To solve the problems, we first need to determine the sample proportion (p̂) and then calculate the confidence intervals for various confidence levels.
Step 1: Calculate the sample proportion (p̂)
From the sample of 500 students, 420 own a personal computer.
\[ \hat{p} = \frac{x}{n} = \frac{420}{500} = 0.84 \]
Where:
- \(x = 420\) (number of students who own a personal computer)
- \(n = 500\) (total number of students in the sample)
Step 2: Calculate the standard error (SE)
The standard error of the sample proportion is calculated using the formula:
\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.84(1 - 0.84)}{500}} = \sqrt{\frac{0.84 \times 0.16}{500}} = \sqrt{\frac{0.1344}{500}} \approx \sqrt{0.0002688} \approx 0.0164 \]
Step 3: Find the confidence intervals
We will use the formula for the confidence interval for proportions:
\[ CI = \hat{p} \pm Z \times SE \]
Where \(Z\) is the Z-score corresponding to the desired confidence level.
7.1: 99% Confidence Interval
For a 99% confidence level, \(Z\) is approximately 2.576.
\[ CI = 0.84 \pm 2.576 \times 0.0164 \] \[ CI = 0.84 \pm 0.0423 \] \[ CI = (0.7977, 0.8823) \]
7.2: 95% Confidence Interval
For a 95% confidence level, \(Z\) is approximately 1.96.
\[ CI = 0.84 \pm 1.96 \times 0.0164 \] \[ CI = 0.84 \pm 0.0321 \] \[ CI = (0.8079, 0.8721) \]
7.3: 90% Confidence Interval
For a 90% confidence level, \(Z\) is approximately 1.645.
\[ CI = 0.84 \pm 1.645 \times 0.0164 \] \[ CI = 0.84 \pm 0.0269 \] \[ CI = (0.8131, 0.8669) \]
Summary of Confidence Intervals
- 99% CI: (0.7977, 0.8823)
- 95% CI: (0.8079, 0.8721)
- 90% CI: (0.8131, 0.8669)
7.4: Standard Error for the 99% CI
The standard error calculated earlier (for the 99% CI) is:
\[ SE \approx 0.0164 \]
Thus, the answers are:
- 7.1: (0.7977, 0.8823)
- 7.2: (0.8079, 0.8721)
- 7.3: (0.8131, 0.8669)
- 7.4: 0.0164