A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 656 babies were born, and 328 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 656 babies were born, and 328 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?
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![## Clinical Trial for Increasing Probability of Conceiving a Girl: An Analysis
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study, 656 babies were born, and 328 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?
### Constructing the Confidence Interval
The problem requires the construction of a 99% confidence interval for the proportion \( p \) of girls born. The interval will be of the form:
\[ \text{Confidence Interval} = \left( \hat{p} - E, \hat{p} + E \right) \]
Where:
- \( \hat{p} \) is the sample proportion (\( \hat{p} = \frac{\text{number of girls}}{\text{total number of babies}} \)).
- \( E \) is the margin of error calculated as \( E = Z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( Z \) is the z-score corresponding to the confidence level and \( n \) is the total number of babies.
### Data Provided:
- **Total number of babies born (n)**: 656
- **Number of girls (x)**: 328
### Calculation
1. **Sample Proportion (\( \hat{p} \))**:
\[
\hat{p} = \frac{328}{656} = 0.5
\]
2. **Z-Score for a 99% Confidence Level**:
The Z-score corresponding to a 99% confidence interval is approximately 2.576.
3. **Margin of Error (E)**:
\[
E = 2.576 \cdot \sqrt{\frac{0.5 \cdot (1 - 0.5)}{656}} = 2.576 \cdot \sqrt{\frac{0.25}{656}} \approx 2.576 \cdot 0.0196 = 0.0505
\]
4. **Confidence Interval**:
\[
\text{Confidence Interval} = (0.5 - 0.0505, 0.5 + 0.0505) = (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e4f48b8-ad6a-4fd5-bfc6-01337edf5aa9%2Fb1453405-e7d0-4065-9d06-8d58ccc53595%2Fxdkgay_processed.png&w=3840&q=75)
Transcribed Image Text:## Clinical Trial for Increasing Probability of Conceiving a Girl: An Analysis
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study, 656 babies were born, and 328 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?
### Constructing the Confidence Interval
The problem requires the construction of a 99% confidence interval for the proportion \( p \) of girls born. The interval will be of the form:
\[ \text{Confidence Interval} = \left( \hat{p} - E, \hat{p} + E \right) \]
Where:
- \( \hat{p} \) is the sample proportion (\( \hat{p} = \frac{\text{number of girls}}{\text{total number of babies}} \)).
- \( E \) is the margin of error calculated as \( E = Z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( Z \) is the z-score corresponding to the confidence level and \( n \) is the total number of babies.
### Data Provided:
- **Total number of babies born (n)**: 656
- **Number of girls (x)**: 328
### Calculation
1. **Sample Proportion (\( \hat{p} \))**:
\[
\hat{p} = \frac{328}{656} = 0.5
\]
2. **Z-Score for a 99% Confidence Level**:
The Z-score corresponding to a 99% confidence interval is approximately 2.576.
3. **Margin of Error (E)**:
\[
E = 2.576 \cdot \sqrt{\frac{0.5 \cdot (1 - 0.5)}{656}} = 2.576 \cdot \sqrt{\frac{0.25}{656}} \approx 2.576 \cdot 0.0196 = 0.0505
\]
4. **Confidence Interval**:
\[
\text{Confidence Interval} = (0.5 - 0.0505, 0.5 + 0.0505) = (
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