Business Analytics Quiz: Multiple Regression

10 questions · exam conditions

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Multiple RegressionQuestion 1 of 10

A delivery company regresses customer satisfaction on average wait time, order size, and the number of employees assigned to an order. The estimated coefficient on wait time is $-1.6$ , with satisfaction measured in points and wait time measured in minutes. The data are observational.

Which interpretation of the wait-time coefficient is most defensible?

An additional minute of wait time causes satisfaction to decline by exactly

1.6

points for every order

Holding order size and staffing constant, one additional minute is associated with

1.6

fewer satisfaction points on average

Reducing average wait time by one minute will increase total company satisfaction by

1.6

points overall

Wait time and satisfaction have a simple bivariate correlation of

-1.6

after the other variables are removed

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Business Analytics Quiz

Business Analytics Quiz: Multiple Regression

Practice Multiple Regression in Business Analytics 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 Multiple Regression, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Analytics.

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

Which interpretation of the wait-time coefficient is most defensible?

An additional minute of wait time causes satisfaction to decline by exactly $1.6$ points for every order
Holding order size and staffing constant, one additional minute is associated with $1.6$ fewer satisfaction points on average (correct answer)
Reducing average wait time by one minute will increase total company satisfaction by $1.6$ points overall
Wait time and satisfaction have a simple bivariate correlation of $-1.6$ after the other variables are removed

Explanation: Whenever you see a regression coefficient in a business-analytics question, ask yourself two things: does the data allow a causal claim, and does the coefficient represent a partial or bivariate relationship? Those two checks will cut through nearly every interpretation trap. Here, the coefficient of

-1.6

on wait time comes from a multiple regression using observational data. Because the regression controls for order size and staffing, the coefficient already holds those variables constant — it captures the partial relationship between wait time and satisfaction. Because the data are observational (not from a controlled experiment), you cannot assert that wait time causes the change. The only defensible language is associational. That makes B correct: "holding order size and staffing constant, one additional minute is associated with

1.6

fewer satisfaction points on average." It respects both the partial nature of the coefficient and the observational limitation. A fails on two counts: it uses causal language ("causes") and drops the critical caveat that other variables are held constant — both unwarranted given observational data. C compounds the causal error by projecting the effect onto "total company satisfaction," which the regression coefficient simply does not address; the coefficient is a per-order average, not an aggregate claim. D mischaracterizes what a multiple regression coefficient is — it is not a simple bivariate correlation with other variables "removed." Partial regression coefficients and simple correlations are fundamentally different quantities. Study tip: When evaluating regression interpretations, mentally run a checklist — causal or associational? partial or bivariate? average or aggregate? Observational data always forces associational language, and multiple regression always forces the "holding other variables constant" qualifier.

Question 2

A retailer fits the multiple regression model $PredictedSales = 12 + 0.8(AdSpend) + 3(Online) - 0.2(PricePremium),$ where quarterly sales and advertising spending are measured in thousands of dollars, $Online = 1$ for an online campaign and $Online = 0$ otherwise, and price premium is measured in dollars.

What quarterly sales does the model predict for an online campaign with advertising spending of $10$ thousand dollars and a price premium of $20$ dollars?

$19$ thousand dollars, after including the online effect and subtracting the price-premium effect (correct answer)
$16$ thousand dollars, after excluding the online indicator but retaining the other model terms
$27$ thousand dollars, after treating the price-premium coefficient as a positive effect
$7$ thousand dollars, after summing the predictor contributions but excluding the intercept

Explanation: When you see a multiple regression equation, your job is straightforward: substitute the given values for each predictor and calculate the result carefully, paying close attention to signs and which indicators apply. Here, the model is

\text{PredictedSales} = 12 + 0.8(\text{AdSpend}) + 3(\text{Online}) - 0.2(\text{PricePremium}).

For an online campaign with AdSpend = 10 and PricePremium = 20, plug in directly:

12 + 0.8(10) + 3(1) - 0.2(20) = 12 + 8 + 3 - 4 = 19.

