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 $a=0.2$ significance level. Based on the sample of 300 people, $74 \%$ workers were employed from internet resume sites. What is the test statistic? Round your answer to two decimal places. What is the $b$-value? Round your answer to four decimal places.

Step 1 ：Given that the sample proportion (p̂) is 0.74, the population proportion (p) is 0.80, and the sample size (n) is 300.

Step 2 ：We calculate the test statistic using the formula for the z-score, which is \((p̂ - p) / \sqrt{(p(1 - p) / n)}\).

Step 3 ：Substituting the given values into the formula, we get a test statistic of approximately -2.60.

Step 4 ：The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. The null hypothesis in this case is that the proportion of workers employed from internet resume sites is 80%.

Step 5 ：Using a standard normal distribution table, we calculate the p-value to be approximately 0.0094.

Step 6 ：Final Answer: The test statistic is \(\boxed{-2.60}\) and the p-value is \(\boxed{0.0094}\).