Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test. \[ \begin{array}{l} H_{0}: p=0.79 \text { versus } H_{1}: p \neq 0.79 \\ n=500, x=390, \alpha=0.1 \end{array} \] Find the P-value. P-value $=\square$ (Round to three decimal places ag-needed.)

Step 1 ：Given values are \(n = 500\), \(x = 390\), \(p_{null} = 0.79\), and \(\alpha = 0.1\).

Step 2 ：Calculate the sample proportion (\(p_{hat}\)) using the formula \(p_{hat} = \frac{x}{n}\). Substituting the given values, we get \(p_{hat} = \frac{390}{500} = 0.78\).

Step 3 ：Calculate the standard error (se) using the formula \(se = \sqrt{\frac{p_{null} \cdot (1 - p_{null})}{n}}\). Substituting the given values, we get \(se = \sqrt{\frac{0.79 \cdot (1 - 0.79)}{500}} = 0.01821537811850196\).

Step 4 ：Calculate the z-score using the formula \(z = \frac{p_{hat} - p_{null}}{se}\). Substituting the given values, we get \(z = \frac{0.78 - 0.79}{0.01821537811850196} = -0.5489866822936099\).

Step 5 ：Calculate the P-value using the formula \(P-value = 2 \cdot (1 - CDF(|z|))\), where CDF is the cumulative distribution function of the standard normal distribution. Substituting the given values, we get \(P-value = 2 \cdot (1 - CDF(|-0.5489866822936099|)) = 0.583014589331385\).

Step 6 ：The P-value is 0.583, which is greater than the significance level of 0.1. This means that we do not have enough evidence to reject the null hypothesis that the population proportion is 0.79.

Step 7 ：Final Answer: The P-value is \(\boxed{0.583}\).