The predicted quarterly sales are 19 thousand dollars, making A correct. Choice B arrives at 16 thousand dollars by setting

Online = 0

instead of 1, ignoring that the campaign is explicitly online. This means excluding the

+3

contribution, which directly contradicts the problem setup. Choice C reaches 27 thousand dollars by treating the price-premium coefficient as positive

(+0.2)

rather than negative

(-0.2)

, turning a cost drag into a benefit. Sign errors are one of the most common mistakes in regression problems — always preserve the coefficient's sign. Choice D gets 7 thousand dollars by omitting the intercept entirely. In regression, the intercept (here, 12) is always part of the prediction equation — it is not optional. Study tip: When substituting into a regression equation, work term by term and write out each product before summing. Double-check indicator variable values (0 vs. 1) and coefficient signs — these are exactly where exam questions hide their traps.

Question 3

A marketing analyst first regresses sales on advertising spending alone and obtains an advertising coefficient of $2.4.$ After adding market size to the model, the advertising coefficient falls to $1.1.$ Market size is positively associated with both advertising spending and sales.

Which explanation of the coefficient change is most appropriate?

The coefficient decline shows that market size is irrelevant because relevant predictors always increase existing coefficients
The second model proves that every dollar of advertising causally generates exactly $1.1$ dollars of sales
Adding market size reverses the sign of omitted-variable bias, so the advertising coefficient must become negative
The first model likely attributed some positive market-size effect to advertising because market size was omitted (correct answer)

Explanation: Whenever you see a regression coefficient change after adding a new variable, think omitted-variable bias (OVB). The core idea: if a variable is left out of a model, its effect gets absorbed by whichever included variables are correlated with it. Here, market size is positively correlated with both advertising spending and sales. In the first model, the regression has no way to "know" that some of the sales variation comes from market size — so it wrongly credits advertising for that effect. The advertising coefficient of

2.4

is inflated because it's doing double duty: capturing the true advertising effect plus the indirect effect of market size. Once you add market size to the model in step two, that borrowed effect is properly assigned, and the advertising coefficient drops to

1.1

. This is exactly what D describes — the first model attributed some positive market-size effect to advertising because market size was omitted. A is wrong because there is no rule that relevant predictors always increase existing coefficients. Adding a variable can raise, lower, or even reverse a coefficient depending on the direction and magnitude of the correlation. B is wrong because regression coefficients represent associations, not proven causal relationships. Establishing causality requires much more than a well-fitted model — it requires careful study design and ruling out confounders. C is wrong because OVB doesn't automatically reverse signs. The direction of the bias depends on the signs of the correlations involved. Since both correlations here are positive, the bias is upward, not sign-reversing. Your takeaway: when a coefficient shrinks after adding a variable, ask yourself whether the omitted variable was positively correlated with both the predictor and the outcome — that's the classic upward OVB signature.

Question 4

A sales model includes price, competitor price, household income, and neighborhood spending. The overall regression has an $F$ -test p-value of $0.003,$ but each individual slope has a two-sided t-test p-value greater than $0.10.$ Several predictors are highly correlated.

What is the most accurate interpretation of these results?

Exactly one slope is nonzero, but the $F$ -test cannot identify which coefficient has the nonzero value
Every slope is individually significant because the overall $F$ -test p-value is below the usual significance threshold
No predictor contributes to the model because each individual t-test fails to reject its corresponding null hypothesis
The model provides evidence that at least one slope is nonzero, although correlated predictors may prevent precise identification of individual effects (correct answer)

Explanation: Whenever you see a regression question where the overall

F

-test is significant but individual

t

-tests are not, your immediate instinct should be multicollinearity. These two facts together are a classic diagnostic pattern, not a contradiction. The

F

-test evaluates whether at least one slope is nonzero by examining all predictors jointly. A p-value of

0.003

gives strong evidence that the model explains meaningful variation in sales — something real is going on. However, when predictors like competitor price, household income, and neighborhood spending are highly correlated, regression struggles to isolate each variable's individual contribution. The coefficients become unstable and their standard errors inflate, causing individual

t

-test p-values to exceed

0.10

even when the predictors collectively matter. Answer D captures this exactly: the model has joint explanatory power, but correlated predictors prevent precise attribution of effects to specific variables. A is wrong because the

F

-test says at least one slope is nonzero — it makes no claim that exactly one is, and it's not designed to identify which predictor is responsible. B reverses the logic entirely: a significant

F

-test does not imply every individual slope is significant. Those are separate hypotheses tested separately. C commits the most dangerous error — concluding that because each

t

-test fails individually, no predictor matters. This ignores that multicollinearity can suppress individual significance while the predictors still jointly explain the outcome. As a study habit, memorize this pattern: significant

F

+ insignificant

t

's + correlated predictors = multicollinearity, not model failure.

