Math 3 Quiz: Fitting Models To Data

18 questions · exam conditions

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Fitting Models To DataQuestion 1 of 18

An economist models income distribution using $f(x) = ax^{-b}$ where $x$ is income level and $f(x)$ represents frequency density. After fitting to data, she obtains $f(x) = 1250x^{-1.8}$ with $R^2 = 0.91$ . The model fits well for incomes above $20,000 but overestimates frequencies for incomes below $15,000. What constraint should be incorporated to improve the model?

Change the exponent to

f(x) = 1250x^{-1.6}

to reduce the rate of decrease at low incomes

Use a piecewise model with different parameters below and above the $15,000 threshold

Multiply by a correction factor:

f(x) = 0.85 \cdot 1250x^{-1.8}

to reduce all predictions uniformly

Add a minimum income threshold:

f(x) = 1250(x - x_{min})^{-1.8}

for

x \geq x_{min}

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Math 3 Quiz

Math 3 Quiz: Fitting Models To Data

Practice Fitting Models To Data in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Fitting Models To Data, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Change the exponent to $f(x) = 1250x^{-1.6}$ to reduce the rate of decrease at low incomes
Use a piecewise model with different parameters below and above the $15,000 threshold
Multiply by a correction factor: $f(x) = 0.85 \cdot 1250x^{-1.8}$ to reduce all predictions uniformly
Add a minimum income threshold: $f(x) = 1250(x - x_{min})^{-1.8}$ for $x \geq x_{min}$ (correct answer)

Explanation: When you encounter a power law model that fits well in one range but systematically overestimates in another, you need to identify what's causing the breakdown at the problematic range. The current model

f(x) = 1250x^{-1.8}

works well above $20,000 but overestimates frequencies below $15,000. This suggests the model predicts unrealistically high frequency densities for very low incomes. The issue is that as income approaches zero, $

x^{-1.8}

$ approaches infinity, which doesn't reflect real-world income distributions where there's typically a practical minimum income level. Adding a minimum income threshold creates

f(x) = 1250(x - x_{min})^{-1.8}

for

x \geq x_{min}

. This prevents the frequency density from exploding as income approaches the threshold, since the model now depends on income above the minimum rather than absolute income. This constraint directly addresses the overestimation problem in the low-income range while preserving the good fit at higher incomes. Option A changes the exponent, but this would alter the model's behavior across all income levels, potentially ruining the good fit above $20,000. Option B suggests a piecewise approach, which adds complexity and may create discontinuities without addressing the underlying mathematical issue. Option C applies a uniform scaling factor that would reduce accuracy in the well-fitting higher income range while not specifically targeting the low-income problem. Remember: when a power law model breaks down at extreme values, look for constraints that address the mathematical behavior causing the breakdown, not just parameter adjustments that affect the entire model.

Question 2

An ecologist fits a logistic growth model $P(t) = \frac{L}{1 + ae^{-rt}}$ to population data and obtains $P(t) = \frac{800}{1 + 15e^{-0.12t}}$ , where $t$ is time in months. The model has $R^2 = 0.94$ . If environmental conditions change such that the growth rate increases by 25%, but the carrying capacity decreases by 10%, what would be the new model?

$P(t) = \frac{720}{1 + 15e^{-0.15t}}$ (correct answer)
$P(t) = \frac{720}{1 + 18.75e^{-0.15t}}$
$P(t) = \frac{800}{1 + 11.25e^{-0.15t}}$
$P(t) = \frac{890}{1 + 15e^{-0.135t}}$

Explanation: The carrying capacity

L = 800

decreases by 10% to

720

. The growth rate

r = 0.12

increases by 25% to

0.15

. The parameter

a = 15

represents initial conditions and doesn't change with environmental conditions. Choice B incorrectly modifies the

a

parameter. Choice C doesn't adjust the carrying capacity. Choice D incorrectly calculates both the carrying capacity increase and growth rate.

Question 3

A chemist models the concentration of a reactant over time using $C(t) = C_0e^{-kt}$ , where $C_0 = 2.5$ M and $k = 0.032$ min $^{-1}$ . The model fits the first 30 minutes of data well ( $R^2 = 0.96$ ), but subsequent measurements show the actual concentration decreasing more slowly than predicted. What is the most likely explanation and appropriate model adjustment?

