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Correlation

Learning goals

  • Computes or interprets the correlation coefficient.
  • Describes direction and strength of linear association.
  • Recognizes that correlation is unitless and sensitive to outliers.
  • Avoids using correlation to establish causation.

Click problems to add them to your worksheet.

Correlation 1
For a sample measuring weekly rehearsal hours and tempo-accuracy score, the correlation is \(r = 0.91\). Interpret the value and state one conclusion that is not justified by correlation alone.

Hints

- Use the sign of \(r\) to determine whether the variables tend to move in the same or opposite directions. - Use how close \(|r|\) is to \(1\) to judge linear strength. - Distinguish a statement about a tendency in the sample from a claim that rehearsal causes the change.

Solution

1. Because \(r\) is positive, greater weekly rehearsal time tends to be associated with higher tempo-accuracy scores. 2. Because \(|r| = 0.91\) is close to \(1\), the linear association is strong. 3. Correlation alone does not show that increasing rehearsal time causes a higher tempo-accuracy score.

Answer

There is a strong positive linear association: performers who rehearse more tend to have higher tempo-accuracy scores. The correlation alone does not establish that more rehearsal causes higher accuracy.
Correlation 2
For a sample measuring air temperature and natural-gas use, the correlation is \(r = -0.74\). Interpret the value and state one conclusion that is not justified by correlation alone.

Hints

- Read the negative sign as movement in opposite directions. - Judge strength from the magnitude \(0.74\), not from the sign. - Keep an association statement separate from a claim that temperature itself causes all changes in gas use.

Solution

1. Because \(r\) is negative, higher air temperatures tend to be associated with lower natural-gas use. 2. Because \(|r| = 0.74\), the linear association is moderately strong. 3. Correlation alone does not prove that changing air temperature directly causes the observed change in natural-gas use.

Answer

There is a moderately strong negative linear association: higher temperatures tend to accompany lower natural-gas use. The correlation alone does not establish a cause-and-effect relationship.
Correlation 3
For a sample measuring tree trunk diameter and tree age, the correlation is \(r = 0.18\). Interpret the value and state one conclusion that is not justified by correlation alone.

Hints

- Use the sign of \(0.18\) for direction. - Compare the magnitude \(0.18\) with \(0\) and \(1\) to judge linear strength. - Avoid turning a sample association between tree measurements into a causal claim.

Solution

1. The positive sign means larger trunk diameters tend to be associated with greater tree ages in this sample. 2. Because \(|r| = 0.18\) is close to \(0\), the linear association is weak. 3. The correlation does not prove that increasing trunk diameter causes a tree to become older or that age alone causes the observed diameter.

Answer

There is a weak positive linear association between trunk diameter and tree age. The correlation alone does not establish a cause-and-effect relationship.
Correlation 4
For a sample measuring ticket price and number of tickets sold, the correlation is \(r = -0.96\). Interpret the value and state one conclusion that is not justified by correlation alone.

Hints

- Interpret the negative sign as opposite movement between price and sales. - Use the closeness of \(0.96\) to \(1\) to judge strength. - Consider whether factors such as event popularity could affect sales as well as price.

Solution

1. The negative sign means higher ticket prices tend to be associated with fewer tickets sold. 2. Because \(|r| = 0.96\) is close to \(1\), the linear association is strong. 3. Correlation alone does not prove that changing price caused the entire change in ticket sales; other factors may differ across observations.

Answer

There is a strong negative linear association: higher ticket prices tend to accompany fewer tickets sold. The correlation alone does not prove that price caused the change in sales.
Correlation 5
For a sample measuring river speed and downstream canoe distance traveled in a fixed time, the correlation is \(r = 0.63\). Interpret the value and state one conclusion that is not justified by correlation alone.

Hints

- Use the positive sign to determine whether river speed and distance tend to increase together. - Treat \(0.63\) as the magnitude when judging strength. - Separate the observed tendency from a claim that river speed is the only cause of the distance traveled.

Solution

1. The positive sign means faster river speeds tend to be associated with greater downstream canoe distances in the fixed time. 2. Because \(|r| = 0.63\), the linear association is moderate. 3. Correlation alone does not prove that river speed by itself caused the observed distances; paddling effort and other conditions may also vary.

Answer

There is a moderate positive linear association: faster river speeds tend to accompany greater downstream distances in the fixed time. The correlation alone does not establish causation.
Correlation 6
For a sample measuring the number of practice attempts and completion time, the correlation is \(r=0.02\). Interpret the value and state one conclusion that is not justified by correlation alone.

Hints

- Think about the magnitude of \(r\), not just its sign. - What kind of relationship does correlation measure? - Which other information would you need to claim more than a sample linear association?

Solution

1. The sample correlation is very close to zero, so the observed linear association is extremely weak. 2. Its positive sign describes the sample's direction, but the size tells us that the linear pattern is negligible. Without the sample size, we cannot determine statistical significance from \(r\) alone. 3. Correlation also does not rule out a nonlinear pattern or establish causation.

Answer

The sample has almost no linear association. This value alone does not show causation or rule out a nonlinear relationship; statistical significance would also require the sample size.
Correlation 7
Consider the statement: “Since \(r=0\) for two quantitative variables, the variables have no relationship of any kind.” Identify and correct the error.

Hints

- Recall that correlation measures only straight-line association. - Think of a symmetric U-shaped pattern whose positive and negative linear tendencies cancel. - Replace “no relationship” with the narrower claim supported by \(r=0\).

Solution

1. A correlation of \(r=0\) means the data have no linear association. 2. Correlation does not measure nonlinear association. 3. The variables could still have a curved or some other nonlinear relationship.

Answer

The error is treating zero correlation as no relationship. \(r=0\) indicates no linear association, but a nonlinear relationship may still be present.
Correlation 8
Consider the statement about distance run and recovery heart rate: “The correlation is \(1.08\), showing an extremely strong association.” Identify and correct the error.