Question 5

In a randomized store-level promotion test, treated stores had pre-promotion sales averaging $10$ thousand dollars more than control stores by chance. The unadjusted difference in post-promotion mean sales was $12.5$ thousand dollars. A regression of post-promotion sales on pre-promotion sales and a treatment dummy produces $PredictedPostSales = 20 + 0.85(PreSales) + 4(Treatment),$ where sales are measured in thousands of dollars.

Which interpretation best reconciles the unadjusted difference with the regression result?

The adjusted treatment effect is $12.5$ thousand dollars because randomization requires the regression and raw estimates to match
The treatment accounts for $8.5$ thousand dollars of the raw gap, leaving a baseline effect of $4$ thousand dollars
The baseline sales imbalance accounts for $8.5$ thousand dollars of the raw gap, leaving an adjusted treatment effect of $4$ thousand dollars (correct answer)
The adjusted treatment effect is $16.5$ thousand dollars because the treatment coefficient must be added to the raw difference

Explanation: Whenever you see a question combining a raw difference-in-means with a regression-adjusted estimate, your job is to decompose where the raw gap comes from — specifically, how much is explained by pre-existing imbalances versus the actual treatment effect. Here, treated stores started with

\$10

thousand higher pre-promotion sales by chance. The regression coefficient on pre-sales is

0.85

, meaning each dollar of pre-sale advantage predicts

\$0.85

in post-sales. So the baseline imbalance alone predicts a post-sales gap of

0.85 \times 10 = \$8.5

thousand, even without any promotion. The remaining

12.5 - 8.5 = \$4

thousand is the true, covariate-adjusted treatment effect — exactly what the regression's treatment coefficient reports. That's why C is correct: the

\$8.5

thousand is attributable to the pre-sales imbalance, leaving a clean

\$4

thousand treatment effect. A is wrong because randomization does not guarantee that raw and adjusted estimates match — it prevents systematic bias on average, but chance imbalances still occur in any single experiment, which is precisely what happened here. B reverses the logic: it assigns

\$8.5

thousand to the treatment and calls

\$4

thousand a "baseline effect," which is backwards. The

\$8.5

thousand is the baseline artifact, and

\$4

thousand is the treatment effect. D invents an additive rule that has no statistical basis — you don't add the treatment coefficient to the raw difference. As a strategy, always check whether a covariate imbalance times its regression coefficient explains the gap between raw and adjusted estimates. That multiplication (imbalance × slope) is the adjustment doing the work.

Question 6

Using $50$ observations, an analyst fits a base model with $4$ predictors and obtains $R^2 = 0.64.$ An expanded model with $8$ predictors produces $R^2 = 0.66.$ The analyst will choose the model with the higher adjusted $R^2,$ calculated as $AdjustedR^2 = 1 - (1-R^2)(n-1)/(n-p-1),$ where $p$ is the number of predictors.

Which model should the analyst choose under this rule?

The expanded model, because its adjusted $R^2$ is approximately $0.651$ and exceeds that of the base model
The base model, because its adjusted $R^2$ is approximately $0.608$ versus $0.594$ for the expanded model (correct answer)
The expanded model, because any increase in ordinary $R^2$ guarantees an increase in adjusted $R^2$
Either model, because the two adjusted $R^2$ values must equal their respective ordinary $R^2$ values

Explanation: Whenever you see a question comparing models by adjusted

R^2

, remember that adjusted

R^2

penalizes for each additional predictor added. Unlike ordinary

R^2

, which can only stay the same or increase as you add variables, adjusted

R^2

can actually decrease if the new predictors don't contribute enough explanatory power to justify their inclusion. The formula is

\text{Adj.} R^2 = 1 - (1 - R^2)\frac{n-1}{n-p-1}

. For the base model (

n=50, p=4, R^2=0.64

): the multiplier is

49/45 \approx 1.0889

, giving

1 - (0.36)(1.0889) \approx 1 - 0.392 = 0.608

. For the expanded model (

p=8, R^2=0.66

): the multiplier is

49/41 \approx 1.1951

, giving

1 - (0.34)(1.1951) \approx 1 - 0.406 = 0.594

. The base model wins with

\approx 0.608

versus

\approx 0.594

, confirming B is correct. A is wrong because it reports an adjusted

R^2

0.651

for the expanded model — this number is simply incorrect given the formula and inputs. C is a classic misconception: ordinary

R^2

rising does not guarantee adjusted

R^2

rises, because the penalty term grows with each added predictor. D is wrong because adjusted and ordinary

R^2

are only equal in trivial cases; the adjustment always lowers

R^2

when

p > 0

. As a study tip: when comparing models, always compute both adjusted values from scratch rather than assuming the model with higher

R^2

automatically wins. The penalty for complexity is the whole point of the adjustment.