Temperature effects are slowing the reaction; adjust to $C(t) = 2.5e^{-0.025t}$ with a smaller rate constant
The reaction is approaching equilibrium; use $C(t) = C_{eq} + (C_0 - C_{eq})e^{-kt}$ with non-zero equilibrium concentration (correct answer)
Measurement precision is decreasing at low concentrations; weight early data points more heavily in the fit
The reaction mechanism is changing; switch to a second-order model $C(t) = \frac{C_0}{1 + C_0kt}$

Explanation: When an exponential decay model initially fits well but then overpredicts the rate of decrease, it typically indicates the system is approaching an equilibrium rather than going to zero. Adding an equilibrium term accounts for this. Choice A changes the rate but doesn't explain why the early fit was good. Choice C addresses measurement issues but not the systematic deviation. Choice D assumes a different reaction mechanism without justification.

Question 4

A researcher fits an exponential model $y = ab^x$ to a dataset using technology and obtains $y = 12.7(1.23)^x$ . The technology also reports that the coefficient of determination is $R^2 = 0.847$ . If the researcher transforms the data using logarithms to create a linear model, what should be the approximate y-intercept of this linearized model?

$\ln(12.7) \approx 2.54$ (correct answer)
$\ln(1.23) \approx 0.207$
$12.7$
$\log(12.7) \approx 1.10$

Explanation: When transforming the exponential model

y = ab^x

using natural logarithms, we get

\ln(y) = \ln(a) + x\ln(b)

. This creates a linear relationship where the y-intercept is

\ln(a) = \ln(12.7) \approx 2.54

. Choice B gives the slope of the linearized model. Choice C uses the original coefficient without transformation. Choice D uses the common logarithm instead of natural logarithm.

Question 5

A physics student models the relationship between pressure and volume of a gas using the function $P(V) = \frac{k}{V}$ , where $k$ is a constant. After fitting this model to experimental data, she obtains $P(V) = \frac{24.7}{V}$ with $R^2 = 0.88$ . To improve the fit, she considers the model $P(V) = \frac{k}{V} + c$ . What does the parameter $c$ most likely represent in the physical context?

Atmospheric pressure that affects all measurements equally regardless of volume changes (correct answer)
Measurement error in the pressure sensor that scales proportionally with pressure readings
Temperature effects that become more significant at higher pressures and lower volumes
Container volume corrections that account for the finite size of gas molecules

Explanation: The constant

c

represents a baseline pressure that doesn't depend on volume, which physically corresponds to atmospheric pressure affecting all measurements. Choice B describes proportional error, which wouldn't be a constant term. Choice C describes temperature effects that would vary with conditions, not be constant. Choice D describes molecular size effects, but these would depend on volume and wouldn't appear as a simple additive constant.

Question 6

A biologist models bacterial growth using the function $P(t) = 250e^{0.08t}$ , where $P(t)$ is the population after $t$ hours. After collecting additional data, the biologist determines that the model overestimates the actual population by approximately 15% at $t = 10$ hours. What adjustment to the model would best correct this systematic error?

Change the growth rate to $P(t) = 250e^{0.068t}$ to reduce exponential growth
Change the initial population to $P(t) = 212.5e^{0.08t}$ to scale down uniformly
Change to $P(t) = 250e^{0.08t} - 67.6$ to subtract the overestimate at $t = 10$
Change the model to $P(t) = 0.85 \cdot 250e^{0.08t}$ to reduce all predictions proportionally (correct answer)

Explanation: If the model consistently overestimates by 15%, multiplying by 0.85 (which is 1 - 0.15) corrects this proportional error across all time values. Choice A changes the growth rate but doesn't directly address the 15% overestimate. Choice B only adjusts the initial value but won't fix the overestimate at later times. Choice C subtracts a fixed amount, which doesn't account for the exponential nature of the error.

Question 7

A data analyst fits a polynomial model $y = 2.1x^3 - 15.2x^2 + 28.7x + 45.3$ to a dataset with $R^2 = 0.97$ . The data spans $x$ values from 1 to 8. When using this model to interpolate at $x = 5.5$ , which limitation should be the primary concern?

