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?
7.4 what is the standard error of the interval of question number 7.1

Answer 1

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