Hints

- Recall the complete allowable interval for a correlation coefficient. - Compare \(1.08\) with the largest possible value. - An invalid coefficient cannot support a valid strength interpretation.

Solution

1. Every correlation coefficient must satisfy \(-1\le r\le1\). 2. Because \(1.08>1\), the reported value cannot be a valid correlation. 3. The data or calculation must be checked before the association's direction or strength can be interpreted.

Answer

\(1.08\) is not a possible correlation because \(-1\le r\le1\). The correlation must be recomputed before interpreting the association.
Correlation 9
The paired observations are \((1, 9)\), \((2, 6)\), and \((3, 3)\). Consider the claim: “The correlation cannot be negative because every \(x\)-value and every \(y\)-value is positive.” Compute the correlation and correct the reasoning.

Hints

- Plot or inspect the ordered pairs from left to right. - Correlation direction is about how the variables change together, not about the signs of the raw measurements. - Check whether the points lie exactly on a straight line.

Solution

1. The three points lie exactly on the decreasing line \(y=12-3x\). 2. Therefore, the correlation is \(r=-1\). 3. The sign of correlation describes whether larger-than-average values of one variable tend to occur with larger-than-average or smaller-than-average values of the other; it does not depend on whether the raw measurements themselves are positive or negative.

Answer

\(r=-1\). Correlation is negative because the variables have a perfect decreasing linear relationship, even though all of the observed values are positive.
Correlation 10
Study A reports a correlation of \(r=0.64\) between two quantitative variables. Study B reports \(r=-0.64\) for a different pair of quantitative variables measured in completely different units. Compare the strength and direction of the two linear associations.

Hints

- Separate the magnitude of a correlation from its sign. - Decide which part describes strength and which part describes direction. - Recall whether correlation carries measurement units.

Solution

1. The magnitude \(|r|\) describes the strength of the linear association, while the sign describes its direction. 2. Both correlations have magnitude \(0.64\), so they represent the same linear-association strength. 3. Study A has a positive direction, while Study B has a negative direction. 4. Because correlation is unitless, the different measurement units do not prevent this comparison of linear strength.

Answer

The two associations have equal linear strength because both have \(|r|=0.64\). Study A is positive, while Study B is negative.
Correlation 11
When one student's study time is above its mean, the student's quiz score tends to be above its mean as well. Is the covariance of study time and score likely positive or negative?

Hints

- Think about the signs of the two deviations from their respective means. - What sign does the product of two deviations with the same sign have?

Solution

1. Study-time and score deviations from their means tend to have the same sign. 2. Products of deviations with the same sign are positive, so covariance tends to be positive.

Answer

Positive covariance.
Correlation 12
Two quantitative variables have covariance \(-8\). Their standard deviations are both positive. Must their correlation be positive or negative? Explain without calculating it.

Hints

- Write the relationship between correlation, covariance, and standard deviations. - Consider whether the denominator can reverse the numerator's sign.

Solution

1. Correlation is covariance divided by the product of two standard deviations. 2. The denominator is positive, so correlation has the same sign as covariance. 3. The correlation is negative.

Answer

Negative, because dividing the negative covariance by a positive number keeps its sign.
Correlation 13
Two plots compare practice sessions with the average number of correct answers on repeated six-question quizzes. Which plot shows the stronger positive linear association, a) or b)?
Figure for problem 563069

Hints

- Compare the scatter around a possible increasing straight line. - Strength describes how closely the points follow a line, not just the line's steepness.

Solution

1. Both sets of points rise as the input grows. 2. Points in panel a) lie much closer to a straight increasing line; panel b) is more scattered. 3. Panel a) has the stronger positive linear association.

Answer

Panel a).
Correlation 14
The correlation between distance from a motion sensor in feet and recorded signal strength is \(r=0.29\). The explanatory variable is converted from feet to meters. What is the new correlation? Explain.

Hints

- Write the unit conversion as multiplying every distance by the same positive number. - Correlation is computed from standardized values, which are unchanged by positive rescaling. - Check whether either the direction or the relative spacing of the points changes.

Solution

1. Converting feet to meters multiplies every explanatory-variable value by the same positive constant. 2. Positive rescaling does not change standardized values or the direction and strength of a linear association. 3. Therefore, the new correlation is \(r=0.29\).

Answer

\(r=0.29\). Converting feet to meters is a positive rescaling, so correlation does not change.
Correlation 15
The correlation between elevation and average air temperature is \(r=-0.64\). The response variable is converted from degrees Celsius to degrees Fahrenheit. What is the new correlation? Explain.

Hints

- Express the Fahrenheit conversion as a multiplication followed by an addition. - Determine whether the multiplier is positive or negative. - Correlation is unchanged by shifts and positive rescaling of either variable.

Solution

1. Converting Celsius to Fahrenheit uses \(F=\frac{9}{5}C+32\). 2. Adding a constant and multiplying by a positive constant do not change standardized values or correlation. 3. Therefore, the new correlation is \(r=-0.64\).

Answer

\(r=-0.64\). The Celsius-to-Fahrenheit conversion is a positive affine transformation, so correlation is unchanged.
Correlation 16
The correlation between fabric weight and shipping cost is \(r=0.27\). Every response value is multiplied by \(-1\). What is the new correlation? Explain.

Hints

- Decide what reflecting all response values across zero does to an upward or downward trend. - The transformation changes direction but not how tightly the points follow the pattern. - Keep the magnitude \(0.27\) and determine the new sign.

Solution

1. Multiplying one variable by a negative constant reverses the direction of the linear association. 2. The transformation does not change the magnitude of correlation. 3. Therefore, the new correlation is \(r=-0.27\).

