A certain drug is used to treat asthma. In a clinical trial of the drug, 20 of 300 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 10% of treate subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level. What is the final conclusion? 1-PropzTest prop <0.1 z= -1.924500897 p= 0.0271459142 p= 0.0666666667 n= 300

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**Transcription and Explanation of the Statistical Analysis**

**Context:**
A clinical trial was conducted to test the effects of a certain drug used to treat asthma. Out of 300 treated subjects, 20 experienced headaches. The manufacturer claims that less than 10% of treated subjects experience headaches. A hypothesis test is conducted to verify this claim using a significance level of 0.05 and normal distribution as an approximation to the binomial distribution.

**Hypothesis Test:**
The text shows the results from a 1-Proportion Z Test with the following details:

- **Null Hypothesis (\( H_0 \)):** The proportion of treated subjects experiencing headaches is 10% or more.
  
- **Alternative Hypothesis (\( H_a \)):** The proportion of treated subjects experiencing headaches is less than 10%.

**Computer Output Results:**

1. **Test Type:** 1-PropZTest
2. **Proportion Claim:** less than 0.1 (10%)
3. **Z-Score (\( z \)):** 1.924500897 
   - This is the standardized test statistic.
4. **P-Value (\( p \)):** 0.0271459142
   - This is the probability of observing a test statistic as extreme as the one observed, assuming the null hypothesis is true.
5. **Sample Proportion (\( \hat{p} \)):** 0.0666666667
   - Calculated as the number of subjects with headaches divided by the total number of subjects (20/300).
6. **Sample Size (\( n \)):** 300

**Explanation of Results:**

- The **p-value** (0.0271) is less than the significance level of 0.05.
- Because the p-value is less than 0.05, we reject the null hypothesis.

**Conclusion:**
There is sufficient statistical evidence at the 0.05 significance level to support the claim that less than 10% of treated subjects experience headaches when using the drug.
Transcribed Image Text:**Transcription and Explanation of the Statistical Analysis** **Context:** A clinical trial was conducted to test the effects of a certain drug used to treat asthma. Out of 300 treated subjects, 20 experienced headaches. The manufacturer claims that less than 10% of treated subjects experience headaches. A hypothesis test is conducted to verify this claim using a significance level of 0.05 and normal distribution as an approximation to the binomial distribution. **Hypothesis Test:** The text shows the results from a 1-Proportion Z Test with the following details: - **Null Hypothesis (\( H_0 \)):** The proportion of treated subjects experiencing headaches is 10% or more. - **Alternative Hypothesis (\( H_a \)):** The proportion of treated subjects experiencing headaches is less than 10%. **Computer Output Results:** 1. **Test Type:** 1-PropZTest 2. **Proportion Claim:** less than 0.1 (10%) 3. **Z-Score (\( z \)):** 1.924500897 - This is the standardized test statistic. 4. **P-Value (\( p \)):** 0.0271459142 - This is the probability of observing a test statistic as extreme as the one observed, assuming the null hypothesis is true. 5. **Sample Proportion (\( \hat{p} \)):** 0.0666666667 - Calculated as the number of subjects with headaches divided by the total number of subjects (20/300). 6. **Sample Size (\( n \)):** 300 **Explanation of Results:** - The **p-value** (0.0271) is less than the significance level of 0.05. - Because the p-value is less than 0.05, we reject the null hypothesis. **Conclusion:** There is sufficient statistical evidence at the 0.05 significance level to support the claim that less than 10% of treated subjects experience headaches when using the drug.
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