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?

1 answer

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: \( (0.8131, 0.8669) \)