Given two independent random samples with the following results: \[ \begin{array}{ll} n_{1}=195 & n_{2}=408 \\ \hat{p}_{1}=0.66 & \hat{p}_{2}=0.78 \end{array} \] Use this data to find the $80 \%$ confidence interval for the true difference between the population proportions. Step 2 of 3: Find the value of the standard error. Round your answer to three decimal places.

Step 1 ：Given two independent random samples with the following results: \[\begin{array}{ll} n_{1}=195 & n_{2}=408 \\ \hat{p}_{1}=0.66 & \hat{p}_{2}=0.78 \end{array}\]

Step 2 ：The standard error for the difference in proportions can be calculated using the formula: \[SE = \sqrt{\frac{{\hat{p}_1(1-\hat{p}_1)}}{{n_1}} + \frac{{\hat{p}_2(1-\hat{p}_2)}}{{n_2}}}\] where: \(\hat{p}_1\) and \(\hat{p}_2\) are the sample proportions, \(n_1\) and \(n_2\) are the sample sizes.

Step 3 ：Substitute the given values into this formula to find the standard error: \[n1 = 195, n2 = 408, p1_hat = 0.66, p2_hat = 0.78\]

Step 4 ：Calculate the standard error: \[SE = 0.03964035148763628\]

Step 5 ：The standard error for the difference in proportions, rounded to three decimal places, is \(\boxed{0.040}\).