Question
A mail-order company has two warehouses, one on the West Cost and one on the East Coast. The policy of the company is to mail all orders placed with it within 48 hrs. A recently taken sample of 450 orders placed with the warehouse on the West Coast showed that 370 of them mailed within 48 hrs. Another sample of 370 orders placed with the warehouse on the East Coast showed 280 of them were mailed within 48 hrs. Construct a 90% confidence interval for the difference between the population proportions. (p_1-p_2)
Answers
GPT-4o mini
To construct a 90% confidence interval for the difference between the population proportions (p1 - p2), we follow these steps:
1. **Define the sample proportions and sizes.**
- For the West Coast warehouse (Sample 1):
- \( n_1 = 450 \)
- \( x_1 = 370 \)
- Sample proportion \( \hat{p}_1 = \frac{x_1}{n_1} = \frac{370}{450} \approx 0.8222 \)
- For the East Coast warehouse (Sample 2):
- \( n_2 = 370 \)
- \( x_2 = 280 \)
- Sample proportion \( \hat{p}_2 = \frac{x_2}{n_2} = \frac{280}{370} \approx 0.7568 \)
2. **Calculate the difference between the sample proportions.**
- \( \hat{p}_1 - \hat{p}_2 \approx 0.8222 - 0.7568 \approx 0.0654 \)
3. **Calculate the standard error (SE) for the difference in proportions.**
\[
\text{SE} = \sqrt{\frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2}}
\]
- Calculate \( \hat{p}_1(1 - \hat{p}_1) \):
\[
\hat{p}_1(1 - \hat{p}_1) = 0.8222 \times (1 - 0.8222) = 0.8222 \times 0.1778 \approx 0.1461
\]
- Calculate \( \hat{p}_2(1 - \hat{p}_2) \):
\[
\hat{p}_2(1 - \hat{p}_2) = 0.7568 \times (1 - 0.7568) = 0.7568 \times 0.2432 \approx 0.1845
\]
- Substitute into the SE formula:
\[
\text{SE} = \sqrt{\frac{0.1461}{450} + \frac{0.1845}{370}} \approx \sqrt{0.0003258 + 0.0004986} \approx \sqrt{0.0008244} \approx 0.0287
\]
4. **Determine the critical z-value for a 90% confidence interval.**
For a 90% confidence level, the critical z-value (z*) can be found using a z-table:
- \( z^* \approx 1.645 \)
5. **Calculate the margin of error (ME).**
\[
\text{ME} = z^* \times \text{SE} = 1.645 \times 0.0287 \approx 0.0472
\]
6. **Construct the confidence interval.**
\[
\text{Confidence interval} = (\hat{p}_1 - \hat{p}_2) \pm \text{ME} = 0.0654 \pm 0.0472
\]
- Lower limit: \( 0.0654 - 0.0472 \approx 0.0182 \)
- Upper limit: \( 0.0654 + 0.0472 \approx 0.1126 \)
Thus, the 90% confidence interval for the difference between the population proportions (p1 - p2) is:
\[
\boxed{(0.0182, 0.1126)}
\]
1. **Define the sample proportions and sizes.**
- For the West Coast warehouse (Sample 1):
- \( n_1 = 450 \)
- \( x_1 = 370 \)
- Sample proportion \( \hat{p}_1 = \frac{x_1}{n_1} = \frac{370}{450} \approx 0.8222 \)
- For the East Coast warehouse (Sample 2):
- \( n_2 = 370 \)
- \( x_2 = 280 \)
- Sample proportion \( \hat{p}_2 = \frac{x_2}{n_2} = \frac{280}{370} \approx 0.7568 \)
2. **Calculate the difference between the sample proportions.**
- \( \hat{p}_1 - \hat{p}_2 \approx 0.8222 - 0.7568 \approx 0.0654 \)
3. **Calculate the standard error (SE) for the difference in proportions.**
\[
\text{SE} = \sqrt{\frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2}}
\]
- Calculate \( \hat{p}_1(1 - \hat{p}_1) \):
\[
\hat{p}_1(1 - \hat{p}_1) = 0.8222 \times (1 - 0.8222) = 0.8222 \times 0.1778 \approx 0.1461
\]
- Calculate \( \hat{p}_2(1 - \hat{p}_2) \):
\[
\hat{p}_2(1 - \hat{p}_2) = 0.7568 \times (1 - 0.7568) = 0.7568 \times 0.2432 \approx 0.1845
\]
- Substitute into the SE formula:
\[
\text{SE} = \sqrt{\frac{0.1461}{450} + \frac{0.1845}{370}} \approx \sqrt{0.0003258 + 0.0004986} \approx \sqrt{0.0008244} \approx 0.0287
\]
4. **Determine the critical z-value for a 90% confidence interval.**
For a 90% confidence level, the critical z-value (z*) can be found using a z-table:
- \( z^* \approx 1.645 \)
5. **Calculate the margin of error (ME).**
\[
\text{ME} = z^* \times \text{SE} = 1.645 \times 0.0287 \approx 0.0472
\]
6. **Construct the confidence interval.**
\[
\text{Confidence interval} = (\hat{p}_1 - \hat{p}_2) \pm \text{ME} = 0.0654 \pm 0.0472
\]
- Lower limit: \( 0.0654 - 0.0472 \approx 0.0182 \)
- Upper limit: \( 0.0654 + 0.0472 \approx 0.1126 \)
Thus, the 90% confidence interval for the difference between the population proportions (p1 - p2) is:
\[
\boxed{(0.0182, 0.1126)}
\]