Simple Linear Regression - Business Analytics

1. Which pair correctly gives the model's coefficient of determination and residual sum of squares?
   Answer: $R^2=0.49$ and the residual sum of squares is $408$ Whenever you see a question linking correlation to regression fit, remember two key formulas: the coefficient of determination equals the square of the correlation coefficient ($R^2 = r^2$), and the residual sum of squares equals the total sum of squares minus the explained variation ($SS_{res} = SS_{tot} - SS_{reg}$, where $SS_{reg} = R^2 \times SS_{tot}$). Here, $r = -0.70$, so $R^2 = (-0.70)^2 = 0.49$. This tells you the model explains 49% of the variation in satisfaction. The regression (explained) sum of squares is $0.49 \times 800 = 392$, leaving a residual sum of squares of $800 - 392 = 408$. That matches answer A, which is correct. Answer B is wrong on two counts: $R^2$ can never be negative (it's a squared value), and the residual sum of squares of 1192 exceeds the total, which is mathematically impossible. Answer C confuses $R^2$ with $|r|$ itself — a very common trap. Using $r = 0.70$ instead of $r^2 = 0.49$ overstates explanatory power and produces an incorrect residual of 240. Answer D gets $R^2$ right but swaps the two components: it reports $SS_{reg} = 392$ as the residual rather than the explained portion, mixing up what's "left over" versus what's "accounted for." A quick memory trick: $R^2$ is always the square of $r$, always positive, and always between 0 and 1. The residual SS is what the model didn't explain — subtract the explained portion from the total, never add.
2. Which interval should the analyst use, and how should its width generally compare with other regression intervals?
   Answer: Use a prediction interval at $480{,}000$; it should exceed the corresponding mean-response interval and usually widen away from the sample mean Whenever regression questions ask about estimating a single future observation versus an average, the interval type and its behavior relative to the sample mean are both being tested simultaneously. For a single future store, you need a prediction interval, not a confidence interval for mean response. The prediction interval accounts for two sources of uncertainty: uncertainty in the estimated regression line itself, plus the natural variability of an individual observation around that line. This makes it inherently wider than the corresponding confidence interval for mean revenue at the same $x$ value. Because management wants to estimate revenue for one specific future store — not the average revenue across many stores at population $480{,}000$ — a prediction interval is the correct tool. Additionally, both interval types widen as $x$ moves away from the sample mean $\bar{x} = 300{,}000$, because the leverage factor $\frac{(x^* - \bar{x})^2}{S_{xx}}$ grows with distance from the mean. Since $480{,}000$ is far from $300{,}000$, the interval at that point is generally wider than at the mean. This confirms D. A reverses the relationship — a confidence interval for mean response is narrower than a prediction interval, not wider. B targets the wrong population value ($300{,}000$ instead of $480{,}000$) and uses the wrong interval type; a mean-response confidence interval does not measure uncertainty for one future store. C correctly identifies the prediction interval but gets the width comparison backwards — intervals widen, not narrow, as you move away from $\bar{x}$. Study tip: Always ask two questions: "One observation or average?" (prediction vs. confidence interval) and "How far from $\bar{x}$?" (wider farther away). These two checks will resolve most regression-interval questions quickly.
3. How does this change of units affect the fitted equation and its predictions?
   Answer: The equation becomes $y_{pred}=18+2.4z$, while predictions remain unchanged for equivalent spending levels When you rescale a predictor variable in regression, the slope must adjust to keep predictions identical — the model is describing the same real-world relationship, just expressed differently. The key question to ask yourself is: how does the new unit relate to the old one, and how must the slope compensate? Since $z$ is spending in hundreds of dollars, $x = 100z$, meaning one unit of $z$ equals 100 units of $x$. Substituting directly into the original equation gives you the new form: $y_{pred} = 18 + 0.024(100z) = 18 + 2.4z$ So the intercept stays at 18, the slope becomes 2.4, and — critically — predictions are unchanged for any equivalent spending level. If you spend $500, that's $x = 500$ or $z = 5$: the original gives $18 + 0.024(500) = 30$, and the new gives $18 + 2.4(5) = 30$. Same result. This confirms D is correct. A is wrong because it multiplies the intercept by 100 instead of the slope — the intercept has no units of $x$, so it doesn't change. B divides the slope by 100 rather than multiplying, which would be the correct adjustment if you were going the other direction (dollars to hundredths of dollars). C leaves the slope unchanged entirely, which would produce different predictions — a fundamental error that violates the purpose of rescaling. A useful rule of thumb: when your predictor's unit gets larger by a factor of $k$, the slope gets larger by the same factor $k$, and the intercept never changes.
