Answer: Test statistic: 1.768; Reject H₀

## Explanation ### Hypothesis Formulation - **H₀**: μ = 120 (null hypothesis - average IQ is 120) - **H₁**: μ > 120 (alternative hypothesis - average IQ is greater than 120) - This is a **one-tailed test** since we're testing for an increase only ### Test Statistic Calculation Given: - Sample mean (x̄) = 125 - Population mean (μ) = 120 - Population standard deviation (σ) = 20 - Sample size (n) = 50 Test statistic formula: $$z = \frac{(\bar{x} - \mu)}{(\sigma / \sqrt{n})}$$ Calculation: $$z = \frac{(125 - 120)}{(20 / \sqrt{50})} = \frac{5}{(20 / 7.071)} = \frac{5}{2.828} = 1.768$$ ### Decision Making **Critical Value Approach:** - For 5% significance level (one-tailed), critical z-value = 1.6449 - Since 1.768 > 1.6449, we **reject H₀** **P-value Approach:** - P(Z > 1.768) = 1 - P(Z < 1.768) = 1 - 0.96147 = 0.03853 (3.853%) - Since p-value (0.03853) < significance level (0.05), we **reject H₀** ### Conclusion There is sufficient statistical evidence at the 5% significance level to conclude that the average IQ of FRM candidates is greater than 120.

Author: Tanishq Prabhu