A human resources representative claims that the proportion of employees earning more than $\$ 50,000$ is greater than $40 \%$. To test this claim, a random sample of 700 employees is taken and 305 employees are determined to earn more than $\$ 50,000$ The following is the setup for this hypothesis test: \[ \begin{array}{l} H_{0}: p=0.40 \\ H_{a}: p>0.40 \end{array} \] Find the $p$-value for this hypothesis test for a proportion and round your answer to 3 decimal places. The following table can be utilized which provides areas under the Standard Normal Curve:

Step 1 ：First, we calculate the sample proportion \(\hat{p}\), which is the number of employees earning more than $50,000 divided by the total number of employees in the sample. In this case, \(\hat{p} = \frac{305}{700} = 0.4357142857142857\).

Step 2 ：Next, we substitute the values into the formula for the z-score and calculate it. The z-score is calculated as \(z = \frac{\hat{p} - p0}{\sqrt{\frac{p0 * (1 - p0)}{n}}} = \frac{0.4357142857142857 - 0.4}{\sqrt{\frac{0.4 * (1 - 0.4)}{700}}} = 1.928791874526148\).

Step 3 ：Then, we calculate the p-value. The p-value is the probability that we would observe a result as extreme as, or more extreme than, the result we observed, given that the null hypothesis is true. We can find the p-value by subtracting the cumulative distribution function (CDF) of the z-score from 1. In this case, the p-value is \(1 - CDF(1.928791874526148) = 0.027\).

Step 4 ：Final Answer: The p-value for this hypothesis test for a proportion is \(\boxed{0.027}\).