Polynomial models are inherently unstable and small data changes cause large coefficient changes
The high $R^2$ value suggests overfitting, making interpolation unreliable even within the data range
Cubic polynomials can oscillate significantly, potentially giving unrealistic values between data points (correct answer)
Interpolation at $x = 5.5$ is too close to the boundary conditions to be mathematically valid

Explanation: High-degree polynomials can exhibit oscillatory behavior between data points, even with good overall fit. This makes interpolation risky despite being within the data range. Choice A overstates instability concerns. Choice B misunderstands that high

R^2

with polynomial models isn't necessarily overfitting. Choice D incorrectly suggests that

x = 5.5

is near a boundary when the data spans 1 to 8.

Question 8

A company's revenue data over 6 quarters is fitted with the model $R(t) = 45 + 12t - 1.5t^2$ , where $R$ is revenue in thousands of dollars and $t$ is quarters since the start of data collection. The model has $R^2 = 0.93$ . Management wants to use this model to predict revenue for quarter 8. What is the most significant limitation of this approach?

The $R^2$ value is too low to make reliable predictions beyond the data range
The quadratic model predicts decreasing revenue after quarter 4, which may not continue realistically
Extrapolating 2 quarters beyond the data range exceeds reasonable bounds for this model type (correct answer)
The model coefficients suggest the revenue pattern is too volatile for accurate long-term forecasting

Explanation: Extrapolating beyond the data range is always risky, especially with polynomial models that can exhibit unrealistic behavior outside the fitted domain. While the model shows decreasing revenue after quarter 4 (Choice B), this could be realistic business behavior. The

R^2

of 0.93 is quite good (Choice A). Choice D misinterprets the coefficients - they show a reasonable quadratic trend, not excessive volatility.

Question 9

A quality control engineer models defect rates using $D(t) = D_0(1 - e^{-\lambda t})$ , where $D(t)$ is cumulative defect percentage and $t$ is time in hours. The fitted model is $D(t) = 0.15(1 - e^{-0.08t})$ with $R^2 = 0.93$ . Management asks what this model implies about the defect rate (the derivative $D'(t)$ ) at $t = 10$ hours. What is the most accurate interpretation?

The defect rate at $t = 10$ is approximately $0.15 \times 0.08 = 0.012$ or 1.2% per hour
The defect rate at $t = 10$ is $0.15 \times 0.08 \times e^{-0.8} \approx 0.0054$ or 0.54% per hour (correct answer)
The defect rate approaches zero since $e^{-0.8}$ is small, indicating quality improvement over time
The defect rate cannot be determined from the cumulative model without additional rate parameters

Explanation: The derivative of

D(t) = 0.15(1 - e^{-0.08t})

D'(t) = 0.15 \times 0.08 \times e^{-0.08t}

. At

t = 10

, this gives

0.15 \times 0.08 \times e^{-0.8} \approx 0.0054

. Choice A omits the exponential term. Choice C correctly notes the rate is decreasing but doesn't calculate the specific value. Choice D incorrectly suggests the derivative can't be found.

Question 10

A marketing analyst fits two models to sales data: Model A (linear): $S = 120 + 45t$ with $R^2 = 0.81$ , and Model B (exponential): $S = 135e^{0.28t}$ with $R^2 = 0.87$ , where $S$ is sales in thousands and $t$ is years. Both models predict similar values for the next 2 years, but diverge significantly after year 5. What additional information would be most crucial for model selection?

The residual patterns and whether they show systematic trends or random scatter around zero
Market saturation limits and whether unlimited exponential growth is realistic long-term (correct answer)
The confidence intervals around each model's predictions to assess uncertainty levels
Cross-validation results using held-out data to test each model's predictive accuracy

Explanation: Since both models fit reasonably well initially but diverge dramatically, the key consideration is which long-term behavior is realistic. Market context and growth limits are essential for choosing between linear and exponential models. Choice A (residuals) is useful but secondary to realistic constraints. Choice C (confidence intervals) and Choice D (cross-validation) are good practices but don't address the fundamental issue of realistic long-term behavior.

Question 11

A researcher fits a sinusoidal model $y = A\sin(Bt + C) + D$ to cyclical data and obtains $y = 12\sin(0.52t + 1.3) + 25$ with $R^2 = 0.88$ . The data represents monthly measurements over 3 years. If the researcher suspects there's also a gradual long-term trend in addition to the cyclical pattern, which model modification would be most appropriate?