Answer

\(r=-0.27\). Multiplying one variable by \(-1\) reverses the sign of correlation but leaves its magnitude unchanged.
Correlation 17
The correlation between rainfall and reservoir inflow is \(r=0.45\). The explanatory and response variables are interchanged. What is the new correlation? Explain.

Hints

- Compare the standardized-product formula before and after swapping the variables. - Multiplication is commutative, so each paired product is unchanged. - Swapping axes does not reflect or rescale either variable.

Solution

1. Correlation treats the two variables symmetrically: \(r_{xy}=r_{yx}\). 2. Interchanging the axes changes neither the direction nor the strength of the linear association. 3. Therefore, the new correlation is \(r=0.45\).

Answer

\(r=0.45\). Correlation is symmetric, so interchanging the explanatory and response variables does not change it.
Correlation 18
The correlation between stage-light angle and illuminated area is \(r=0.77\). Every explanatory-variable value is increased by \(100\). What is the new correlation? Explain.

Hints

- Decide whether adding \(100\) changes the spacing between any two angle values. - A horizontal shift does not reflect or stretch the scatterplot. - Correlation depends on standardized positions, not the variable's zero point.

Solution

1. Adding \(100\) shifts every explanatory value by the same amount. 2. A shift changes the mean but leaves all standardized values and the scatterplot's shape unchanged. 3. Therefore, the new correlation is \(r=0.77\).

Answer

\(r=0.77\). Adding a constant to one variable does not change correlation.
Correlation 19
A coffee roaster records roasting time \(x\) and bean mass \(y\), with correlation \(r=-0.73\). The measurements are recoded as numbers \(u=-2x\) and \(v=-3y\). What is the correlation between \(u\) and \(v\)? Explain.

Hints

- Imagine the effect of reversing just one numerical axis. - Then consider reversing the other one too. - Is correlation measured in physical units, or is it unchanged by a rescaling?

Solution

1. Multiplying one variable by a negative constant reverses the sign of correlation. 2. Multiplying both by negative constants reverses the sign twice, returning it to the original sign. 3. The magnitude does not change under nonzero rescaling, so \(r=-0.73\).

Answer

\(r=-0.73\). Reflecting both numerical scales preserves the original correlation.
Correlation 20
Use the scatterplot of arm span and height. Is \(r\) most plausibly near \(-1\), near \(0\), or near \(1\)? Explain.
Figure for problem 539445

Hints

- Use the left-to-right direction of the plotted points to choose the sign of \(r\). - Judge the magnitude by how tightly the points follow a straight line. - A value near \(1\) represents a very strong positive linear association.

Solution

1. The points rise from left to right, so the correlation is positive. 2. The points cluster tightly around an increasing straight line, indicating a strong linear association. 3. Therefore, \(r\) is most plausibly near \(1\).

Answer

\(r\) should be near \(1\) because the points follow a strong positive linear pattern.
Correlation 21
Use the scatterplot of trail steepness and hiking speed. Is \(r\) most plausibly near \(-1\), near \(0\), or near \(1\)? Explain.
Figure for problem 539446

Hints

- Determine the sign from whether hiking speed rises or falls as trail steepness increases. - Judge the magnitude by how tightly the plotted points follow a straight line. - Combine the negative direction with a magnitude close to \(1\).

Solution

1. The points fall from left to right, so the correlation is negative. 2. The points cluster tightly around a decreasing straight line, indicating a strong linear association. 3. Therefore, \(r\) is most plausibly near \(-1\).

Answer

\(r\) should be near \(-1\) because the points follow a strong negative linear pattern.
Correlation 22
Use the scatterplot of shoe length and height. Is \(r\) most plausibly a large negative value, near \(0\), or a large positive value? Explain.
Figure for problem 539447

Hints

- Look for the overall left-to-right direction rather than requiring every point to rise. - Use the amount of vertical scatter to judge whether the magnitude is near \(0\) or near \(1\). - Combine the slight positive direction with a small magnitude.

Solution

1. The points have only a slight upward tendency, so the direction is weakly positive. 2. The points form a diffuse cloud rather than clustering tightly around a line. 3. Therefore, \(r\) should be a small positive value, which is closer to \(0\) than to \(1\).

Answer

\(r\) should be small and positive. The direction is slightly positive, but the wide scatter makes the linear association weak.
Correlation 23
Use technology to compute the correlation for the paired data on crowd size in hundreds and concession wait in minutes. Round to three decimal places, then interpret its direction and strength. <table> <thead><tr><th>Crowd size in hundreds</th><th>Concession wait in minutes</th></tr></thead> <tbody> <tr><td>1</td><td>12.30</td></tr> <tr><td>2</td><td>14.23</td></tr> <tr><td>3</td><td>16.83</td></tr> <tr><td>4</td><td>18.67</td></tr> <tr><td>5</td><td>21.07</td></tr> <tr><td>6</td><td>23.17</td></tr> </tbody> </table>

Hints

- Keep each crowd size paired with the wait time in the same table row. - Use the correlation command rather than fitting a nonlinear model. - After calculating \(r\), use its sign for direction and its distance from \(0\) for linear strength.

Solution

1. Enter the crowd-size values and corresponding wait times in paired technology lists. 2. The computed correlation is \(r = 0.9993316\ldots\), which rounds to \(r \approx 0.999\). 3. The positive sign and magnitude close to \(1\) indicate a very strong positive linear association.

