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.