Business Statistics Quiz: Correlation And Simple Regression

20 questions · exam conditions

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Correlation And Simple RegressionQuestion 1 of 20

A marketing manager models the relationship between monthly advertising spend in dollars (X) and monthly sales in dollars (Y). The resulting regression equation is $\hat{Y} = 5000 + 2.5X$ , with a correlation coefficient $r = 0.70$ . If the manager decides to re-run the analysis with advertising spend measured in thousands of dollars (let's call it $X^*$ ), what will be the new slope and correlation coefficient?

The new slope will be 0.0025, and the correlation will be 0.70.

The new slope will be 2500, and the correlation will be 0.70.

The new slope will be 2500, and the correlation will be 0.07.

The new slope will be 2.5, and the correlation will be 0.0007.

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Business Statistics Quiz

Business Statistics Quiz: Correlation And Simple Regression

Practice Correlation And Simple Regression in Business Statistics 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 Correlation And Simple Regression, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Statistics.

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

The new slope will be 0.0025, and the correlation will be 0.70.
The new slope will be 2500, and the correlation will be 0.70. (correct answer)
The new slope will be 2500, and the correlation will be 0.07.
The new slope will be 2.5, and the correlation will be 0.0007.

Explanation: Correlation is a unitless measure and is unaffected by linear transformations (like changing units from dollars to thousands of dollars). Thus, the correlation remains 0.70. The slope, however, will change. Since the new predictor

X^*

represents thousands of dollars,

X = 1000X^*

. Substituting this into the original equation gives

\hat{Y} = 5000 + 2.5(1000X^*) = 5000 + 2500X^*

. The new slope is 2500.

Question 2

In a simple linear regression analysis of sales revenue (Y) versus advertising budget (X), the hypothesis test for the slope coefficient ( $\beta_1$ ) yields a p-value of 0.03. Using a significance level of $\alpha = 0.05$ , what is the correct conclusion?

There is a 3% chance that the null hypothesis is true.
The model is a good fit, explaining at least 95% of the variance in sales revenue.
There is statistically significant evidence of a linear relationship between advertising budget and sales revenue. (correct answer)
We are 97% confident that a linear relationship exists between advertising budget and sales revenue.

Explanation: The p-value represents the probability of observing a sample slope as extreme as the one obtained, assuming the null hypothesis (that the true slope is zero) is true. Since the p-value (0.03) is less than the significance level (0.05), we reject the null hypothesis. This means we have sufficient statistical evidence to conclude that a linear relationship exists between the variables.

Question 3

A data analyst finds that for a simple linear regression with $n = 25$ observations, the sum of squared residuals is $SSE = 180$ and the total sum of squares is $SST = 500$ . What percentage of the variation in the dependent variable is explained by the regression model?

$36\%$
$64\%$ (correct answer)
$78\%$
$91\%$

Explanation: The coefficient of determination is

R^2 = 1 - \frac{SSE}{SST} = 1 - \frac{180}{500} = 1 - 0.36 = 0.64 = 64\%

. Choice A gives

\frac{SSE}{SST} = 36\%

, which represents unexplained variation. Choice C incorrectly calculates

\sqrt{0.64} \times 100\%

. Choice D uses an incorrect formula altogether.

Question 4

A logistics company analyzes the relationship between the weight of a package (X, in pounds) and its delivery time (Y, in hours). The analysis provides the following summary statistics: Standard deviation of package weight ( $s_x$ ) = 5 pounds, Standard deviation of delivery time ( $s_y$ ) = 10 hours, and the slope of the regression line ( $b_1$ ) = 1.6. What is the coefficient of determination ( $R^2$ ) for this model?

0.32
0.64 (correct answer)
0.80
1.28

Explanation: This is a two-step problem. First, find the correlation coefficient,

r

, using the formula

b_1 = r \frac{s_y}{s_x}

. Rearranging gives

r = b_1 \frac{s_x}{s_y} = 1.6 \times \frac{5}{10} = 1.6 \times 0.5 = 0.8

. Second, find the coefficient of determination by squaring

r

R^2 = r^2 = (0.8)^2 = 0.64

Question 5

An analyst creates a simple linear regression model. To check the model's validity, they produce a plot of the residuals versus the predicted values. The plot shows the points forming a distinct horizontal funnel shape, wider on the right than on the left. What does this pattern in the residual plot indicate?

The relationship between the variables is non-linear.
The variance of the errors is not constant (heteroscedasticity). (correct answer)
The errors are not normally distributed.
There is a significant outlier in the data.

