A marketing survey involves product recognition in New York and California. Of 702 New Yorkers surveyed, 220 knew of the product while 379 of 924 Californians knew of the product. Construct a $99 \%$ confidence interval for the difference between the two population proportions, $p_{1}-p_{2}$ (where $p_{1}$ is the proportion of New Yorkers who knew of the product, and $p_{2}$ is the proportion of Californians who knew of the product).
Give your answers as decimals, rounded to 3 places after the decimal point (if necessary).

Step 1 ：First, we calculate the sample proportions for New Yorkers and Californians. The sample proportion for New Yorkers, denoted as \(p_{1}\), is calculated as \(\frac{220}{702} = 0.313\). The sample proportion for Californians, denoted as \(p_{2}\), is calculated as \(\frac{379}{924} = 0.410\).

Step 2 ：Next, we calculate the difference in sample proportions, \(p_{1} - p_{2}\), which is \(0.313 - 0.410 = -0.097\).

Step 3 ：We then calculate the standard error for the difference in proportions. The standard error, denoted as SE, is calculated as \(\sqrt{\frac{p_{1}(1-p_{1})}{702} + \frac{p_{2}(1-p_{2})}{924}} = 0.024\).

Step 4 ：We use the Z-score for a 99% confidence interval to calculate the margin of error. The Z-score for a 99% confidence interval is approximately 2.576. The margin of error, denoted as ME, is calculated as \(Z \times SE = 2.576 \times 0.024 = 0.061\).

Step 5 ：Finally, we subtract and add this margin of error from the difference in sample proportions to get the confidence interval. The lower limit of the confidence interval is \(-0.097 - 0.061 = -0.158\), and the upper limit of the confidence interval is \(-0.097 + 0.061 = -0.035\).

Step 6 ：So, the 99% confidence interval for the difference between the two population proportions, \(p_{1}-p_{2}\), is \(\boxed{[-0.158, -0.035]}\).