Question 3 rom 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 $\alpha=0.05$ significance level. Preliminary: a. Is it safe to assume that $n \leq 0.05$ of all.subjects in the population? Yes No b. Verify $n \hat{p}(1-\widehat{p}) \geq 10$. Round your answer to one decimal place. \[ n \widehat{p}(1+\widehat{p})= \]

Step 1 ：The first part of the question is asking whether it is safe to assume that the sample size (n) is less than or equal to 5% of the total population. Since we don't have information about the total population, we can't answer this question. So, the answer to the first part of the question is 'No'.

Step 2 ：The second part of the question is asking us to verify whether the product of the sample size (n), the sample proportion (p-hat), and one minus the sample proportion is greater than or equal to 10. This is another rule of thumb used in statistics to ensure that the sample size is large enough for the Central Limit Theorem to apply, which allows us to make inferences about the population from the sample.

Step 3 ：In this case, we know that n = 100 and p-hat = 0.12 (from the 12% of workers who are satisfied with their job). We can plug these values into the formula and calculate the result.

Step 4 ：\(n = 100\)

Step 5 ：\(\hat{p} = 0.12\)

Step 6 ：\(n \hat{p}(1-\hat{p}) = 100 * 0.12 * (1 - 0.12) = 10.56\)

Step 7 ：The result of the calculation is 10.56, which is greater than 10. This means that the sample size is large enough for the Central Limit Theorem to apply, and we can make inferences about the population from the sample.

Step 8 ：Therefore, the final answer is \(\boxed{No, 10.56}\).