Answer

\(r \approx 0.999\); the data show a very strong positive linear association.
Correlation 24
Use technology to compute the correlation for the paired data on sunlight hours and net electricity drawn from the grid in kilowatt-hours. Negative grid values mean electricity was exported. Round to three decimal places, then interpret the correlation’s direction and strength. <table> <thead><tr><th>Sunlight hours</th><th>Net grid electricity in kilowatt-hours</th></tr></thead> <tbody> <tr><td>2</td><td>5.40</td></tr> <tr><td>3</td><td>2.47</td></tr> <tr><td>4</td><td>0.87</td></tr> <tr><td>5</td><td>-2.27</td></tr> <tr><td>6</td><td>-4.27</td></tr> <tr><td>7</td><td>-6.87</td></tr> </tbody> </table>

Hints

- Treat negative electricity values as valid numerical observations representing net export. - Preserve the row-by-row pairing when entering the two lists. - Interpret the negative sign separately from the magnitude near \(1\).

Solution

1. Enter the sunlight-hour values and corresponding net-grid values as paired data. 2. The computed correlation is \(r = -0.9977906\ldots\), which rounds to \(r \approx -0.998\). 3. The negative sign and magnitude close to \(1\) indicate a very strong negative linear association.

Answer

\(r \approx -0.998\); the data show a very strong negative linear association.
Correlation 25
In one study, trail steepness and hiking speed have correlation \(r_1=-0.82\). In another, hours of study and exam score have correlation \(r_2=0.76\). Which association is stronger? Compare their directions too.

Hints

- Which part of a correlation value indicates direction? - How is strength compared for correlations with different signs? - Describe each study after deciding which magnitude is larger.

Solution

1. Compare the magnitudes: \(|r_1|=0.82\) and \(|r_2|=0.76\). 2. The hiking study has the slightly stronger linear association because \(0.82>0.76\). 3. Its association is negative, while the study-time association is positive.

Answer

The hiking study has the stronger linear association, and it is negative. The study-time association is positive.
Correlation 26
A least-squares regression of light intensity on distance from a lamp has \(r^2 = 0.79\). Both variables are converted to different units using positive linear conversions. Explain what happens to \(r^2\).

Hints

- Think about whether changing units changes the relative positions of the observations. - Positive shifts and rescalings leave standardized values unchanged. - Use the relationship between \(r\) and \(r^2\).

Solution

1. A positive linear conversion shifts and/or rescales a variable without changing its standardized values. 2. Therefore, the correlation \(r\) is unchanged. 3. Because \(r^2\) is the square of the correlation, \(r^2\) remains \(0.79\).

Answer

\(r^2\) remains \(0.79\) because positive linear unit conversions do not change the correlation.
Correlation 27
Across \(20\) schools, the correlation between each school's average homework time and average math score is \(r=0.88\). A principal concludes that within every school, students who do more homework must score higher. Explain why the school-level coefficient does not justify that conclusion.

Hints

- Identify what one point in the scatter plot would represent. - Distinguish a group average from a person-level measurement. - Ask what data would be needed to support the individual claim.

Solution

1. Each point represents a school average, not an individual student. 2. A strong association among school averages can result from differences between schools, such as resources, course offerings, or student populations. 3. The individual-student relationship within a school could be weaker, absent, or different in direction, so individual data are needed.

Answer

The conclusion is not justified. The coefficient describes an association among school averages and does not determine the relationship among individual students within each school.
Correlation 28
A data set contains paired measurements, but every value of the explanatory variable is exactly \(5.0\). Consider the statement: “The correlation must be \(0\) because there is no upward or downward trend.” Is the statement correct? Explain.

Hints

- Check whether both variables have nonzero spread before interpreting correlation. - Think about what happens when all horizontal coordinates are identical. - Distinguish an undefined statistic from a statistic whose value is zero.

Solution

1. Correlation requires variation in both quantitative variables. 2. Here, the explanatory variable has standard deviation \(0\). 3. Because the correlation calculation divides by the explanatory-variable standard deviation, the correlation is undefined rather than \(0\).

Answer

No. The correlation is undefined because the explanatory variable has no variation.
Correlation 29
Software gives a correlation of \(r=0.9964\) for two quantitative variables. A report rounds this to \(r=1.00\) and then states, “The data have a perfect positive linear association.” Evaluate the statement.

Hints

- Use the unrounded value when deciding whether a boundary value has actually been reached. - Recall what a correlation of exactly \(1\) means geometrically. - Distinguish “very close to perfect” from “perfect.”

Solution

1. The unrounded correlation \(0.9964\) is less than \(1\), so the association is not mathematically perfect. 2. Rounding to two decimal places hides the small difference from \(1\). 3. The data have an extremely strong positive linear association, but a perfect association would require every point to lie exactly on an increasing line.

Answer

The statement is too strong. The association is extremely strong and positive, but \(r=0.9964\) is not a perfect correlation.
Correlation 30
A data set has correlation \(r=0.98\) between two quantitative variables. A student says, “That means a fitted line is \(98\%\) accurate, so predictions will be within about \(2\%\) of the actual values.” Explain the mistake.

Hints

- What information does correlation summarize? - Is \(r\) defined as a percentage of predictions that are correct? - Which quantity compares an actual response with its predicted value?

Solution

1. A correlation of \(0.98\) means the sample has a very strong positive linear association. 2. Correlation does not measure prediction error as a percentage of actual values. 3. To judge individual prediction errors, compare actual responses with the line's predictions by examining residuals.

Answer

The student is wrong. \(r=0.98\) describes a strong positive linear association, not \(98\%\) prediction accuracy. Individual errors must be checked using residuals.
Correlation 31
Sample A contains \(12\) paired observations and has correlation \(r=0.65\). Sample B contains \(400\) paired observations and also has correlation \(r=0.65\). Consider the claim: “The association in Sample B is stronger because the sample is much larger.” Evaluate the claim, distinguishing the strength of the observed linear association from the amount of evidence available about a population relationship.

Hints

- Separate what the value of \(r\) describes from what sample size affects. - Ask whether the numerical correlation changes just because more observations are present. - Think about precision of population conclusions as a different issue from descriptive association strength.

