Consumer Reports (January 2005) indicates that profit margins on extended warranties are much greater than on the purchase of most products. In this exercise we consider a major electronics retailer that wishes to increase the proportion of customers who buy extended warranties on digital cameras. Historically, 20 percent of digital camera customers have purchased the retailer’s extended warranty. To increase this percentage, the retailer has decided to offer a new warranty that is less expensive and more comprehensive. Suppose that three months after starting to offer the new warranty, a random sample of 535 customer sales invoices shows that 169 out of 535 digital camera customers purchased the new warranty. Find a 95 percent confidence interval for the proportion of all digital camera customers who have purchased the new warranty. Are we 95 percent confident that this proportion exceeds .20? (Round your answers to 3 decimal places.)

To solve this problem, we need to construct a 95% confidence interval for the proportion of all digital camera customers who purchased the new warranty. The steps involved are as follows:

### Step 1: Define the sample proportion
The sample proportion \( \hat{p} \) is given by:

\[
\hat{p} = \frac{x}{n}
\]

Where:
- \( x = 169 \) (the number of customers who purchased the new warranty)
- \( n = 535 \) (the total number of customers in the sample)

\[
\hat{p} = \frac{169}{535} \approx 0.316
\]

### Step 2: Find the standard error of the proportion
The standard error (SE) for the proportion is calculated using the formula:

\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]

Substituting the values:

\[
SE = \sqrt{\frac{0.316(1 - 0.316)}{535}} = \sqrt{\frac{0.316 \times 0.684}{535}} \approx 0.020
\]

### Step 3: Find the critical value for a 95% confidence interval
For a 95% confidence interval, the critical value (z-value) from the standard normal distribution is 1.96.

### Step 4: Calculate the margin of error
The margin of error (ME) is given by:

\[
ME = z \times SE = 1.96 \times 0.020 \approx 0.039
\]

### Step 5: Construct the confidence interval
The 95% confidence interval for the proportion is:

\[
\hat{p} \pm ME = 0.316 \pm 0.039
\]

Thus, the confidence interval is:

\[
(0.316 - 0.039, 0.316 + 0.039) = (0.277, 0.355)
\]

### Step 6: Interpret the result
We are 95% confident that the true proportion of all digital camera customers who have purchased the new warranty is between 0.277 and 0.355.

### Step 7: Compare with 0.20
Since the lower bound of the confidence interval (0.277) is greater than 0.20, we are 95% confident that the proportion of customers who purchased the new warranty is greater than 0.20.

### Conclusion:
Yes, we are 95% confident that the proportion of customers who purchased the new warranty exceeds 0.20.

Would you like any more details on the process or have any questions?

---

Here are 5 related questions you might consider:
1. What is the formula for a confidence interval for proportions, and why is it used?
2. How would the confidence interval change if we used a different confidence level (e.g., 90%)?
3. What is the impact of sample size on the width of a confidence interval?
4. How can we test whether the increase in warranty purchases is statistically significant?
5. What assumptions do we make when constructing a confidence interval for proportions?

**Tip:** Increasing the sample size reduces the margin of error, which leads to a narrower confidence interval.

Learn how to calculate a 95% confidence interval for the proportion of digital camera customers purchasing a new extended warranty. The problem involves key concepts like sample proportion, margin of error, and standard error, and is suitable for students in Grades 11-12 or college statistics courses.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation