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Errors and Power: Detecting a Difference Without Claiming Certainty — Solutions and Rubrics

Use these solutions after completing the independent application. The rubrics describe what the requested action must demonstrate; equivalent correct reasoning is welcome.

Original generated practice for the Fall 2026 course. Return to the lesson and independent checks.

A complete power model for packet mass: worked answer

This newly specified packet-mass model samples five packets independently from a Normal population with known SD 0.010 kg. Target mean is 18.000 kg. A two-sided rule rejects if sample mean is below 17.9912346 or above 18.0087654 kg, equal to 18.000±1.96(0.010/√5). The sample-mean SD is σ/√n. For a true mean 18.012 kg, the standardized acceptance endpoints are -4.6432816 and -0.7232816; their normal CDF values are 0.000001715 and 0.234753456. Under the null Φ(1.96)=0.975002105. This supplied probability model illustrates power and does not add a required unknown-σ mean method.

Part a

Define a Type I error and Type II error in this context, and explain the actual true mean and decision involved in each. State what power measures.

Type I: the population mean is really 18.000 kg, but the sample triggers rejection and apparent evidence of a different mean. Type II: the population mean differs from 18.000 kg, but the rule fails to reject, missing the shift. Power at a specified shifted mean is the probability of correctly rejecting the false target null. A small p-value can cause the Type I decision even though the alternative is false; a large p-value can cause the Type II decision even though the alternative is true.

Part a scoring criteria
Evidence for creditPoints
States correct truth/decision combination for both errors.1
Defines power as correctly rejecting a false null at a stated alternative.1

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Part b

Recompute the standard deviation of the sample mean and both acceptance boundaries from the model. Find the Type I probability under the target and the Type II probability and power when μ=18.012. Show the interval probability before taking the complement.

SE=0.010/√5=0.00447214 kg. The half-width is 1.96SE=0.00876539 kg; failure region is [17.9912346,18.0087654]. Under μ=18.000, α=2[1−Φ(1.96)]≈0.0499958. At μ=18.012, β=P(lower≤x̄≤upper)=Φ(-0.7232816)−Φ(-4.6432816)≈0.234751741. Power=1−β≈0.765248259. The interval probability is a missed shift under the specified false null; the complement is correct detection.

Part b scoring criteria
Evidence for creditPoints
Uses sigma/sqrt(n) and both correct boundaries.1
Computes beta as difference of CDFs and power as complement; distinguishes null and alternative conditioning.1

Return to this independent task

Part c

Interpret β and power in repeated quality checks. Explain why reversing these two labels or calling failure to reject “accepting a true target” would misstate the result.

If the true population mean is 18.012 kg, about 23.475% of five-packet checks fail to flag the shift and about 76.525% flag it. Both assume the target null is false at this alternative. Failure to reject does not establish a true target; here it misses a false target. Alpha instead concerns rejection when the target mean is true.

Part c scoring criteria
Evidence for creditPoints
Interprets both rates under the actual true mean 18.012 kg: β≈0.234751741 and power≈0.765248259.1
Explains why failure to reject is not proof or acceptance of truth.1

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False alarms and missed contamination: worked answer

A food-testing agency treats a shipment’s population contamination proportion p=0.02 as the baseline and tests Hₐ:p>0.02 with a prespecified valid procedure. At the baseline its rejection probability is 0.01. At a harmful proportion p=0.035 its power is 0.76. Rejecting a safe shipment destroys it; failing to flag a contaminated shipment can expose customers. A proposed research change lowers α while leaving sample size, variability, and the harmful alternative fixed. Another proposal increases sample size at the same α.

Part a

Describe the two possible errors using the shipment decisions and population truth. Calculate the Type I probability, the Type II probability at p=0.035, and the probability of detecting that harmful alternative. Explain why the 0.24 Type II probability cannot be attached to every possible contaminated proportion.

Type I: reject a shipment whose true p is at the baseline 0.02; probability α=0.01. Type II at p=0.035: fail to flag a shipment with that harmful population contamination; β=1−0.76=0.24. Detection probability is power=0.76 at p=0.035. At a different true p, the distance from the null and therefore β/power generally differ; α is specified under the null, not as the Type II probability.

Part a scoring criteria
Evidence for creditPoints
Correct shipment truth/decision errors and alpha=0.01.1
Beta=0.24, power=0.76, each tied to p=0.035.1

Return to this independent task

Part b

Explain which error may have more serious consequences here, and how that judgment should affect planning before data collection. Describe the consequences of lowering α at fixed n and increasing n at fixed α; address both Type I and Type II concerns.

Missed contamination can harm customers, so Type II consequences may be more serious here, though false destruction also has costs. A different setting could prioritize false alarms; compare consequences before the study. Lowering α makes false rejection at the baseline less likely but makes rejection harder, tending to increase β and reduce power for the fixed harmful alternative. Increasing n reduces sampling variation and tends to reduce β/increase power at the same α; sample-size planning can address missed harm without simply raising the false-alarm rate. Type I seriousness informs α; Type II seriousness informs n.

Part b scoring criteria
Evidence for creditPoints
Compares consequences before study and connects Type I to alpha, Type II to n.1
Correct directions for lowered alpha and larger n, with fixed-alternative qualifiers.1

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Which design has the best chance to detect a shift?: worked answer

A supplied one-sided Normal sampling model tests θ=0 versus θ>0. At true θ, the standardized statistic has mean θ/SE and SD 1; reject when Z>c. At α=.05 c=1.644854; at α=.10 c=1.281552. Baseline: θ=.12, SE=.10. A doubles n so SE=.10/√2. B reduces population variability at original n so SE=.08. C uses true θ=.22 at original SE. D uses α=.10 at original θ/SE. All other stated settings stay fixed. Use power=Φ(θ/SE−c). For computation: Φ(-0.4448540)=0.32821263; Φ(0.0522023)=0.52081624; Φ(-0.1448540)=0.44241307; Φ(0.5551460)=0.71060260; Φ(-0.0815520)=0.46750149.

Part a

Calculate baseline power and Type II probability, then calculate power and Type II probability for A–D. Identify the relevant factor in each comparison, and explain in words how it makes correct rejection more likely. Explain why sample size and standard error are linked mechanisms.

Baseline power=0.32821263, β=0.67178737. A: power=0.52081624, β=0.47918376; larger n reduces SE. B: power=0.44241307, β=0.55758693; less variability gives smaller SE at the same n. C: power=0.71060260, β=0.28939740; true value farther from the null is easier to distinguish. D: power=0.46750149, β=0.53249851; higher alpha lowers the rejection threshold. Each β=1−power. Other relevant stated settings are fixed in each comparison. Larger n commonly acts through reduced SE, so n and SE are connected mechanisms rather than independent physical switches.

Part a scoring criteria
Evidence for creditPoints
Correct baseline and four power/beta pairs from model.1
Explains all four directions, fixed settings, and n/SE linkage.1

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Part b

A teammate says “The largest power guarantees the null is false, and increasing α is always the best choice.” Assess both claims and choose a defensible design change if false alarms are expensive.

Power is a probability of rejection assuming a specified false null, not evidence that any current null is false. No test guarantees a correct individual decision. Increasing α trades more detection for more Type I error under the null. If false alarms are expensive, larger n or lower measurement variability may improve power while retaining α=0.05; feasibility and cost still matter. The numerical model alone does not select a universal best policy.

Part b scoring criteria
Evidence for creditPoints
Rejects power-as-truth-probability and recognizes individual error.1
Proposes n/precision change and compares error costs before study.1

Return to this independent task

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