Solution

1. The sample correlation \(r\) describes the direction and strength of the observed linear association, so both samples have the same reported sample linear strength: \(r=0.65\). 2. The larger sample size does not make the numerical sample correlation stronger. 3. A larger sample can provide more precise and stable information about a population relationship, so Sample B can support stronger statistical evidence even though the observed correlation coefficient is the same.

Answer

The claim is incorrect about sample association strength: both samples have the same reported linear association, \(r=0.65\). The larger sample in B can provide more precise evidence about the population, but it does not make \(r=0.65\) a stronger sample correlation.
Correlation 32
Software reports \(r^2=0.64\) for a simple linear regression, and the fitted slope is negative. Find the sample correlation \(r\). Explain why \(r^2\) alone is not enough to determine the sign.

Hints

- Recover the magnitude of the correlation from the coefficient of determination. - Use the fitted line’s direction to determine the missing sign. - Remember what information is lost when a signed quantity is squared.

Solution

1. Since \(r^2=0.64\), the magnitude of the correlation is \(|r|=\sqrt{0.64}=0.80\). 2. In simple linear regression, the correlation and the least-squares slope have the same sign. 3. Because the fitted slope is negative, \(r=-0.80\). 4. Squaring removes the sign, so \(r^2\) alone gives only the magnitude of \(r\), not its direction.

Answer

\(r=-0.80\). The value \(r^2=0.64\) gives \(|r|=0.80\), and the negative slope determines that the correlation is negative.
Correlation 33
Data Sets A and B both have sample correlation \(r=0.70\). Consider the claim: “Because the correlations are equal, the two data sets have the same association.” Evaluate the claim using the scatterplots.
Figure for problem 548039

Hints

- Compare the shapes of the point clouds on the matching scales. - Does correlation describe curvature and isolated points, or mainly a straight-line trend? - Decide whether one number can capture all features of both plots.

Solution

1. Correlation describes the direction and strength of a linear association, not the full shape of the data. 2. Data Set A follows a roughly straight increasing pattern. 3. Data Set B follows a curved pattern for most points, with one isolated point high on the right. 4. Equal correlations do not guarantee the same shape or the same unusual observations.

Answer

The claim is false. The correlations are approximately equal, but A is roughly linear while B is curved with an isolated point.
Correlation 34
Consider the paired data <table> <thead><tr><th>\(x\)</th><th>\(y\)</th></tr></thead> <tbody> <tr><td>\(0\)</td><td>\(1\)</td></tr> <tr><td>\(1\)</td><td>\(2\)</td></tr> <tr><td>\(2\)</td><td>\(4\)</td></tr> <tr><td>\(3\)</td><td>\(8\)</td></tr> <tr><td>\(4\)</td><td>\(16\)</td></tr> </tbody> </table> Use technology to compute the correlation and round to three decimals. Then explain why the result does not mean the data follow an exact linear relationship.

Hints

- Compute the linear correlation from the paired values. - Look at how the response changes from one row to the next. - A strong correlation can occur for a systematic pattern that is not exactly linear.

Solution

1. The correlation for the five pairs is \(r\approx0.933\), indicating a strong positive linear association. 2. The response values double whenever \(x\) increases by \(1\), so the points follow the exact exponential pattern \(y=2^x\). 3. The association is therefore perfectly systematic but nonlinear, which is why the correlation is strong yet less than \(1\).

Answer

\(r\approx0.933\). The association is strong and positive, but the points follow the exact exponential pattern \(y=2^x\), not a straight line.
Correlation 35
A data set of paired quantitative measurements has correlation \(r=-0.71\). The rows are randomly shuffled while the two values in each row stay together. What is the correlation after the row order is shuffled? Explain.

Hints

- Ask whether any \(x\)-value has been paired with a different \(y\)-value. - Distinguish changing row order from changing the pairings themselves. - A summary of paired data does not depend on which row is listed first.

Solution

1. Correlation depends on the set of paired observations, not on the order in which the rows are stored or displayed. 2. Shuffling whole rows preserves every original ordered pair. 3. Therefore, all deviations, products, and standardized pairings used in the correlation are unchanged as a collection.

Answer

The correlation remains \(r=-0.71\). Reordering complete paired observations does not change the association.
Correlation 36
Two quantitative variables have correlation \(r=-0.43\). Every \(x\)-value is converted to its z-score, and every \(y\)-value is converted to its z-score. What is the correlation between \(z_x\) and \(z_y\)? Explain.

Hints

- Break standardization into a shift and a rescaling. - Check the sign of the scaling factor used for a standard deviation. - Recall which linear changes leave correlation unchanged.

Solution

1. Standardizing \(x\) subtracts \(\bar x\) and divides by the positive standard deviation \(s_x\). 2. Standardizing \(y\) likewise subtracts \(\bar y\) and divides by the positive standard deviation \(s_y\). 3. Shifts and positive rescalings do not change correlation, so the direction and strength of the linear association remain the same.

Answer

The correlation remains \(r=-0.43\).
Correlation 37
Two data sets use the same four \(x\)-values, \(1, 2, 3, 4\), and the same four \(y\)-values, \(1, 2, 3, 4\). Data set A pairs them as \((1, 1), (2, 2), (3, 3), (4, 4)\). Data set B pairs them as \((1, 4), (2, 3), (3, 2), (4, 1)\). Compare the correlations. What does this show about using the separate distributions of \(x\) and \(y\) to determine correlation?

Hints

- Plot or mentally arrange the ordered pairs for each data set. - Check whether each set lies on an increasing or decreasing straight line. - Notice what changed between the data sets even though the individual value lists did not.

Solution

1. In data set A, all points lie exactly on an increasing line, so \(r=1\). 2. In data set B, all points lie exactly on a decreasing line, so \(r=-1\). 3. The two data sets have identical separate lists of \(x\)-values and \(y\)-values, but different pairings. 4. Therefore, the marginal distributions alone do not determine correlation; the pairing of the measurements matters.

