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 484 customer sales invoices shows that 151 out of 484 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 find the 95 percent confidence interval for the proportion of all digital camera customers who have purchased the new warranty, we can follow these steps: 1. **Identify the known values:** - Sample size (\(n\)) = 484 - Number of customers who purchased the warranty (\(x\)) = 151 - Sample proportion (\(\hat{p}\)) = \(\frac{x}{n} = \frac{151}{484}\) 2. **Calculate the sample proportion:** \[ \hat{p} = \frac{151}{484} \approx 0.311 \] 3. **Calculate the standard error (SE) of the sample proportion:** \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.311(1 - 0.311)}{484}} \] 4. **Determine the z-score for a 95% confidence level:** The z-score for a 95% confidence level is approximately 1.96. 5. **Calculate the margin of error (ME):** \[ ME = z \times SE \] 6. **Construct the confidence interval:** \[ \text{Confidence Interval} = \hat{p} \pm ME \] Now, let's perform the calculations step by step. ### Step 3: Calculate the sample proportion \[ \hat{p} = \frac{151}{484} \approx 0.311 \] ### Step 4: Calculate the standard error (SE) \[ SE = \sqrt{\frac{0.311(1 - 0.311)}{484}} = \sqrt{\frac{0.311 \times 0.689}{484}} \approx \sqrt{\frac{0.214199}{484}} \approx \sqrt{0.000442} \approx 0.0210 \] ### Step 5: Calculate the margin of error (ME) \[ ME = 1.96 \times 0.0210 \approx 0.0412 \] ### Step 6: Construct the confidence interval \[ \text{Confidence Interval} = 0.311 \pm 0.0412 \] Calculating the lower and upper bounds: - Lower bound: \(0.311 - 0.0412 \approx 0.2708\) - Upper bound: \(0.311 + 0.0412 \approx 0.3522\) ### Final Confidence Interval Rounding to three decimal places, the 95% confidence interval for the proportion of all digital camera customers who have purchased the new warranty is: \[ (0.271, 0.352) \] ### Conclusion To determine if we are 95% confident that this proportion exceeds 0.20, we can see that both bounds of the confidence interval (0.271 and 0.352) are greater than 0.20. Therefore, we can conclude that we are 95% confident that the proportion of customers who purchased the new warranty exceeds 0.20.