
* **Hypothesis test:** Left-tailed z-test for proportions.
* **p-value:** approximately 0.1246.
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