Business Analytics Quiz: Simple Linear Regression

10 questions · exam conditions

0:00

Simple Linear RegressionQuestion 1 of 10

An online retailer predicts a customer's monthly spending, in dollars, from the number of prior orders using $y_{pred}=75+48x$ . A customer with $8$ prior orders actually spends $510$ dollars.

What is this customer's residual, and what does its sign indicate?

The residual is

-51

dollars, indicating that the model underpredicted the customer's spending

The residual is

51

dollars, indicating that the model underpredicted the customer's spending

The residual is

126

dollars, indicating that the model omitted the fixed spending component

The residual is

-126

dollars, indicating that the model overpredicted the customer's spending

← Back to quizzes

Business Analytics Quiz

Business Analytics Quiz: Simple Linear Regression

Practice Simple Linear 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 Simple Linear 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

An online retailer predicts a customer's monthly spending, in dollars, from the number of prior orders using $y_{pred}=75+48x$ . A customer with $8$ prior orders actually spends $510$ dollars.

What is this customer's residual, and what does its sign indicate?

The residual is $-51$ dollars, indicating that the model underpredicted the customer's spending
The residual is $51$ dollars, indicating that the model underpredicted the customer's spending (correct answer)
The residual is $126$ dollars, indicating that the model omitted the fixed spending component
The residual is $-126$ dollars, indicating that the model overpredicted the customer's spending

Explanation: 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.

Question 2

A company models weekly website conversions, $y$ , from advertising spending in dollars, $x$ , using $y_{pred}=18+0.024x$ . An analyst redefines the predictor as $z$ , the same spending measured in hundreds of dollars, so $x=100z$ .

How does this change of units affect the fitted equation and its predictions?

The equation becomes $y_{pred}=1800+0.024z$ , while predictions remain unchanged for equivalent spending levels
The equation becomes $y_{pred}=18+0.00024z$ , while predictions remain unchanged for equivalent spending levels
The equation remains $y_{pred}=18+0.024z$ , so predicted conversions become smaller at every positive spending level
The equation becomes $y_{pred}=18+2.4z$ , while predictions remain unchanged for equivalent spending levels (correct answer)

Explanation: 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

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.

Question 3

A logistics company regresses savings in thousands of dollars on each additional automation unit installed. A $95\%$ confidence interval for the slope is $[0.2,1.8]$ . Management considers the program operationally worthwhile only if the true average savings are at least $1.5$ thousand dollars per additional unit.

Which conclusion is most defensible from the confidence interval?

The association is statistically positive at the $5\%$ level, but the interval includes values below $1.5$ , so operational worthiness is not established (correct answer)
The point estimate of the slope equals $1.5$ , confirming both statistical significance and operational worthiness
Because $1.5$ lies inside the interval, the analysis confirms that the true slope meets management's threshold
Because zero is excluded, each automation unit necessarily causes savings of between $0.2$ and $1.8$ thousand dollars

Explanation: 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

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.

Question 4

A franchise analyst models first-year revenue from local population using stores with populations between $100{,}000$ and $500{,}000$ . The sample mean population is $300{,}000$ . Management wants an interval estimate for the revenue of one future store in a market with population $480{,}000$ .

Which interval should the analyst use, and how should its width generally compare with other regression intervals?

Use a confidence interval for mean revenue at $480{,}000$ ; it should be wider than a prediction interval at that population
Use a confidence interval for mean revenue at $300{,}000$ ; it directly measures uncertainty for the future store
Use a prediction interval at $480{,}000$ ; it should generally be narrower there than at the sample mean
Use a prediction interval at $480{,}000$ ; it should exceed the corresponding mean-response interval and usually widen away from the sample mean (correct answer)

Explanation: 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.

Question 5

An analyst fits a simple linear regression predicting order-processing time from order size. The residuals average approximately zero overall. However, average residuals are positive for small orders, negative for medium orders, and positive again for large orders.