Question 7

A bank predicts loan-processing time using application length, number of uploaded documents, and number of pages submitted. These predictors are strongly correlated. The model has a high $R^2$ and good validation accuracy, but the individual coefficients change substantially when a small number of records are removed.

Which conclusion best explains these results?

Multicollinearity is likely inflating coefficient uncertainty even though combined predictions within the observed predictor range may remain useful (correct answer)
Heteroskedasticity is necessarily causing biased fitted values even though all three predictors have strong individual effects
The high $R^2$ proves that each coefficient is stable and can be interpreted independently as a causal effect
The model must be underfit because correlated predictors always reduce both training fit and validation accuracy

Explanation: Whenever you see a question where predictors are strongly correlated, a model fits well overall, but individual coefficients are unstable, you're looking at a multicollinearity problem — and understanding what multicollinearity does (and doesn't) break is the key to this question. Multicollinearity means predictors share so much information that the model can't reliably separate each one's individual contribution. The coefficients become highly sensitive to small data changes — exactly the instability described in the passage. However, the combined prediction within the observed data range can still be accurate, which is why

R^2

stays high and validation accuracy holds up. The model knows collectively what the predictors explain; it just can't disentangle who deserves credit. Answer A captures this nuance precisely and is correct. Answer B is wrong because heteroskedasticity refers to non-constant variance in residuals — it doesn't cause the coefficient instability described here, and it doesn't necessarily bias fitted values when predictors are well-measured. Nothing in the passage points to heteroskedasticity. Answer C is a classic trap. A high

R^2

tells you the model predicts well overall, but it says nothing about coefficient stability or causal interpretation. In fact, high

R^2

alongside multicollinearity is precisely the scenario where individual coefficients are least trustworthy. Answer D is wrong because correlated predictors don't automatically reduce fit. Underfitting would show poor

R^2

and poor validation — the opposite of what's described. Study tip: Remember that multicollinearity harms coefficient interpretability, not necessarily predictive accuracy. If an exam question separates these two outcomes, multicollinearity is almost always the culprit.

Question 8

A subscription business estimates $PredictedRevenue = 40 + 5(Discount) + 8(Online) - 2(Discount)(Online),$ where discount is measured in percentage points and $Online = 1$ for online offers. Revenue is measured in dollars per customer.

Which statement correctly compares online and non-online offers under this model?

A three-point discount increase raises revenue by $15$ dollars for both channels, and online always adds $8$ dollars
A three-point discount increase raises online revenue by $9$ dollars, and at a five-point discount online revenue is $2$ dollars lower (correct answer)
A three-point discount increase lowers online revenue by $6$ dollars, and at a five-point discount online revenue is $8$ dollars higher
A three-point discount increase raises online revenue by $15$ dollars, and at a five-point discount online revenue is $10$ dollars lower

Explanation: When a regression model includes an interaction term, the effect of one variable depends on the value of another. Here, the term

-2(\text{Discount})(\text{Online})

means the discount's impact on revenue differs by channel — so you must evaluate each channel separately rather than reading coefficients in isolation. For non-online offers (Online = 0), the model simplifies to

40 + 5(\text{Discount})

, so each additional discount point adds $5. A three-point increase raises revenue by

5 \times 3 = 15

dollars. For online offers (Online = 1), the model becomes

40 + 5(\text{Discount}) + 8 - 2(\text{Discount}) = 48 + 3(\text{Discount})

. A three-point increase raises online revenue by only

3 \times 3 = 9

dollars. At a five-point discount, online revenue =

48 + 15 = 63

and non-online revenue =

40 + 25 = 65

, making online revenue $2 lower. That confirms B is correct. A is wrong because it ignores the interaction — a three-point discount raises online revenue by only $9, not $15, and the online premium isn't a flat $8 (it shrinks as discount grows). C is wrong because it claims the discount lowers online revenue, but the net coefficient for online is still positive (+3). D is wrong on both counts: it applies the non-online slope to online, and overstates the gap at five points as $10 rather than $2. When you spot an interaction term, always substitute each group's indicator value first to find the effective slope before doing any arithmetic.