$y = 12\sin((0.52 + Et) + 1.3) + 25$ where $E$ represents frequency change over time
$y = 12\sin(0.52t + 1.3) + 25e^{Et}$ where $E$ represents exponential trend growth
$y = (12 + Et)\sin(0.52t + 1.3) + 25$ where $E$ represents amplitude change over time
$y = 12\sin(0.52t + 1.3) + 25 + Et$ where $E$ represents the linear trend coefficient (correct answer)

Explanation: When analyzing cyclical data with sinusoidal models, you need to understand how each parameter affects the overall pattern. The original model

y = 12\sin(0.52t + 1.3) + 25

captures the periodic behavior well with

R^2 = 0.88

, but the researcher suspects an additional long-term trend that gradually shifts the data over time. A long-term trend represents a consistent directional change in the baseline level of your data. The most straightforward way to model this is by adding a linear trend term to your existing sinusoidal model. Option D accomplishes this by adding

Et

to the constant term, creating

y = 12\sin(0.52t + 1.3) + 25 + Et

. This preserves the cyclical pattern while allowing the entire curve to drift upward or downward over time at rate

E

. Option A modifies the frequency over time, which would create a pattern that speeds up or slows down rather than showing a consistent directional trend. Option B introduces exponential growth through

25e^{Et}

, which represents accelerating change rather than the gradual, steady trend the researcher suspects. Option C changes the amplitude over time with

(12 + Et)

, meaning the cyclical swings would get larger or smaller, but this doesn't address a shift in the overall level. Remember that when you suspect a trend in time series data, start with the simplest modification first. Adding a linear trend term (

Et

) to the baseline is usually the most appropriate initial approach before considering more complex modifications like changing frequencies or exponential components.

Question 12

A researcher collects data on the relationship between hours of sleep per night ( $x$ ) and test scores ( $y$ ) for 15 students. Using technology, she fits three different models to the data: Linear: $y = 12.3x + 42.1$ ( $R^2 = 0.73$ ), Quadratic: $y = -2.1x^2 + 25.6x + 8.9$ ( $R^2 = 0.89$ ), Exponential: $y = 45.2(1.18)^x$ ( $R^2 = 0.68$ ). If the quadratic model predicts that test scores are maximized when students get approximately 6.1 hours of sleep, what does this suggest about the constraint on the domain for this application?

The domain should be restricted to $0 ≤ x ≤ 6.1$ because test scores decrease after 6.1 hours of sleep according to the model
The domain should be restricted to $4 ≤ x ≤ 10$ because these represent realistic sleep amounts regardless of the mathematical model's behavior (correct answer)
The domain should be unrestricted because the quadratic model has the highest $R^2$ value and therefore most accurately represents the relationship
The domain should be restricted to $x ≥ 6.1$ because the model is only valid after the maximum point where the relationship becomes linear

Explanation: When fitting models to real-world data, the domain should be constrained by practical considerations, not just mathematical properties. While the quadratic model suggests test scores peak at 6.1 hours and then decrease, this doesn't mean we should exclude data beyond 6.1 hours - it means we should consider realistic sleep ranges (typically 4-10 hours for students) when interpreting and applying the model. The mathematical maximum doesn't define the appropriate domain; real-world constraints do.

Question 13

A company tracks the relationship between advertising spending ( $x$ thousand dollars) and monthly revenue ( $y$ thousand dollars) and fits the model $y = 45 + 3.2x - 0.08x^2$ . The marketing manager notes that this model has a maximum revenue point and asks what this means for budget planning. What is the most appropriate interpretation of the constraint this maximum imposes?

Advertising spending should never exceed the amount that maximizes revenue because any additional spending reduces total revenue
Advertising spending should target exactly the maximum point because this represents the optimal return on investment for the company
The maximum indicates diminishing returns, so spending decisions should consider profit margins and other business factors beyond just revenue (correct answer)
The maximum suggests the model is invalid for business applications because revenue should increase indefinitely with advertising spending

Explanation: The quadratic model shows diminishing returns - revenue increases at a decreasing rate and eventually decreases. However, the optimal business decision isn't necessarily to maximize revenue; it should maximize profit (revenue minus costs). The maximum revenue point provides important information about diminishing returns, but spending decisions should consider advertising costs, profit margins, and strategic goals beyond just revenue maximization.