Explanation: A plot of residuals versus predicted values is used to detect patterns in the errors. A funnel or fan shape, where the vertical spread of the residuals changes as the predicted values change, is a classic sign of heteroscedasticity, which is the violation of the constant variance assumption.

Question 6

A regression analysis is performed on 50 data points, showing a moderate positive correlation. A 51st data point is then added. This new point has an x-value far from the mean of the other x-values and falls well below the established regression line. Which of the following is the most likely impact of adding this influential point?

It will decrease the y-intercept and decrease the slope.
It will have a negligible effect as it is only one point out of 51.
It will decrease the slope but increase the correlation coefficient.
It will increase the y-intercept and decrease the slope. (correct answer)

Explanation: When you encounter regression problems involving influential points or outliers, focus on how extreme data points can "pull" the regression line toward them, especially when they're far from the mean of the x-values. This new point is described as having an x-value far from the mean and falling "well below" the established regression line. Since the regression line will be pulled toward this influential point, imagine how the line must rotate and shift to get closer to this below-the-line point. The correct answer is D because the line must rotate downward (decreasing the slope) to move toward the low y-value point. When a line rotates downward around data that's mostly to the right of the y-axis, the y-intercept typically increases - think of it as the left end of the line lifting up as the right end drops down to approach the outlier. Answer A incorrectly suggests both the intercept and slope decrease. While the slope does decrease, the y-intercept actually increases due to the rotation pattern. Answer B falls into the trap of thinking one point can't matter much - but influential points with extreme x-values have disproportionate impact on regression lines, especially with moderate sample sizes like 50. Answer C correctly identifies that the slope decreases but wrongly claims the correlation coefficient increases. Adding an outlier that doesn't follow the established pattern typically weakens the correlation. Remember: influential points with extreme x-values act like leverage points that can dramatically alter regression lines. The farther a point is from the x-mean, the more it can "pull" the line toward itself.

Question 7

A researcher regresses employee job satisfaction (Y) on the number of hours worked per week (X) and finds a regression line $\hat{Y} = b_0 + b_1 X$ and a coefficient of determination $R^2$ . If the researcher decides to switch the variables, regressing hours worked (now Y') on job satisfaction (now X'), which of the following statements about the new model is true?

The new slope, $b'_1$ , will be the reciprocal of the original slope ( $1/b_1$ ).
The sign of the new slope, $b'_1$ , will be the opposite of the sign of the original slope, $b_1$ .
The new coefficient of determination, $R'^2$ , will be identical to the original $R^2$ . (correct answer)
The p-value for testing the significance of the new slope will be different from the original p-value.

Explanation: The coefficient of determination,

R^2

, is symmetric. It measures the proportion of variance in one variable that is explained by the other, regardless of which is designated as the dependent variable. Correlation (r) is also symmetric. The slope is not symmetric, and the p-value for the slope will be identical because the underlying t-statistic for the relationship is the same.

Question 8

A regression model predicting monthly utility costs for a factory based on production volume has a standard error of the estimate ( $S_e$ ) of $450. The model predicts a utility cost of $12,000 for a month with a production volume of 1,000 units. Which is the best interpretation of this information?

There is a 95% probability that the actual utility cost for this month will be between $11,100 and $12,900.
The standard deviation of the monthly utility costs in the original sample data is $450.
The model's predictions have a typical error magnitude of about $450. (correct answer)
The model's predictions are, on average, off by $450 for every 1,000 units of production.

Explanation: The standard error of the estimate (

S_e

), also known as the residual standard error, represents the typical or average size of the residuals (prediction errors). It is a measure of the typical distance between the observed values and the values predicted by the regression line, in the units of the dependent variable.

Question 9

A business analyst is preparing to run a simple linear regression. Which of the following conditions is generally considered the least critical for the primary purpose of estimating the regression coefficients (slope and intercept), although it is more important for hypothesis testing and interval estimation?

The relationship between the variables should be approximately linear.
The observations should be independent of one another.
The residuals should follow a normal distribution. (correct answer)
The variability of the residuals should be constant for all values of the predictor variable.

Explanation: The normality assumption of the residuals is primarily required for the validity of hypothesis tests (t-tests) and the construction of confidence and prediction intervals. The regression line's coefficients (slope and intercept) can be calculated and used for point prediction even if normality is violated, especially in large samples. The other conditions (linearity, independence, constant variance) are more fundamental to the model being a meaningful and unbiased representation of the relationship.