4. What is the most appropriate interpretation of this residual behavior?
   Answer: The systematic pattern suggests curvature, so the linear model may miss part of the relationship despite zero mean residuals Whenever you see a residual plot question in regression analysis, your job is to look for patterns — because patterns in residuals reveal what your model is missing. A well-specified linear model should produce residuals that are randomly scattered with no systematic structure. Here, the residuals follow a clear U-shaped pattern: positive for small orders, negative for medium orders, then positive again for large orders. This wave-like behavior is a textbook signal of curvature — the true relationship between order size and processing time bends in a way the straight regression line cannot capture. Even though residuals average zero overall, that global zero mean masks a systematic local structure. Answer C correctly identifies this: the linear model is misspecified, and a quadratic or other nonlinear term should be considered. A is wrong because a zero overall mean does not rule out meaningful patterns — it only means the model isn't biased on average. Predictive structure can absolutely exist within residuals that sum to zero, as this example proves. B is wrong because the pattern described is about the sign of residuals changing across predictor values, not about their spread increasing. That would suggest heteroscedasticity, which is a different problem entirely requiring different remedies. D is wrong because it confuses sign patterns with slope estimates. A nonzero slope is perfectly consistent with positive residuals at both extremes — in fact, that's exactly what a U-shaped residual pattern looks like around a fitted line. Study tip: Memorize the phrase "patterns in residuals = model misspecification." If residuals aren't random, your model's functional form is likely wrong — regardless of what the overall mean or R² says.
5. Which conclusion is best supported by this information?
   Answer: The removed week can be influential despite its zero residual because it has an extreme predictor value and changes the fitted line Whenever you see a question about regression diagnostics, distinguish carefully between outliers (unusual y values) and influential points (observations that substantially change the fitted line when removed). These are different concepts, and an observation can be one without being the other. Here, the removed week has $x = 20$ and $y = 320$. Under the full model, $\hat{y} = 80 + 12(20) = 320$, confirming a residual of exactly zero — the point sits perfectly on the original line. Yet removing it shifts the model dramatically: the intercept rises from 80 to 95 and the slope drops from 12 to 8. That's a meaningful change in both coefficients, which is the definition of an influential observation. Why can a zero-residual point be influential? Because influence is driven primarily by leverage — how extreme the predictor value is. An observation at $x = 20$ (likely far from the mean of $x$) acts like a anchor pulling the regression line toward it. Remove that anchor, and the line pivots noticeably. This makes C the correct conclusion. A is wrong because it treats zero residual as proof of non-influence — a common misconception. Residuals and influence measure different things entirely. B is wrong because "outcome outlier" refers to a large residual in the current model; under the full-model, the residual is zero, so it is not a response outlier by definition. D is wrong because both the slope and intercept change ($12 \to 8$ and $80 \to 95$), so the claim that only the intercept shifts is factually false. Study tip: Always ask two separate questions — "Is this an outlier?" (look at residuals) and "Is this influential?" (look at leverage and what happens when you remove it). A point can be both, neither, or just one.
6. How should the model's prediction at the proposed price be interpreted?
   Answer: The equation gives $790$ subscriptions, but this is an extrapolation whose reliability is not established by the observed data Whenever a regression model is applied to a value outside the range of data used to build it, you're dealing with extrapolation — and recognizing its limits is a core business analytics concept. Here, the model was fitted using prices from $p = 20$ to $p = 50$, so $p = 10$ falls outside that observed range. Plugging $p = 10$ into the equation gives $\hat{y} = 900 - 11(10) = 900 - 110 = 790$ subscriptions. The math works — but the key issue is whether that number is trustworthy. The relationship between price and demand was only validated within $[20, 50]$. Outside that window, the true relationship could bend, flatten, or shift entirely. That makes B the correct answer: the equation produces $790$, but it's an extrapolation with no empirical support from the data. A is tempting but wrong. Being within the mathematical domain of an equation (which technically extends to all real numbers) is irrelevant — what matters is the data's domain. The model has no observed evidence beyond its fitting range. C contains a calculation error: $900 - 11(10) = 790$, not $1{,}010$. It also incorrectly asserts the slope must reverse — that's speculation, not a statistical principle. D is false. Regression equations are mathematically defined for any input value; the issue is reliability and validity, not computational ability. Your study tip: always distinguish between mathematical possibility and statistical validity. A model can always produce a number — the real question is whether that number means anything given the data behind it.
7. According to the fitted regression, what revenue is predicted for a week with $52$ sales-call hours?
   Answer: $218$ thousand dollars, using a fitted slope of $1.5$ and an intercept of $140$ When you see a simple linear regression question, your first move should always be to compute the slope and intercept before touching the prediction formula. The slope is calculated as $\hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} = \frac{120}{80} = 1.5$. Then, because the regression line always passes through the point of sample means, you find the intercept by plugging in $\bar{x} = 40$ and $\bar{y} = 200$: $\hat{\beta}_0 = \bar{y} - \hat{\beta}_1\bar{x} = 200 - 1.5(40) = 200 - 60 = 140$. Your fitted model is $\hat{y} = 140 + 1.5x$. Predicting at $x = 52$: $\hat{y} = 140 + 1.5(52) = 140 + 78 = 218$ thousand dollars. That confirms A is correct. Choice B uses a negative slope of $-1.5$, which would mean more sales-call hours reduce revenue — mathematically wrong because $S_{xy} = 120$ is positive, so the slope must be positive. Choice C skips the intercept entirely and computes $1.5(52) = 78$, then adds it directly to $\bar{y} = 200$, giving 278. This confuses predicting at $x = 52$ with predicting the change from the mean. Choice D claims predictions always equal $\bar{y}$, which is only true when predicting at $x = \bar{x}$; for any other $x$, the prediction shifts along the line. A reliable study tip: always derive slope first (sign and magnitude), then use the mean-point property to anchor the intercept. These two steps prevent every common trap shown here.
