You wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.01$. \[ \begin{array}{l} H_{o}: p_{1}=p_{2} \\ H_{a}: p_{1} \neq p_{2} \end{array} \] You obtain a sample from the first population with 56 successes out of 488 trials. You obtain a sample from the second population with 64 successes and 425 trials. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic $=$ What is the p-value for this sample? (Report answer accurate to four decimal places.) \[ \text { p-value }= \]

Step 1 ：Given data: number of successes in the first sample \(x_1 = 56\), size of the first sample \(n_1 = 488\), number of successes in the second sample \(x_2 = 64\), size of the second sample \(n_2 = 425\).

Step 2 ：Calculate the sample proportions: \(p_{1_{hat}} = \frac{x_1}{n_1} = 0.11475409836065574\), \(p_{2_{hat}} = \frac{x_2}{n_2} = 0.15058823529411763\).

Step 3 ：Calculate the pooled sample proportion: \(p_{hat} = \frac{x_1 + x_2}{n_1 + n_2} = 0.13143483023001096\).

Step 4 ：Calculate the test statistic: \(z = \frac{p_{1_{hat}} - p_{2_{hat}}}{\sqrt{p_{hat} * (1 - p_{hat}) * (\frac{1}{n_1} + \frac{1}{n_2})}} = -1.5984884295812334\).

Step 5 ：Calculate the p-value: \(p_{value} = 2 * (1 - \text{cdf of normal distribution at } |z|) = 0.10993431840242684\).

Step 6 ：\(\boxed{\text{Final Answer: The test statistic for this sample is } -1.598 \text{ and the p-value for this sample is } 0.1099}\)