The numbers are almost too clean, which made me second-guess myself mid-calculation.
Set up a one-sample z-test for the population mean with a known standard deviation. State the null and alternative hypotheses, compute the z-statistic, and compare the p-value to a chosen significance level (e.g., 0.05). Conclude whether to reject the PM's claim based on the statistical evidence.
Pro tip: Always clarify the significance level and whether the test is one-tailed or two-tailed before diving into calculations. In a business context, also consider the practical significance of the difference, not just statistical significance.
Define the null hypothesis (H0: μ = $50) and the alternative hypothesis (H1: μ > $50) since you believe the true average is higher. This sets up a one-tailed test.
Select a significance level (commonly α = 0.05). Since the population standard deviation is known, use a z-test. Compute the z-statistic using the formula: z = (x̄ - μ0) / (σ / √n).
Plug in the values: x̄ = 85, μ0 = 50, σ = 20, n = 100. Compute z = (85 - 50) / (20 / 10) = 35 / 2 = 17.5.
Find the p-value corresponding to z = 17.5. Since this is extremely large, the p-value is essentially 0. Compare it to α = 0.05; if p < α, reject H0. Here, reject H0 and conclude the average is significantly higher than $50.
Explain that the sample provides strong evidence that the true average monthly spend is greater than $50. Discuss any practical implications or limitations (e.g., sample representativeness).
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