From previous studies, it is concluded that $80 \%$ of workers were employed from internet resume sites. A researcher claims that the proportion is significantly different than $80 \%$ and decides to survey 300 working adults. Test the researcher's claim at the $\alpha=0.2$ significance level.
Preliminary:
a. Is it safe to assume that $n \leq 0.05$ of all subjects in the population?
No
Yes
$\sigma^{6}$
b. Verify $n p(1-p) \geq 10$. Round your answer to one decimal place.
\[
n p(1-p)=
\]

Step 1 ：This is a hypothesis testing problem for a proportion. The null hypothesis is that the proportion of workers employed from internet resume sites is 0.80, and the alternative hypothesis is that the proportion is not 0.80. We are given a sample size of 300 and a significance level of 0.2.

Step 2 ：The first question asks whether it is safe to assume that the sample size is less than or equal to 5% of the population. Since we are not given any information about the population size, we cannot answer this question definitively. However, in practice, a sample size of 300 is often considered to be sufficiently large for the Central Limit Theorem to apply, so we might assume that this condition is met. So, the answer is Yes.

Step 3 ：The second question asks us to verify that the product of the sample size, the assumed proportion, and one minus the assumed proportion is greater than or equal to 10. This is a condition for the normal approximation to the binomial distribution to be valid.

Step 4 ：We can calculate this value using the given information. Let n = 300 and p = 0.8. Then, \(n p(1-p) = 300 * 0.8 * (1 - 0.8) = 48.0\)

Step 5 ：The calculated value of \(n p(1-p)\) is approximately 48, which is greater than 10. Therefore, the condition for the normal approximation to the binomial distribution is met. So, the answer is \(\boxed{48.0}\)