Question 14

A pharmacologist studying drug concentration over time fits the model $C(t) = 250e^{-0.12t}$ where $C$ is concentration in mg/L and $t$ is time in hours since administration. The data used for fitting spans 0-24 hours. If the model is used to determine when concentration falls below 5 mg/L for safety protocols, what additional validation should be considered?

Check whether the exponential decay rate 0.12 matches known pharmacokinetic properties for this specific drug type
Verify that the initial concentration 250 mg/L represents the actual administered dose rather than peak blood concentration
Confirm that exponential decay continues to be valid for the time period needed to reach 5 mg/L concentration (correct answer)
Ensure that the $R^2$ value for the fitted model exceeds 0.90 before using it for safety-critical predictions

Explanation: To find when C(t) = 5: 5 = 250e^(-0.12t), solving gives t ≈ 32.6 hours. Since this extends beyond the 24-hour data range used for fitting, the most critical validation is confirming that exponential decay continues to be the appropriate model for this extended time period. Drug elimination might follow different kinetics (e.g., shift to linear elimination) at very low concentrations, making extrapolation potentially unsafe for critical protocols.

Question 15

A biologist studying bacterial growth fits the model $P(t) = 150e^{0.23t}$ to population data, where $P$ is the population and $t$ is time in hours. The parameter 0.23 in this model represents the growth rate. If the biologist later discovers that measurements were taken every 30 minutes but recorded as if they were hourly measurements, how should the growth rate parameter be reinterpreted?

The actual growth rate is $0.23 \times 2 = 0.46$ per hour because measurements were twice as frequent as recorded
The actual growth rate is $0.23 \div 2 = 0.115$ per hour because the time scale was compressed by a factor of 2 (correct answer)
The actual growth rate remains 0.23 per hour, but the initial population should be adjusted to $150 \times 2 = 300$
The actual growth rate is $0.23 \div 2 = 0.115$ per 30 minutes, which equals 0.23 per hour, so no adjustment is needed

Explanation: If measurements taken every 30 minutes were recorded as hourly data, then what the model interprets as 1 hour of growth actually represents 0.5 hours of real growth. The fitted parameter 0.23 represents growth per recorded time unit (30 minutes), so the actual hourly growth rate is 0.23 ÷ 2 = 0.115 per hour. The time compression means the actual growth rate is slower than the model suggests.

Question 16

A researcher studying learning curves fits the model $S(t) = 95(1 - e^{-0.15t})$ to data where $S$ represents skill level (0-100 scale) and $t$ represents practice time in hours. According to this model, what does the parameter 95 represent in the context of learning, and what constraint does this place on the model's validity?

The parameter 95 represents initial skill level, constraining the model to learners who start with high proficiency levels
The parameter 95 represents average skill level after extensive practice, constraining the model to normally distributed learning outcomes
The parameter 95 represents the learning rate, constraining the model to skills that can be acquired relatively quickly
The parameter 95 represents the theoretical maximum skill level, constraining the model's validity to skills where perfect mastery isn't achievable (correct answer)

Explanation: When you encounter exponential growth or decay models like

S(t) = 95(1 - e^{-0.15t})

, start by analyzing what happens to the function's behavior over time. As

t

approaches infinity,

e^{-0.15t}

approaches zero, making the expression

(1 - e^{-0.15t})

approach 1. This means

S(t)

approaches

95 \times 1 = 95

. The parameter 95 represents the horizontal asymptote—the theoretical maximum skill level that can never quite be reached but can be approached indefinitely. In learning contexts, this reflects the reality that complete mastery (100%) is often unattainable due to human limitations, measurement constraints, or the nature of the skill itself. The model is only valid for situations where perfect mastery isn't expected or possible. Choice A incorrectly identifies 95 as the initial skill level. At

t = 0

, we get

S(0) = 95(1 - e^0) = 95(1 - 1) = 0

, so learners actually start at zero skill level. Choice B misinterprets 95 as an average rather than a maximum, and incorrectly introduces normal distribution concepts that aren't relevant to this exponential model. Choice C confuses the asymptote (95) with the learning rate parameter (-0.15 in the exponent), which actually controls how quickly the skill level approaches the maximum. Remember: In exponential models of the form

A(1 - e^{-kt})

, the coefficient

A

always represents the limiting value or theoretical maximum, not the initial condition or rate of change.