Question 10

A company-wide analysis shows a strong positive correlation between employees' years of experience and their performance review scores. However, when the data is analyzed separately for each of the company's three divisions, each division shows a slight negative correlation. What is the best explanation for this apparent contradiction?

The overall positive correlation is correct, and the divisional analyses are flawed due to smaller sample sizes.
The analysis must have used the wrong variables; experience should have been the dependent variable.
The negative correlations within divisions are spurious and likely due to measurement error in the performance scores.
A confounding variable (the division) is creating an artificial positive trend when the data are aggregated. (correct answer)

Explanation: When you encounter contradictory correlation patterns between aggregated and subgroup data, you're looking at a classic case of Simpson's Paradox. This occurs when a confounding variable masks the true relationship within groups but creates a misleading pattern when data are combined. Here's what's happening: Each division likely has different baseline characteristics - perhaps Division A hires mostly entry-level employees who receive intensive training, Division B focuses on mid-career hires, and Division C recruits senior executives. Within each division, more experienced employees might actually perform slightly worse (negative correlation), possibly due to complacency or outdated skills. However, when you combine all divisions, the higher-performing divisions happen to have more experienced employees overall, creating an artificial positive correlation. Answer D correctly identifies that the division itself is the confounding variable creating this paradox. The aggregated positive correlation is misleading because it ignores important structural differences between divisions. Answer A incorrectly assumes sample size is the issue - Simpson's Paradox can occur regardless of sample size. Answer B misunderstands the relationship direction; the dependent/independent variable assignment isn't the problem. Answer C wrongly attributes the pattern to measurement error when it's actually a statistical phenomenon caused by data aggregation. Study tip: Whenever you see opposing correlation patterns between grouped and ungrouped data, immediately think Simpson's Paradox and look for a lurking confounding variable. This is a favorite exam scenario because it tests your understanding of when correlation can be misleading.

Question 11

For a dataset, the mean of the independent variable ( $\bar{x}$ ) is 40 and the mean of the dependent variable ( $\bar{y}$ ) is 100. The slope of the simple linear regression line is 1.5. What is the y-intercept ( $b_0$ ) of the regression line?

-20
Cannot be determined without the correlation coefficient.
60
40 (correct answer)

Explanation: This question tests your understanding of the fundamental relationship between a regression line and the means of your variables. In simple linear regression, there's a key property: the regression line always passes through the point

(\bar{x}, \bar{y})

, which represents the intersection of the mean values of both variables. Using the regression equation

y = b_0 + b_1x

, you can substitute the known values. Since the line passes through

(\bar{x}, \bar{y}) = (40, 100)

and the slope

b_1 = 1.5

, you have:

100 = b_0 + 1.5(40)

100 = b_0 + 60

b_0 = 100 - 60 = 40

Let's examine why the other options are incorrect. Option A (-20) would result if you mistakenly subtracted the mean values:

40 - 100 = -60

, though even this doesn't yield -20, making it a clear distractor. Option B suggests you need the correlation coefficient, but this is false—the y-intercept can always be calculated when you know the slope and the means of both variables, regardless of correlation strength. Option C (60) represents the common error of calculating

b_1 \times \bar{x} = 1.5 \times 40 = 60

, which gives you the wrong component of the equation. Remember this key principle: the regression line always passes through

(\bar{x}, \bar{y})

. Whenever you're given the slope and both means, you can immediately find the y-intercept using this relationship. This property makes regression problems much more manageable than they initially appear.

Question 12

A manufacturer models the number of defective units per day (Y) as a function of the production line speed (X, measured in units per minute). The fitted regression model is $\widehat{Defects} = 3.2 + 0.05 \times (Speed)$ . How should the slope of 0.05 be interpreted?

For each one unit-per-minute increase in line speed, the model predicts an average increase of 0.05 defective units per day. (correct answer)
For each one unit-per-minute increase in line speed, the number of defective units increases by 5%.
For every 20 units-per-minute increase in line speed, the model predicts an average increase of one defective unit per day.
On average, 5% of all units produced per day are defective.

Explanation: The slope of a regression line represents the predicted change in the dependent variable (Y) for a one-unit increase in the independent variable (X). In this context, Y is 'defective units per day' and X is 'speed in units per minute'. Therefore, the slope of 0.05 means that for each additional unit per minute of speed, the predicted number of defects per day increases by 0.05. Choice C is a correct restatement but not the direct interpretation of the slope value itself. Choice B incorrectly interprets the slope as a percentage change.

