Answer: z = 2.60, and reject the null hypothesis.

## Explanation **Step 1: Determine the appropriate test statistic** - Population size: 3,000,000 (very large) - Sample size: n = 81 - Since the population is very large and n > 30, we can use the z-test (normal distribution) rather than the t-test **Step 2: Calculate the test statistic** Given: - Sample mean (x̄) = 46.3 - Hypothesized population mean (μ₀) = 45 - Sample standard deviation (s) = 4.5 - Sample size (n) = 81 z-statistic formula: \[ z = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \] \[ z = \frac{46.3 - 45}{4.5/\sqrt{81}} = \frac{1.3}{4.5/9} = \frac{1.3}{0.5} = 2.60 \] **Step 3: Make decision at 1% significance level** - Null hypothesis: H₀: μ ≤ 45 - Alternative hypothesis: H₁: μ > 45 - This is a one-tailed test - Critical z-value for α = 0.01 (one-tailed): z = 2.326 Since calculated z = 2.60 > critical z = 2.326, we **reject the null hypothesis** **Step 4: Verify the correct answer** - z = 2.60 matches option B - Decision to reject null hypothesis matches option B Therefore, the correct answer is **B. z = 2.60, and reject the null hypothesis.**

Author: LeetQuiz .