Question
7. The president of a university would like to estimate the proportion of the student population that owns a personal computer. In a sample of 500 students, 420 own a personal computer.
7.1 Find the 99% confidence interval for the proportion of the student population who own a personal computer?
7.2 Find the 95% confidence interval for the proportion of the student population who own a personal computer?
7.3 Find the 90% confidence interval for the proportion of the student population who own a personal computer?
7.1 Find the 99% confidence interval for the proportion of the student population who own a personal computer?
7.2 Find the 95% confidence interval for the proportion of the student population who own a personal computer?
7.3 Find the 90% confidence interval for the proportion of the student population who own a personal computer?
Answers
GPT-4o mini
To calculate the confidence intervals for the proportion of the student population that owns a personal computer based on the provided sample data, we can use the formula for confidence intervals for a population proportion:
\[
\text{Confidence Interval} = \hat{p} \pm z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
Where:
- \(\hat{p}\) is the sample proportion
- \(n\) is the sample size
- \(z\) is the z-score associated with the desired confidence level
Given:
- Number of students in the sample \(n = 500\)
- Number of students who own a personal computer = 420
- Sample proportion \(\hat{p} = \frac{420}{500} = 0.84\)
### Step 1: Calculate the sample proportion and standard error.
1. **Sample Proportion**:
\(\hat{p} = \frac{420}{500} = 0.84\)
2. **Standard Error (SE)**:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.84(1 - 0.84)}{500}} = \sqrt{\frac{0.84 \cdot 0.16}{500}} \approx \sqrt{\frac{0.1344}{500}} \approx \sqrt{0.0002688} \approx 0.0164
\]
### Step 2: Find the confidence intervals for various confidence levels.
#### 7.1: 99% Confidence Interval
For a 99% confidence level, the z-score is approximately 2.576.
\[
\text{Margin of Error} = z \cdot SE = 2.576 \cdot 0.0164 \approx 0.0424
\]
\[
\text{Confidence Interval} = 0.84 \pm 0.0424 \Rightarrow (0.7976, 0.8824)
\]
#### 7.2: 95% Confidence Interval
For a 95% confidence level, the z-score is approximately 1.96.
\[
\text{Margin of Error} = 1.96 \cdot SE = 1.96 \cdot 0.0164 \approx 0.0322
\]
\[
\text{Confidence Interval} = 0.84 \pm 0.0322 \Rightarrow (0.8078, 0.8722)
\]
#### 7.3: 90% Confidence Interval
For a 90% confidence level, the z-score is approximately 1.645.
\[
\text{Margin of Error} = 1.645 \cdot SE = 1.645 \cdot 0.0164 \approx 0.0269
\]
\[
\text{Confidence Interval} = 0.84 \pm 0.0269 \Rightarrow (0.8131, 0.8669)
\]
### Final Results:
- **99% Confidence Interval**: \( (0.7976, 0.8824) \)
- **95% Confidence Interval**: \( (0.8078, 0.8722) \)
- **90% Confidence Interval**: \( (0.8131, 0.8669) \)
\[
\text{Confidence Interval} = \hat{p} \pm z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
Where:
- \(\hat{p}\) is the sample proportion
- \(n\) is the sample size
- \(z\) is the z-score associated with the desired confidence level
Given:
- Number of students in the sample \(n = 500\)
- Number of students who own a personal computer = 420
- Sample proportion \(\hat{p} = \frac{420}{500} = 0.84\)
### Step 1: Calculate the sample proportion and standard error.
1. **Sample Proportion**:
\(\hat{p} = \frac{420}{500} = 0.84\)
2. **Standard Error (SE)**:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.84(1 - 0.84)}{500}} = \sqrt{\frac{0.84 \cdot 0.16}{500}} \approx \sqrt{\frac{0.1344}{500}} \approx \sqrt{0.0002688} \approx 0.0164
\]
### Step 2: Find the confidence intervals for various confidence levels.
#### 7.1: 99% Confidence Interval
For a 99% confidence level, the z-score is approximately 2.576.
\[
\text{Margin of Error} = z \cdot SE = 2.576 \cdot 0.0164 \approx 0.0424
\]
\[
\text{Confidence Interval} = 0.84 \pm 0.0424 \Rightarrow (0.7976, 0.8824)
\]
#### 7.2: 95% Confidence Interval
For a 95% confidence level, the z-score is approximately 1.96.
\[
\text{Margin of Error} = 1.96 \cdot SE = 1.96 \cdot 0.0164 \approx 0.0322
\]
\[
\text{Confidence Interval} = 0.84 \pm 0.0322 \Rightarrow (0.8078, 0.8722)
\]
#### 7.3: 90% Confidence Interval
For a 90% confidence level, the z-score is approximately 1.645.
\[
\text{Margin of Error} = 1.645 \cdot SE = 1.645 \cdot 0.0164 \approx 0.0269
\]
\[
\text{Confidence Interval} = 0.84 \pm 0.0269 \Rightarrow (0.8131, 0.8669)
\]
### Final Results:
- **99% Confidence Interval**: \( (0.7976, 0.8824) \)
- **95% Confidence Interval**: \( (0.8078, 0.8722) \)
- **90% Confidence Interval**: \( (0.8131, 0.8669) \)