What is the most appropriate interpretation of this residual behavior?

The linear specification is adequate because residuals that average zero cannot contain meaningful predictive structure
The pattern proves that residual variance increases with order size, so only a variance correction is needed
The systematic pattern suggests curvature, so the linear model may miss part of the relationship despite zero mean residuals (correct answer)
The fitted slope must equal zero because positive residuals occur at both low and high predictor values

Explanation: 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.

Question 6

For a simple linear regression with an intercept, the correlation between delivery distance and customer satisfaction is $-0.70$ . The total sum of squares for satisfaction, measured around its sample mean, is $800$ .

Which pair correctly gives the model's coefficient of determination and residual sum of squares?

$R^2=0.49$ and the residual sum of squares is $408$ (correct answer)
$R^2=-0.49$ and the residual sum of squares is $1192$
$R^2=0.70$ and the residual sum of squares is $240$
$R^2=0.49$ and the residual sum of squares is $392$

Explanation: 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.

Question 7

A wholesaler fits a simple linear regression relating weekly sales-call hours, $x$ , to weekly revenue in thousands of dollars, $y$ . The sample means are $40$ hours and $200$ thousand dollars. The sum of squared deviations for call hours is $S_{xx}=80$ , and the sum of cross-products is $S_{xy}=120$ .

According to the fitted regression, what revenue is predicted for a week with $52$ sales-call hours?

$218$ thousand dollars, using a fitted slope of $1.5$ and an intercept of $140$ (correct answer)
$182$ thousand dollars, using a fitted slope of $-1.5$ and an intercept of $260$
$278$ thousand dollars, adding $1.5(52)$ directly to the mean revenue
$200$ thousand dollars, because regression predictions remain centered at the sample mean

Explanation: 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.

Question 8

A customer-service manager models satisfaction score, $y$ , from average wait time in minutes, $x$ . The fitted equation is $y_{pred}=92-3.4x$ . Wait times in the data ranged from $2$ to $8$ minutes.

Which interpretation of the fitted model is most appropriate?

A two-minute increase causes satisfaction to decrease by $3.4$ points, and $92$ is the observed maximum score
A two-minute increase is associated with a $3.4$ -point decrease, and $92$ is the average observed score
A two-minute increase is associated with a $6.8$ -point decrease, while the intercept has limited practical meaning (correct answer)
A two-minute increase is associated with a $6.8$ -point increase, while the intercept describes zero-minute waits

Explanation: 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

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.

Question 9

A subscription company fits monthly demand, $y$ , against price in dollars, $p$ , using prices from $20$ through $50$ . The fitted equation is $y_{pred}=900-11p$ . Management is considering a price of $10$ , which was not represented in the data.

How should the model's prediction at the proposed price be interpreted?

The model establishes demand of $790$ subscriptions because a price of $10$ is within the equation's mathematical domain
The equation gives $790$ subscriptions, but this is an extrapolation whose reliability is not established by the observed data (correct answer)
The equation gives $1{,}010$ subscriptions, but the estimate is unreliable because the slope must reverse below the observed range
The model cannot produce a numerical prediction because regression equations are undefined outside the predictor's sample range

Explanation: 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

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.

Question 10

A retailer regresses weekly profit, $y$ , on the number of promotional emails sent, $x$ . With all observations included, the fitted line is $y_{pred}=80+12x$ . After removing one week with $x=20$ and $y=320$ , the fitted line is $y_{pred}=95+8x$ . Under the all-observation model, the removed week has a residual of zero.

Which conclusion is best supported by this information?

The removed week cannot be influential because an observation with a zero residual never affects fitted coefficients
The removed week is necessarily an outcome outlier because the reduced-data model misses its profit by $65$
The removed week can be influential despite its zero residual because it has an extreme predictor value and changes the fitted line (correct answer)
The removed week changes only the intercept because the relationship's marginal effect remains statistically identical

Explanation: 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.