We are tasked with finding a 95% confidence interval for the difference between two population proportions, $(p_1 - p_2)$, where:

• $p_1$ is the proportion of defective bolts from machine A.
• $p_2$ is the proportion of defective bolts from machine B.

Given:

• Sample size from machine A ($n_1$) = 275 bolts
• Defective bolts from machine A ($x_1$) = 22
• Sample size from machine B ($n_2$) = 250 bolts
• Defective bolts from machine B ($x_2$) = 17

Step 1: Calculate the sample proportions

$\hat{p}_1 = \frac{x_1}{n_1} = \frac{22}{275} \approx 0.080$ $\hat{p}_2 = \frac{x_2}{n_2} = \frac{17}{250} = 0.068$

Step 2: Calculate the standard error of the difference between two proportions

The formula for the standard error of the difference between two proportions is:

$SE = \sqrt{\frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2}}$

Substituting the values:

$SE = \sqrt{\frac{0.080(1 - 0.080)}{275} + \frac{0.068(1 - 0.068)}{250}}$ $SE = \sqrt{\frac{0.080(0.920)}{275} + \frac{0.068(0.932)}{250}} \approx \sqrt{\frac{0.0736}{275} + \frac{0.0633}{250}} \approx \sqrt{0.000267 + 0.000253} \approx \sqrt{0.00052} \approx 0.0228$

Step 3: Find the critical value for a 95% confidence interval

For a 95% confidence interval, the critical value $z$ corresponding to 95% is 1.96.

Step 4: Calculate the margin of error

The margin of error is given by:

$ME = z \times SE = 1.96 \times 0.0228 \approx 0.0447$

Step 5: Compute the confidence interval

The difference in sample proportions is:

$\hat{p}_1 - \hat{p}_2 = 0.080 - 0.068 = 0.012$

Now, we calculate the lower and upper limits of the confidence interval:

$\text{Lower limit} = (\hat{p}_1 - \hat{p}_2) - ME = 0.012 - 0.0447 = -0.0327$ $\text{Upper limit} = (\hat{p}_1 - \hat{p}_2) + ME = 0.012 + 0.0447 = 0.0567$

Final Answer:

• Lower limit: -0.033 (rounded to three decimal places)
• Upper limit: 0.057 (rounded to three decimal places)

Would you like any additional details or clarifications?

---

Here are 5 related questions to consider:

1. How would the confidence interval change if the sample sizes were larger?
2. What happens to the margin of error if we use a 99% confidence level instead of 95%?
3. How do we interpret the confidence interval in terms of defective bolts?
4. Can we say machine A produces significantly more defective bolts than machine B based on this interval?
5. How do we calculate confidence intervals for proportions in small sample sizes?

Tip: When constructing confidence intervals, larger sample sizes result in smaller margins of error, leading to more precise estimates.