Answer

Data set A has \(r=1\), while data set B has \(r=-1\). The separate distributions of \(x\) and \(y\) are not enough to determine correlation because correlation depends on how the values are paired.
Correlation 38
For \(n=5\) paired observations, each variable has been converted to sample z-scores. The sum of the five products \(z_{x,i}z_{y,i}\) is \(2.8\). Find the sample correlation.

Hints

- Use the relationship between correlation and the products of paired sample z-scores. - Remember that sample standardization leads to a denominator of \(n-1\). - Check that the resulting value lies between \(-1\) and \(1\).

Solution

1. With sample z-scores, the sample correlation is \(r=\frac{1}{n-1}\sum z_{x,i}z_{y,i}\). 2. Here, \(n-1=4\), so \(r=\frac{2.8}{4}=0.70\).

Answer

The sample correlation is \(r=0.70\).
Correlation 39
A distance variable is supposed to be recorded in miles, but half of its entries were accidentally entered in kilometers without conversion. The correlation is then computed between this mixed-unit column and travel time. Why is the resulting correlation not meaningful as a description of the intended data?

Hints

- Distinguish converting an entire variable from converting only some observations. - Correlation is unit-invariant only under one consistent linear rescaling of the whole variable. - Ask whether equal numerical differences currently represent equal physical distance across all rows.

Solution

1. Correlation is unchanged by a single consistent positive unit conversion applied to every value of a variable. 2. Here, however, only part of the column uses a different scale, so the recorded numbers do not represent one consistent quantitative variable. 3. The mixed units distort relative distances among observations and can change the apparent paired pattern with travel time. 4. The entries must be converted to one common unit before computing correlation.

Answer

The correlation is not meaningful because the distance column mixes two scales within one variable. Convert all distances to a common unit first, then compute the correlation.
Correlation 40
Two quantitative variables have sample correlation \(r=0.80\), with sample standard deviations \(s_x=3\) and \(s_y=5\). Find the sample covariance between \(x\) and \(y\).

Hints

- Relate covariance to correlation and the two sample standard deviations. - Rearrange the standardization formula to isolate covariance. - Unlike correlation, covariance carries units from both variables.

Solution

1. Correlation standardizes covariance: \(r=\frac{s_{xy}}{s_xs_y}\). 2. Therefore, \(s_{xy}=rs_xs_y=0.80\cdot3\cdot5=12\).

Answer

The sample covariance is \(12\) in the product units of \(x\) and \(y\).
Correlation 41
Three paired observations have \(x\)-values \(1,2,3\) and matching \(y\)-values \(1,3,5\). Find their sample covariance using the divisor \(n-1\).

Hints

- Find the mean of each variable first. - Pair deviations from their own means before multiplying. - Remember which divisor defines sample covariance.

Solution

1. The means are \(\bar x=2\) and \(\bar y=3\). 2. The products of paired deviations are \((-1)(-2)=2\), \(0\), and \((1)(2)=2\), totaling \(4\). 3. Divide by \(3-1=2\) to get \(s_{xy}=2\).

Answer

\(s_{xy}=2\).
Correlation 42
A set of paired data has sample covariance \(s_{xy}=3\) and standard deviations \(s_x=2\) and \(s_y=3\). Find the correlation \(r\).

Hints

- Look for the definition connecting covariance and correlation. - The denominator uses the product of the two standard deviations. - Check that your answer is within the possible correlation range.

Solution

1. Use \(r=\frac{s_{xy}}{s_xs_y}\). 2. Substitute the values: \(r=\frac{3}{2\cdot3}=0.5\).

Answer

\(r=0.5\).
Correlation 43
Three garden beds were watered for \(10\), \(20\), and \(30\) minutes. Their matching soil-dryness scores were \(7\), \(5\), and \(3\). Is the sample covariance of watering time and dryness positive or negative? Show the products of deviations from the means for the first and third beds.

Hints

- Find the mean watering time and the mean dryness score separately. - For each of the two outer beds, compare both values with their means. - Multiply the paired deviations, keeping their signs.

Solution

1. The mean watering time is \(20\) minutes, and the mean dryness score is \(5\). 2. For the first bed, the product of deviations is \((10-20)(7-5)=-20\). For the third bed, it is \((30-20)(3-5)=-20\). The middle bed contributes zero. 3. The sum of the products is negative, so the sample covariance is negative.

Answer

Negative. The first and third products of deviations are both \(-20\), so their sum is negative.
Correlation 44
The scatterplot relates outside temperature, in degrees Celsius, to a building's daily energy used for heating and cooling. Is the correlation \(r\) most likely near \(-1\), near \(0\), or near \(1\)? Explain why a strong association can have a small \(|r|\).
Figure for problem 539444

Hints

- Describe the shape before guessing a correlation value. - Compare energy use in cool, moderate, and hot weather. - Does one straight line summarize the entire curved pattern?

Solution

1. Energy use is high in cold and hot weather and lower near moderate temperatures, giving a clear U-shaped pattern. 2. Correlation measures straight-line association. The downward direction on the cooler side is balanced by the upward direction on the warmer side. 3. Thus \(r\) is near \(0\), even though temperature and daily energy use are strongly related in a curved way.

Answer

\(r\) is near \(0\). The two sides of the strong U-shaped pattern have opposite directions, which cancel in a measure of linear association.
Correlation 45
Use the scatterplot of fertilizer amount and days to germination. Would \(r\) necessarily be close to \(-1\)? Explain why the association can be strong even when \(|r|\) is not close to \(1\).
Figure for problem 539448

Hints

- Determine both the direction and the form of the plotted pattern. - Decide whether one straight line would fit the entire curve equally well. - Correlation measures the strength of a linear pattern, not closeness to every possible curve.