Question 13

A marketing analyst finds that the correlation coefficient between advertising expenditure (in thousands of dollars) and monthly sales revenue (in thousands of dollars) is $r = 0.72$ . If the standard deviation of advertising expenditure is $s_x = 15$ and the standard deviation of sales revenue is $s_y = 45$ , what is the slope of the least squares regression line for predicting sales revenue from advertising expenditure?

$2.16$ (correct answer)
$0.24$
$3.33$
$10.8$

Explanation: The slope of the regression line is

b_1 = r \cdot \frac{s_y}{s_x} = 0.72 \cdot \frac{45}{15} = 0.72 \cdot 3 = 2.16

. Choice B incorrectly uses

\frac{s_x}{s_y}

in the denominator. Choice C uses

\frac{s_y}{s_x}

without multiplying by the correlation. Choice D incorrectly multiplies

r

by both standard deviations.

Question 14

A firm is evaluating two different simple linear regression models to forecast sales. Model A uses 'Advertising Spend' as the predictor and has an $R^2$ of 0.49 and a standard error of the estimate ( $S_e$ ) of $10,000. Model B uses 'Number of Sales Calls' as the predictor and has an $R^2$ of 0.64 and an $S_e$ of $8,000. Which statement is the most accurate comparison of the two models?

Model A is preferable because a lower $R^2$ indicates less risk of overfitting the data.
Model B is preferable as it explains more variance in sales and has smaller typical prediction errors. (correct answer)
The models are equally good because Model A's correlation (0.7) and Model B's correlation (0.8) are both strong.
The models cannot be compared without knowing the p-values for their slope coefficients.

Explanation: When comparing regression models for the same dependent variable, a better model generally has a higher

R^2

and a lower standard error of the estimate (

S_e

). Model B has a higher

R^2

(0.64 > 0.49), meaning it explains a larger proportion of the variance in sales. It also has a lower

S_e

($8,000 < $10,000), meaning its predictions are typically more accurate. Therefore, Model B is the superior model based on these metrics.

Question 15

An analyst is comparing two potential linear models to describe a dataset. Model A is the least-squares regression line, $\hat{y} = 10 + 2x$ . Model B is an alternative line, $\hat{y} = 11 + 1.9x$ , proposed by a manager based on industry experience. If both lines are applied to the same dataset, what can be definitively concluded about their sum of squared residuals (SSE)?

The SSE for Model A will be strictly less than the SSE for Model B.
The SSE cannot be compared without knowing the $R^2$ for each model.
The SSE for Model B will be smaller because its intercept is higher.
The SSE for Model A will be less than or equal to the SSE for Model B. (correct answer)

Explanation: When you encounter questions comparing regression models, remember that the least-squares regression line has a special mathematical property: it minimizes the sum of squared residuals (SSE) for any given dataset. The least-squares method finds the unique line that produces the smallest possible SSE by minimizing

\sum(y_i - \hat{y}_i)^2

. This is its defining characteristic. Since Model A is the least-squares line for this dataset, it must have an SSE that is less than or equal to any other possible line fitted to the same data, including Model B. Why "less than or equal to" rather than "strictly less than"? In the extremely unlikely scenario where the manager's proposed line happens to be identical to the least-squares solution, the SSEs would be equal. Otherwise, Model A will have a smaller SSE. Option A is incorrect because it claims Model A will be "strictly" less than Model B, but we can't rule out the possibility they could be equal if the manager somehow guessed the exact least-squares line. Option B is wrong because

R^2

isn't needed to make this comparison—the least-squares property guarantees the relationship regardless of how well either model fits. Option C reflects a fundamental misunderstanding, as a higher intercept doesn't determine SSE; the overall fit to the data points does. Study tip: Remember that "least-squares" means exactly that—it produces the least sum of squared errors. Any time you compare a least-squares model to any alternative model on the same dataset, the least-squares model will have SSE ≤ the alternative.

Question 16

A real estate analyst uses simple linear regression to model the selling price of houses (in thousands of dollars) based on their square footage. The data includes houses ranging from 1,200 to 4,000 square feet. The resulting regression equation is $\widehat{Price} = 40.5 + 0.15 \times (Square Footage)$ . How should the intercept of 40.5 be interpreted?

The base price of any house, before considering its size, is $40,500.
For every one-unit increase in square footage, the price is expected to increase by $40.50.
The intercept lacks a practical business interpretation because a square footage of zero is outside the relevant range of the data. (correct answer)
The model overestimates the price of the smallest house in the dataset by $40,500.

Explanation: The intercept is the predicted value of Y when X=0. In this context, it would be the predicted price for a house with 0 square feet. Since 0 is far outside the range of the observed data (1,200 to 4,000 sq ft), this is an example of extrapolation. Therefore, the intercept has no meaningful, practical interpretation.

