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 1 of 3: Find the critical value that should be used in constructing the confidence interval.

Step 1 ：First, we need to find the critical value for an 80% confidence interval. This is found using the Z-distribution. The critical value for an 80% confidence interval is the Z-score that leaves 10% in each tail of the distribution, because 100% - 80% = 20%, and this 20% is split evenly between the two tails. Looking up this value in a standard Z-table, we find that the critical value is approximately \(Z = 1.28\).

Step 2 ：Next, we calculate the standard error of the difference in proportions. The formula for this is \(\sqrt{\frac{{\hat{p}_1(1-\hat{p}_1)}}{{n_1}} + \frac{{\hat{p}_2(1-\hat{p}_2)}}{{n_2}}}\). Substituting the given values, we get \(\sqrt{\frac{{0.66(1-0.66)}}{{195}} + \frac{{0.78(1-0.78)}}{{408}}}\) which simplifies to approximately 0.039.

Step 3 ：Finally, we calculate the confidence interval. The formula for this is \((\hat{p}_1 - \hat{p}_2) \pm Z \times \text{standard error}\). Substituting the values we have, we get \((0.66 - 0.78) \pm 1.28 \times 0.039\). This simplifies to \(-0.12 \pm 0.05\). Therefore, the 80% confidence interval for the true difference between the population proportions is \(\boxed{(-0.17, -0.07)}\).