Solution

1. The plotted points follow a tight decreasing curve, so the variables have a strong overall association. 2. The curve flattens as fertilizer amount increases, so a straight line does not match the form perfectly. 3. Therefore, \(r\) should be negative but need not be close to \(-1\), because correlation measures only linear association.

Answer

No. \(r\) should be negative, but it need not be close to \(-1\). The points follow a strong decreasing curve rather than a straight line.
Correlation 46
An observational study finds \(r=-0.81\) between weekly recreational screen time and nightly sleep duration. A report says, “Reducing screen time will therefore increase sleep duration.” Evaluate the claim and name a plausible third variable that could help explain the association.

Hints

- First interpret what the negative correlation says about the observed tendency. - Check whether participants were randomly assigned different amounts of screen time. - Look for a factor that could affect both leisure screen time and the sleep schedule.

Solution

1. The value \(r=-0.81\) describes a strong negative linear association in the observed sample. 2. Because the study is observational, the correlation alone does not establish that changing screen time causes a change in sleep duration. 3. One plausible confounding variable is household bedtime rules: stricter bedtime rules could reduce recreational screen time and also increase nightly sleep duration.

Answer

The causal claim is not justified by the observational correlation alone. Household bedtime rules are a plausible confounding variable because they could be related to both recreational screen time and sleep duration.
Correlation 47
Use the scatterplot of distance from a city center and commute time. What would most likely happen to the positive correlation if the far-right point were removed? Explain.
Figure for problem 539452

Hints

- Compare the far-right observation with the direction of the main group of points. - Consider how much that point expands the range of distance values. - A high-leverage point that follows the trend often increases the magnitude of correlation.

Solution

1. The far-right point has high leverage because its distance value is far from the other values. 2. The point lies along the extension of the increasing trend. 3. Removing a high-leverage point that reinforces the trend would most likely decrease the magnitude of the positive correlation.

Answer

The positive correlation would most likely become weaker because the far-right high-leverage point extends and reinforces the increasing trend.
Correlation 48
Use the scatterplot of paint age and color fading. What would most likely happen to the positive correlation if the far-right point were removed? Explain.
Figure for problem 539453

Hints

- Compare the far-right point with the increasing pattern formed by the main group. - A point far from the other explanatory-variable values can have substantial leverage. - Decide whether removing a point that conflicts with the trend should strengthen or weaken \(r\).

Solution

1. The far-right point has high leverage because its paint-age value is far from the rest. 2. It lies well below and opposes the main increasing trend. 3. Removing this influential point would most likely increase the positive correlation and make the remaining linear pattern appear stronger.

Answer

The positive correlation would most likely increase because removing the far-right high-leverage point eliminates an observation that strongly opposes the trend.
Correlation 49
The scatterplot compares weekly online ad spending with weekly sales. How could including the new observation change the correlation? Explain.
Figure for problem 539454

Hints

- Describe the point cloud before considering the cross. - Compare the new observation's explanatory value with the others. - Would an observation far from most of the data have much effect on the numerical correlation?

Solution

1. The original points have little clear linear association. 2. The new point is far to the right and high above the cloud, so its advertising spending is unusually large compared with the other observations. 3. This single extreme point can make the overall correlation noticeably more positive, even though the original data show little linear trend.

Answer

The new high-spending, high-sales observation can make the sample correlation positive even when the main cloud has little linear association.
Correlation 50
Use the scatterplot of wave height and time to complete a fixed surfing course. The open circle shows a point’s original position on the curve, and the red cross shows the same point after it was moved upward. How would this change most likely affect the correlation? Explain.
Figure for problem 539455

Hints

- Compare the moved point’s horizontal position with the center and ends of the observed range. - Small vertical changes at central x-values usually have less influence than changes to high-leverage points. - Correlation still summarizes only the linear part of the curved pattern.

Solution

1. The moved point has low leverage because its wave-height value is near the middle of the observed range. 2. The vertical move is small relative to the full pattern, so it changes the point’s agreement with the decreasing curve only slightly. 3. The correlation would therefore change only modestly, while the overall nonlinear decreasing form remains.

Answer

The correlation would most likely change only modestly because the point is near the center of the x-range and is moved only a small amount; the nonlinear decreasing pattern remains.
Correlation 51
For a least-squares regression line relating tea steeping time and flavor score, the slope is \(b = 1.875\), \(s_x = 2.0\), and \(s_y = 5.0\). Find the correlation \(r\).

Hints

- Rearrange the regression-slope formula to isolate \(r\). - The sign of \(r\) must match the positive slope. - Check that the result lies within the allowable interval \([-1, 1]\).

Solution

1. Start with \(b = r\frac{s_y}{s_x}\). 2. Solve for correlation: \(r = b\frac{s_x}{s_y}\). 3. Substitute: \(r = 1.875\left(\frac{2.0}{5.0}\right) = 0.750\).

Answer

\(r = 0.750\).
Correlation 52
Plot a) pairs each day's rainfall with the same day's river level. Plot b) pairs each day's rainfall with the next day's river level. Use technology to calculate both correlations to the nearest thousandth and explain the difference.
Figure for problem 544085

Hints

- Read the ordered pairs from each plotted panel. - In plot b), each rainfall measurement is paired with the following day's river level. - Compare which pairing reflects the timing described by the context.

Solution

1. Pairing rainfall and river level from the same day gives \(r\approx -0.231\). 2. Pairing Day \(1\) through Day \(6\) rainfall with Day \(2\) through Day \(7\) river level gives \(r\approx 0.991\). 3. The one-day-lag pairing aligns each rainfall measurement with the following day's river level, revealing a strong positive lagged association that the same-day pairing misses.

