In 2011, a U.S. Census report determined that 71\% of college students are working students. A researcher thinks this percentage has changed and surveys 186 college students. The researcher reports that 123 of the 186 are working students. Is there evidence to support the researcher's claim at the $1 \%$ significance level?
Find the p-value. Round to four decimal places.

Step 1 ：The problem is asking for a p-value, which is a measure of the probability that an observed difference could have occurred just by random chance. In this case, we are testing the claim that the proportion of working students has changed from the 2011 Census report of 71%.

Step 2 ：The p-value is calculated using a hypothesis test for a proportion. The null hypothesis (H0) is that the proportion of working students is still 71%, and the alternative hypothesis (H1) is that the proportion has changed.

Step 3 ：We can use a z-test for proportions to calculate the test statistic, and then find the p-value from the standard normal distribution. The formula for the z-test statistic is: \(z = \frac{{p_{hat} - p0}}{{\sqrt{{(p0 * (1 - p0)) / n}}}}\) where \(p_{hat}\) is the sample proportion, \(p0\) is the proportion under the null hypothesis, and \(n\) is the sample size.

Step 4 ：Once we have the z-score, we can find the p-value by looking up the probability in the standard normal distribution and multiplying by 2 (since this is a two-tailed test).

Step 5 ：Given that \(p0 = 0.71\), \(n = 186\), \(x = 123\), \(p_{hat} = 0.6612903225806451\), and \(z = -1.4640083450190289\), we find that the p-value is 0.1432.

Step 6 ：The p-value is 0.1432, which is greater than the significance level of 0.01. This means that we do not have enough evidence to reject the null hypothesis. Therefore, there is not enough evidence to support the researcher's claim that the proportion of working students has changed from the 2011 Census report of 71%.

Step 7 ：Final Answer: The p-value is \(\boxed{0.1432}\).