The standard deviation of a variable is 20. The sample size is 100. Suppose the status quo is the population mean is 40. Which of the following sample means gives the most evidence against the null hypothesis?

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Question restated: The standard deviation of a variable is 20. The sample size is 100. The status quo (null hypothesis) is that the population mean is 40. Which sample mean gives the most evidence against the null hypothesis?

Key setup: When testing a null hypothesis about the mean with known population standard deviation, the sampling distribution of the sample mean x̄ is normal with mean μ under H0 and standard deviation σ/√n. Here, σ = 20 and n = 100, so the standard error is SE = 20/√100 = 20/10 = 2. This means each unit increase in |x̄ − 40| corresponds to a z-score of |x̄ − 40| / 2 in the standard normal distribution.

Option evaluation approach (in general): The strength of evidence against H0 grows as the observed x̄ moves farther away from 40, in either direction, because the z-statistic |z| = |x̄ − 40| / 2 increases with |x̄ − 40|. Therefore, among a set of potential sample means, the one with the largest absolute deviation from 40 will provide the strongest evidence against H0, all else equal.

Specific numeric reasoning (illustrative, since the actual listed options are not shown):
- If x̄ = 40, then |x̄ − 40| = 0 and z = 0, giving no evidence against H0.
- If x̄ = 42, then |x̄ − 40| = 2 and z = 1, which is modest evidence.
- If x̄ = 46, then |x̄ − 40| = 6 and z = 3, which is substantial evidence against H0.
- If x̄ = 50, then |x̄ − 40| = 10 and z = 5, which would be even stronger evidence than 46.

Why 46 specifically (in the context of the provided answer): Among plausible sample means, 46 is farther from the null mean than smaller deviations like 42 or 38, producing a larger z-statistic and thus stronger deviation from what H0 predicts. With SE = 2, the z for 46 is (46 − 40)/2 = 3, which corresponds to a small tail probability in a standard normal (roughly 0.0013 for a two-sided test), indicating strong evidence.

Why other common nearby values are weaker evidence: A mean closer to 40 (e.g., 40, 41, or 42) yields smaller |x̄ − 40| and thus smaller z-values, translating into weaker statistical evidence against H0. Even a mean like 50 produces more extreme deviation than 46, but among a set of options with 46 as the farthest listed value, 46 would be the stronger point relative to 40.

Note on the missing answer options: Since the actual alternative means being compared aren’t shown in the provided data, the general principle remains: the sample mean with the greatest absolute distance from 40 provides the strongest evidence against H0, given the same variability and sample size. If the options include several values, identify the one with the largest |x̄ − 40|; that will maximize the test statistic and the evidence against the null.

The standard deviation of a variable is 20. The sample size is 100. Suppose the status quo is the population mean is 40. Which of the following sample means gives the most evidence against the null hypothesis?Single choice

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