To address this hypothesis testing problem, we'll proceed step by step as required:

(a) Identify the claim and state $H_0$ and $H_a$:

1. Claim: The research center claims that at least 25% (0.25) of adults in the country think their taxes will be audited. This claim implies: $p \geq 0.25$

2. Null Hypothesis ($H_0$): The null hypothesis is what is claimed, which suggests that the proportion of adults who think their taxes will be audited is at least 25%. Thus: $H_0: p \geq 0.25$

3. Alternative Hypothesis ($H_a$): This suggests the opposite of the claim, that the proportion is less than 25%. Therefore: $H_a: p < 0.25$

(b) Find the critical value(s) and identify the rejection region(s):

Since this is a left-tailed test at $\alpha = 0.10$, find the z-critical value for which the cumulative probability is 0.10.

• Use a z-table or standard normal distribution to find the z-critical value corresponding to 0.10 in the left tail.
• The critical z-value is approximately -1.28.

Rejection Region:

• Reject $H_0$ if the calculated z-value is less than -1.28.

(c) Find the standardized test statistic $z$:

The sample proportion ($\hat{p}$) is given as 20% (0.20), and the sample size ($n$) is 600.

Calculate the z-statistic using the formula: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

Where:

• $\hat{p} = 0.20$
• $p_0 = 0.25$
• $n = 600$

Substitute these values in: $z = \frac{0.20 - 0.25}{\sqrt{\frac{0.25 \times 0.75}{600}}}$ $z = \frac{-0.05}{\sqrt{\frac{0.1875}{600}}}$ $z = \frac{-0.05}{\sqrt{0.0003125}}$ $z = \frac{-0.05}{0.01768} \approx -2.83$

(d) Conclusion:

Since the calculated z-value (-2.83) is less than the critical value (-1.28), we reject the null hypothesis $H_0$.

Conclusion:

There is enough evidence at the 0.10 significance level to reject the research center's claim that at least 25% of adults in the country think their taxes will be audited.