1. **State the problem:** We want to test if the proportion of identity theft complaints in a state is higher than 23% based on a sample of 383 identity theft complaints out of 1450 total complaints, at a 9% significance level. 2. **Parameter:** The parameter $p$ represents the true proportion of all consumer complaints in the state that are for identity theft. 3. **Hypotheses:** - Null hypothesis: $$H_0: p = 0.23$$ - Alternative hypothesis: $$H_A: p > 0.23$$ 4. **Assumptions and conditions:** - Sample size $n = 1450$ - Sample proportion $$\hat{p} = \frac{383}{1450}$$ - Check $np = 1450 \times 0.23 = 333.5 \geq 10$ (condition met) - Check $n(1-p) = 1450 \times (1-0.23) = 1116.5 \geq 10$ (condition met) - Population size $N = 1000000 \geq 20n$ (condition met) 5. **Test statistic formula:** $$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$$ 6. **Calculate sample proportion:** $$\hat{p} = \frac{383}{1450}$$ 7. **Set up test statistic as a fraction:** $$z = \frac{\frac{383}{1450} - \frac{23}{100}}{\sqrt{\frac{\frac{23}{100} \times \left(1 - \frac{23}{100}\right)}{1450}}}$$ 8. **Calculate denominator:** $$\sqrt{\frac{0.23 \times 0.77}{1450}} = \sqrt{\frac{0.1771}{1450}} = \sqrt{0.000122138} \approx 0.01105$$ 9. **Calculate numerator:** $$\frac{383}{1450} - 0.23 = 0.2641 - 0.23 = 0.0341$$ 10. **Calculate z-value:** $$z = \frac{0.0341}{0.01105} \approx 3.09$$ **Final answer:** The test statistic is approximately $$z = 3.09$$. Since $z$ is positive and large, this suggests evidence that the proportion of identity theft complaints is greater than 23% at the 9% significance level.