To compute the 95% confidence interval for the proportion of letters delivered the day after they were mailed, we use the formula for a confidence interval for a population proportion:

$CI = \hat{p} \pm z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

• $\hat{p}$ is the sample proportion of success (letters delivered the next day),
• $z$ is the z-score corresponding to the confidence level (for 95%, $z = 1.96$),
• $n$ is the sample size.

Step 1: Calculate the sample proportion ($\hat{p}$).

$\hat{p} = \frac{177}{250} = 0.708$

Step 2: Calculate the standard error (SE).

$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.708(1 - 0.708)}{250}} = \sqrt{\frac{0.708 \times 0.292}{250}} = \sqrt{\frac{0.206736}{250}} = \sqrt{0.000827} \approx 0.02876$

Step 3: Calculate the margin of error (ME).

$ME = z \times SE = 1.96 \times 0.02876 = 0.05636$

Step 4: Calculate the confidence interval.

$CI = \hat{p} \pm ME = 0.708 \pm 0.05636$

• Lower limit = $0.708 - 0.05636 = 0.6516$ → Rounded to 0.65
• Upper limit = $0.708 + 0.05636 = 0.7644$ → Rounded to 0.76

Final Answer:

• Lower limit: 0.65
• Upper limit: 0.76

Would you like to see more details on any specific step or have any questions?

Here are some follow-up questions:

1. What is the importance of using a confidence interval?
2. How does increasing the sample size affect the confidence interval?
3. Why do we use the z-score in confidence intervals?
4. How would the confidence interval change with a different confidence level, like 99%?
5. How does the sample proportion impact the width of the confidence interval?

Tip: Always ensure that the sample size is sufficiently large when applying the formula for confidence intervals in proportion problems.