Question 17

A study finds a strong positive correlation ( $r = 0.85$ ) between the number of coffee shops per capita in a city and the city's average salary level. An economic development advisor argues that attracting more coffee shop chains to the city will help raise local salaries. What is the most significant flaw in this reasoning?

A correlation of 0.85 is not strong enough to suggest a causal link; a value of 0.95 or higher would be needed.
The positive correlation is likely driven by a lurking variable, such as the overall economic prosperity of the city. (correct answer)
The regression model used to find the correlation might not have a statistically significant p-value for the slope.
The relationship is more likely reversed; higher salaries cause more people to buy coffee, leading to more shops.

Explanation: This is a classic example of confusing correlation with causation. A third, or lurking, variable is the most likely explanation. Cities with high economic activity and wealth tend to have both higher average salaries and a greater demand for amenities like coffee shops. The coffee shops are not causing the high salaries; both are effects of a common cause.

Question 18

A financial analyst develops a simple linear regression model to predict a company's stock price (Y) based on its quarterly earnings per share (X). The model yields a coefficient of determination ( $R^2$ ) of 0.64. Which statement correctly describes a property of this model?

The correlation coefficient between the predicted stock prices and the actual stock prices is 0.80. (correct answer)
The correlation coefficient between the quarterly earnings and the actual stock prices is 0.64.
Sixty-four percent of the actual stock prices fall exactly on the regression line.
The correlation coefficient between the quarterly earnings and the residuals of the model is 0.64.

Explanation: The coefficient of determination,

R^2

, is the square of the correlation coefficient,

r

, between the independent (X) and dependent (Y) variables. It is also the square of the correlation between the observed Y values and the predicted Y values (

\hat{Y}

). Therefore, the correlation between predicted and actual stock prices is

\sqrt{R^2} = \sqrt{0.64} = 0.80

Question 19

A company models its monthly profit based on the number of salespeople it employs. The model is built using data from the past five years, during which the number of salespeople ranged from 20 to 50. The resulting equation is $\widehat{Profit} = \$ 10,000 + $2,000 \times (Salespeople)$. The company is considering reducing its sales force to 10. What is the primary concern when using this model to predict the profit with 10 salespeople?

The prediction is an extrapolation, as the model's reliability is unknown for a number of salespeople outside the range of the original data. (correct answer)
The model will be inaccurate because profit is influenced by more factors than just the number of salespeople.
The calculation may result in a negative profit, which is impossible for this business, indicating a model flaw.
A positive slope is unrealistic, as fewer salespeople should logically lead to lower, not higher, profits.

Explanation: The model was built using data where the number of salespeople (X) ranged from 20 to 50. Predicting for X=10 is outside this range. This is known as extrapolation. The linear relationship observed between 20 and 50 salespeople may not hold for a smaller sales force, making the prediction unreliable.

Question 20

An economist studying income inequality calculates correlation coefficients for various relationships. Which correlation coefficient indicates the strongest linear relationship, regardless of direction?

Education level vs. Income: $r = 0.76$
Urban density vs. Cost of living: $r = 0.69$
Age vs. Technology adoption: $r = -0.82$ (correct answer)
Unemployment rate vs. Consumer confidence: $r = -0.71$

Explanation: When evaluating correlation coefficients, you're measuring the strength of linear relationships between variables. The key insight is that correlation strength depends on the absolute value of the coefficient, not whether it's positive or negative. A correlation of -0.90 indicates a stronger relationship than +0.70 because we're looking at how closely points cluster around a straight line, regardless of the line's direction. To find the strongest relationship, compare the absolute values of all coefficients. Looking at our options: |0.76| = 0.76 for education and income, |0.69| = 0.69 for urban density and cost of living, |-0.82| = 0.82 for age and technology adoption, and |-0.71| = 0.71 for unemployment and consumer confidence. Option C shows the strongest linear relationship at 0.82. The negative sign simply tells us that as age increases, technology adoption tends to decrease, but the relationship itself is very strong. Option A (

r = 0.76

) shows a strong positive relationship, but 0.76 < 0.82 in absolute terms. Option B (

r = 0.69

) represents a moderate relationship that's weaker than the others. Option D (

r = -0.71

) shows a strong negative relationship, but again, 0.71 < 0.82 in absolute value. Remember this key principle: when comparing correlation strengths, always use absolute values. A correlation of -0.85 beats +0.75 every time. The sign tells you direction; the absolute value tells you strength.