8. Which interpretation of the fitted model is most appropriate?
   Answer: A two-minute increase is associated with a $6.8$-point decrease, while the intercept has limited practical meaning When reading a regression equation, you need to track two distinct skills: correctly applying the slope to a given change in $x$, and honestly assessing what the intercept means in context. The slope here is $-3.4$, meaning for every one-minute increase in wait time, satisfaction drops by $3.4$ points. For a two-minute increase, you multiply: $2 \times (-3.4) = -6.8$ points. That's the correct scaled interpretation. Now consider the intercept of $92$. It represents predicted satisfaction when wait time equals zero minutes — but the data only ranged from $2$ to $8$ minutes. Predicting at $x = 0$ is extrapolation outside the observed range, so the intercept carries limited practical meaning. That reasoning makes C the correct answer. Answer A makes two errors: it uses "causes" (implying causation, which regression alone cannot establish) and misidentifies $92$ as the observed maximum score — it's the intercept, not a data point. Answer B correctly avoids causal language and properly scales the slope... but then misidentifies $92$ as the "average observed score," which is factually wrong. The intercept is a model parameter, not a summary statistic of your data. Answer D gets the direction of the effect backwards — a two-minute increase is associated with a decrease, not an increase of $6.8$ points — and while it correctly notes the intercept describes zero-minute waits, that description is misleading without acknowledging it's outside the data range. A reliable strategy: always multiply the slope by the actual change in $x$ stated in the question, flag any causal language as a red flag, and ask whether $x = 0$ falls within your data range before interpreting the intercept.
9. Which conclusion is most defensible from the confidence interval?
   Answer: The association is statistically positive at the $5\%$ level, but the interval includes values below $1.5$, so operational worthiness is not established Whenever you see a confidence interval paired with a practical threshold, you need to ask two separate questions: Is the effect statistically significant? And does the interval guarantee the effect is large enough to matter operationally? These are different standards, and conflating them is the central trap in this question. The 95% confidence interval for the slope is $[0.2, 1.8]$. Because zero is excluded entirely, you can conclude the slope is statistically significantly positive at the $5\%$ level — some positive savings effect is real. However, management's threshold is $1.5$ thousand dollars per unit. Since $1.5$ falls inside the interval, the true slope could plausibly be anywhere from $0.2$ to $1.8$, including values well below $1.5$. The data are consistent with a slope that fails to meet the operational benchmark. Answer A correctly captures both facts: statistical significance is established, but operational worthiness is not. Answer B is wrong because the point estimate is the midpoint of the interval, $\frac{0.2 + 1.8}{2} = 1.0$, not $1.5$ — and even if it were $1.5$, a point estimate alone doesn't "confirm" anything about the true parameter. Answer C is wrong because a threshold inside the interval means the data are compatible with falling short of that threshold — the opposite of confirmation. Answer D is wrong on two counts: confidence intervals address population parameters, not individual outcomes, and excluding zero establishes association, never causation. Study tip: Always separate the statistical question (does zero lie outside the interval?) from the practical question (does the entire interval exceed the required threshold?). Operational thresholds demand the lower bound clears the benchmark, not just the upper bound.
10. What is this customer's residual, and what does its sign indicate?
    Answer: The residual is $51$ dollars, indicating that the model underpredicted the customer's spending Whenever you see a question about residuals, anchor yourself to one formula: $\text{residual} = y_{actual} - y_{predicted}$. The order matters — actual minus predicted, never the reverse. Here, the model predicts spending for a customer with 8 prior orders as $y_{pred} = 75 + 48(8) = 75 + 384 = 459$ dollars. The customer actually spent 510 dollars, so the residual is $510 - 459 = +51$ dollars. A positive residual means the actual value exceeded the prediction — the model underpredicted spending. That confirms B is correct. Choice A gets the sign backwards ($-51$) and then contradicts itself by correctly labeling a negative residual as underprediction — but a negative residual actually means overprediction, so both the number and the interpretation are inconsistent. Choice A likely results from computing $459 - 510$ instead of $510 - 459$. Choice C introduces $126$ dollars, which has no basis in the calculation, and the explanation about a "fixed spending component" is irrelevant — the intercept (75) is already included in the prediction. Choice D also uses $-126$, possibly from subtracting only the slope component ($48 \times 8 = 384$) from 510 while ignoring the intercept, making it an incomplete prediction error, not a true residual. A handy memory rule: positive residual = model fell short = underprediction; negative residual = model overshot = overprediction. On exam questions like this, always compute the full predicted value first before finding the difference — partial calculations are exactly what the wrong answer choices exploit.

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