Question 17

A environmental scientist fits three different models to temperature data over time: Linear: $T = 0.08t + 15.2$ , Quadratic: $T = 0.003t^2 + 0.05t + 15.1$ , and Sinusoidal: $T = 4.2\sin(0.52t) + 18.7$ . All models have similar $R^2$ values (0.76-0.82) when fitted to 5 years of data. For predicting temperatures 20 years into the future, which model characteristic presents the greatest concern?

The quadratic model's accelerating growth could predict unrealistically extreme temperatures outside the data range (correct answer)
The linear model's constant rate of change fails to capture natural temperature variability over extended periods
The sinusoidal model's periodic nature doesn't account for potential long-term climate trends or shifts
All models are equally unreliable for 20-year predictions because their $R^2$ values are too similar to distinguish quality

Explanation: When evaluating mathematical models for long-term predictions, you need to consider how each model's mathematical structure behaves when extrapolated far beyond the original data range. This is especially critical when the fitted period (5 years) is much shorter than the prediction horizon (20 years). The quadratic model

T = 0.003t^2 + 0.05t + 15.1

contains a

t^2

term that creates exponential growth over time. While the coefficient 0.003 seems small, after 20 years this quadratic term dominates the equation's behavior. At

t = 20

, the quadratic contribution alone is

0.003(20)^2 = 1.2

degrees, but this acceleration continues indefinitely. For temperature data, this could predict absurdly high or low temperatures that violate physical reality. Answer A correctly identifies this as the greatest concern. Answer B is wrong because while linear models do miss variability, they at least maintain realistic, bounded predictions over time. Answer C incorrectly suggests the sinusoidal model's periodicity is a flaw—actually, temperature cycles are natural and the periodic behavior keeps predictions within reasonable bounds. Answer D misunderstands

R^2

values. Similar

R^2

values only indicate comparable fit to historical data, not equivalent extrapolation reliability. Study tip: For extrapolation problems, always examine the mathematical form's long-term behavior. Polynomial models with degree ≥ 2 often become unrealistic when projected far beyond the data range, even when they fit historical data well.

Question 18

A sports analyst fits the power model $v = 12.4d^{0.31}$ to relate driving distance ( $d$ yards) and clubhead velocity ( $v$ mph) in golf. The model shows $R^2 = 0.78$ . However, when examining the data range, the analyst notices that all measurements were taken with drives between 180-320 yards. What is the most important consideration when applying this model?

The model should only be applied within the 180-320 yard range because power model behavior outside this range is uncertain (correct answer)
The model can be reliably extrapolated to any driving distance because power relationships are universal in sports applications
The model is invalid because the $R^2$ value of 0.78 is too low for a power relationship to be considered reliable
The model can be extrapolated to shorter distances but not longer ones because power models break down at high values

Explanation: When you encounter regression models in statistics, the most critical principle is understanding the difference between interpolation (predicting within the data range) and extrapolation (predicting outside it). This question tests whether you recognize the fundamental limitation that any statistical model is only reliable within the range of data used to create it. The power model

v = 12.4d^{0.31}

was fitted using data exclusively from drives between 180-320 yards. While the model might appear mathematically elegant and the

R^2 = 0.78

shows a reasonably strong relationship, we have no evidence about how clubhead velocity behaves outside this range. Golf physics could change dramatically at extreme distances due to factors not captured in the limited data range. Answer A correctly identifies this fundamental statistical principle - models should only be applied within their data range because behavior outside is uncertain. Answer B is wrong because no relationship, even power relationships, can be assumed universal without supporting data across all ranges. Answer C incorrectly suggests

R^2 = 0.78

invalidates the model, but this value actually indicates a decent fit explaining 78% of the variance. Answer D makes an unsupported claim about power model behavior at different ranges - there's no mathematical reason power models would fail only at high values but not low ones. Remember this key principle: statistical models are photography, not prophecy. They capture relationships within the observed data range but cannot reliably predict beyond those boundaries without additional validation data.