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Topic 6 Homework (Nonadaptive)
Question 10 of 11 (1 point)|Question Attempt: 1 of Unlimited

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Question 10
A random sample of  275 bolts from machine A contained 22 defective bolts, while an independently chosen, random sample of 250 bolts from machine B contained 17 defective bolts. Let  p1 be the proportion of the population of all bolts from machine A that are defective, and let  p2 be the proportion of the population of all bolts from machine B that are defective. Find a 95% confidence interval for −p1p2. Then find the lower limit and upper limit of the 95% confidence interval.
Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. (If necessary, consult a list of formulas.)

Lower limit:
Upper limit:

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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.

A random sample of 275 bolts from machine A contained 22 defective bolts, while an independently chosen, random sample of 250 bolts from machine B contained 17 defective bolts. Let p1 be the proportion of the population of all bolts from machine A that are defective, and let p2 be the proportion of the population of all bolts from machine B that are defective. Find a 95% confidence interval for −p1p2. Then find the lower limit and upper limit of the 95% confidence interval.

Calculate a 95% confidence interval for the difference between two proportions of defective bolts from two machines. This problem involves statistical concepts such as sample proportions, standard error, and the Central Limit Theorem, suitable for advanced high school students (Grades 10-12).

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