Answer: Test statistic: 1.768; Reject H₀

## Explanation This is a **one-sample z-test** for a population mean with known standard deviation. ### Step 1: Formulate Hypotheses - **Null hypothesis (H₀):** μ = 120 (average IQ is 120) - **Alternative hypothesis (H₁):** μ > 120 (average IQ is greater than 120) This is a **one-tailed test** since we're testing if the mean is greater than 120. ### Step 2: Test Statistic Calculation Given: - Sample mean (μ̄) = 125 - Population standard deviation (σ) = 20 (known) - Sample size (n) = 50 - Hypothesized mean (μ₀) = 120 The test statistic formula for z-test: $$z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$$ $$z = \frac{125 - 120}{20/\sqrt{50}} = \frac{5}{20/7.071} = \frac{5}{2.828} = 1.768$$ ### Step 3: Critical Value Approach For a one-tailed test at 5% significance level (α = 0.05): - Critical value = zₐ = 1.6449 (from standard normal distribution) - Decision rule: Reject H₀ if z > 1.6449 Since 1.768 > 1.6449, we **reject H₀**. ### Step 4: P-value Approach P-value = P(Z > 1.768) = 1 - P(Z < 1.768) = 1 - 0.96147 = 0.03853 or 3.853% Since p-value (0.03853) < α (0.05), we **reject H₀**. ### Step 5: Conclusion We have sufficient evidence at the 5% significance level to conclude that the average IQ of FRM candidates is greater than 120. **Why other options are incorrect:** - **Option B (2.828):** Incorrect test statistic calculation - **Option C (1.768; Fail to reject):** Correct test statistic but wrong conclusion - **Option D (1.0606; Fail to reject):** Incorrect test statistic and wrong conclusion

Author: Nikitesh Somanthe