Question 9

A pricing analyst fits a linear multiple regression of demand on price, advertising, and competitor price. After controlling for advertising and competitor price, residuals for very low and very high prices are mostly negative, while residuals for middle prices are mostly positive. This pattern repeats in both training and validation data.

Which model revision most directly addresses the indicated problem while retaining the existing controls?

Add a dummy variable for positive residuals because it will directly separate underpredictions from overpredictions
Remove price entirely because a patterned residual automatically means the predictor contains no useful information
Add a squared-price term while retaining the linear price, advertising, and competitor-price terms (correct answer)
Standardize all predictors because changing measurement units necessarily eliminates curvature in residuals

Explanation: Whenever residuals form a systematic U-shaped (or inverted-U) pattern against a predictor — negative at the extremes, positive in the middle — your model is missing a nonlinear relationship. This is the classic signal that a quadratic term is needed, not a data-cleaning fix or a variable removal. Here, price's effect on demand bends: demand drops off at both very low and very high prices relative to a straight line. The fix is to add a squared-price term (

\text{Price}^2

) alongside the original linear price term, giving the model the flexibility to capture that curve. Because this revision appears in both training and validation data, it reflects a genuine structural pattern — not overfitting. Option C does exactly this while keeping advertising and competitor price intact, directly targeting the curvature without discarding information. Option A is a trap: creating a dummy variable for "positive residuals" is circular — you'd be using the outcome of the model to build a predictor for itself, which tells you nothing about the underlying price-demand relationship and introduces data leakage. Option B is wrong because a patterned residual means the predictor is informative — it just isn't modeled correctly. Removing price entirely would discard the very signal you're trying to fix. Option D misunderstands standardization. Rescaling predictors changes units and can improve numerical stability, but it cannot remove nonlinear patterns. Curvature in residuals survives any linear transformation of inputs. Study tip: On residual-pattern questions, always ask: Is the pattern systematic against a specific predictor? If yes, that predictor likely needs a polynomial or transformed term — not removal or rescaling.

Question 10

A company models monthly account revenue by region using $North$ as the reference category: $PredictedRevenue = 72 - 4(South) + 6(East) + 2(West),$ where exactly one regional indicator equals $1$ for non-North accounts and revenue is measured in thousands of dollars.

If the analyst refits the model using $East$ as the reference category, what should happen to the coefficient for $West$ and to the fitted regional revenues?

The $West$ coefficient becomes $-4$ and all fitted regional revenues remain unchanged (correct answer)
The $West$ coefficient stays $2$ and the intercept stays $72$ , because switching the reference category does not alter any coefficients
The $West$ coefficient becomes $4$ and every fitted regional revenue increases by $6$ , because the East premium is redistributed
The $West$ coefficient becomes $-6$ and only the fitted revenue for $North$ changes, because North is no longer the reference

Explanation: Whenever you see a question about changing the reference category in a dummy-variable regression, remember the golden rule: the reference category changes how coefficients are defined, but never changes the model's fitted values. With

North

as reference, the intercept 72 represents predicted North revenue, and each coefficient measures a region's revenue relative to North. So the fitted revenues are: North =

72

, South =

68

, East =

78

, West =

74

. When you refit using

East

as the new reference, the intercept shifts to East's fitted value:

78

. Every other coefficient now measures distance from East. West's new coefficient becomes

74 - 78 = -4

. North's coefficient becomes

72 - 78 = -6

. South's becomes

68 - 78 = -10

. The fitted values for all four regions remain exactly

72, 68, 78, 74

— nothing changes about what the model predicts. This confirms A is correct: the West coefficient becomes

-4

and all fitted regional revenues remain unchanged. B is wrong because coefficients absolutely do change when you switch the reference — only fitted values stay constant. C is wrong on both counts: the West coefficient is

-4

, not

4

, and fitted revenues do not shift at all — there is no "redistribution" of the East premium. D is wrong because all coefficients change (not just North's), and North's fitted revenue stays at

72

regardless of reference choice. Study tip: On reference-category questions, always compute the actual fitted values for each group first — they are your anchor. Coefficients are just arithmetic differences from whichever group you chose as the baseline.