From previous studies, it is concluded that $12 \%$ of workers indicate that they are satisfied with their job. A researcher claims it has decreased and decides to survey 100 adults. Test the researcher's claim at the $a=0.05$ significance level. What is the $p$-value? Round to 4 decimals.|

Step 1 ：The problem is asking for a p-value for a hypothesis test. The null hypothesis is that the proportion of workers who are satisfied with their job is 0.12, and the alternative hypothesis is that the proportion has decreased, i.e., it is less than 0.12. The sample size is 100. The significance level is 0.05.

Step 2 ：We are not given the sample proportion (the proportion of the 100 adults surveyed who indicated that they are satisfied with their job). Without this information, we cannot calculate the p-value.

Step 3 ：However, assuming that we have this information, we can calculate the p-value using the formula for the test statistic for a proportion, which is \(Z = \frac{{p_{hat} - p_0}}{{\sqrt{{(p_0 * (1 - p_0)) / n}}}}\) where \(p_{hat}\) is the sample proportion, \(p_0\) is the population proportion under the null hypothesis, and n is the sample size.

Step 4 ：The p-value is then the probability that a standard normal random variable is less than Z.

Step 5 ：Assuming \(p_{hat} = 0.1\), \(p_0 = 0.12\), and n = 100, we calculate Z to be approximately -0.6154574548966634.

Step 6 ：The p-value is then calculated to be approximately 0.2691.

Step 7 ：The p-value is 0.2691, which is greater than the significance level of 0.05. Therefore, we do not reject the null hypothesis that the proportion of workers who are satisfied with their job is 0.12. There is not enough evidence to support the researcher's claim that the proportion has decreased.

Step 8 ：Final Answer: The p-value is \(\boxed{0.2691}\).