Answer

Same-day pairing: \(r\approx -0.231\) One-day-lag pairing: \(r\approx 0.991\) The delayed pairing captures a strong positive lagged association between rainfall and the following day's river level.
Correlation 53
The scatter plot shows two production lines analyzed separately. a) State the correlation within each line. b) Use technology to find the correlation after combining all six pairs. c) Explain the reversal in direction.
Figure for problem 544086

Hints

- Analyze each group before pooling the data. - Compare the locations of the two entire clusters. - Separate the within-group direction from the between-group shift.

Solution

1. Within each line, the points lie exactly on a decreasing line, so each correlation is \(r=-1\). 2. For all six pairs together, \(r\approx 0.686\). 3. Line B has both larger \(x\)-values and larger \(y\)-values than Line A. The separation between the two groups creates a positive combined trend that overrides the negative within-line trends.

Answer

a) Line A: \(r=-1\); Line B: \(r=-1\) b) Combined: \(r\approx 0.686\) c) The group-level separation creates a positive overall pattern even though each group individually decreases.
Correlation 54
For \(Y=X^2\), the three displayed \((X,Y)\) outcomes are equally likely. a) Show that \(X\) and \(Y\) have correlation \(0\). b) Let \(A=\{X>0\}\) and \(B=\{Y>0\}\). Determine whether \(A\) and \(B\) are independent. c) Explain what this example shows about zero correlation and independence.
Figure for problem 547391

Hints

- Compute \(E(X)\), \(E(Y)\), and \(E(XY)\) from the three equally likely values. - List which values of \(X\) satisfy each event. - Compare the joint event probability with the product of the two marginal probabilities.

Solution

1. \(E(X)=0\), \(E(Y)=\frac23\), and \(E(XY)=E(X^3)=0\). Therefore \(\operatorname{Cov}(X,Y)=0-0\cdot\frac23=0\). Both variables have positive variance, so their correlation is \(0\). 2. \(P(A)=\frac13\), \(P(B)=\frac23\), and \(P(A\cap B)=\frac13\). 3. Since \(\frac13\ne\frac13\cdot\frac23=\frac29\), the events are not independent. 4. Zero correlation rules out linear association but does not imply independence of the variables or of events formed from them.

Answer

a) \(\operatorname{Corr}(X,Y)=0\). b) The events are not independent because \(P(A\cap B)=\frac13\ne\frac29=P(A)P(B)\). c) Zero correlation does not imply independence.
Correlation 55
A manufacturing study records temperature deviation from a target setting, \(x\), and a defect score, \(y\), for five runs. <table> <thead><tr><th>Temperature deviation \(x\)</th><th>Defect score \(y\)</th></tr></thead> <tbody> <tr><td>\(-2\)</td><td>\(4\)</td></tr> <tr><td>\(-1\)</td><td>\(1\)</td></tr> <tr><td>\(0\)</td><td>\(0\)</td></tr> <tr><td>\(1\)</td><td>\(1\)</td></tr> <tr><td>\(2\)</td><td>\(4\)</td></tr> </tbody> </table> Use technology to compute the correlation coefficient. Then explain why the result does not mean the variables are unrelated.

Hints

- Compute the statistic from the paired observations rather than judging only from the table. - Think about what type of association the correlation coefficient summarizes. - Look for a pattern that a straight line would fail to capture.

Solution

1. The correlation coefficient for the five ordered pairs is \(r=0.000\). 2. The points follow the exact nonlinear pattern \(y=x^2\). 3. Correlation measures linear association, so \(r=0\) indicates no linear association here, not no relationship of any kind.

Answer

\(r=0.000\). The variables have a perfect U-shaped nonlinear relationship even though their linear correlation is zero.
Correlation 56
A data set has correlation \(r=0.62\). A larger file is created by duplicating every ordered pair exactly once, so each original observation now appears twice. Consider the claim that the correlation should become larger because the row count doubled. Determine the new correlation and evaluate the claim.

Hints

- Ask whether the geometry of the point cloud changes when every point is copied. - Think about what happens to centered sums when every contribution is repeated equally. - Separate sample size from the amount of genuinely new information.

Solution

1. Duplicating every pair leaves the \(x\)- and \(y\)-means unchanged. 2. Every centered product and centered square contribution is duplicated, so the quantities in the correlation calculation are all multiplied by the same factor. 3. Their ratio is unchanged, so the correlation remains \(0.62\). 4. The larger row count does not create new information because the observations were copied rather than newly sampled.

Answer

The correlation remains \(r=0.62\). Duplicating every observation does not strengthen the linear association or add new information.
Correlation 57
Two positive variables have correlation \(r=0.78\). Every \(x\)-value is replaced with \(\sqrt{x}\). Consider the claim: “The correlation must still be \(0.78\) because the transformation preserves the order of the observations.” Evaluate the claim.

Hints

- Distinguish transformations that only shift or rescale an axis from transformations that change spacing nonuniformly. - Correlation measures linear association, not just rank order. - Ask whether the transformed point cloud must have the same shape.

Solution

1. Correlation is unchanged by positive linear transformations such as changing units or adding a constant. 2. The transformation \(x\mapsto\sqrt{x}\) is nonlinear. 3. A nonlinear transformation changes relative spacing and can make a relationship more or less linear, so the new correlation cannot be determined from the original \(r\) alone.

Answer

The claim is incorrect. The new correlation need not equal \(0.78\); a nonlinear transformation can change correlation even if it preserves order.
Correlation 58
For a data set with nonconstant explanatory values, the correlation between \(x\) and the observed response \(y\) is \(r=-0.62\). A least-squares regression line with a negative slope is fitted, and each observation is assigned its predicted value \(\hat y\). What is the correlation between \(x\) and the predicted values \(\hat y\)? Explain why it differs from the correlation between \(x\) and the observed responses.

Hints

- Think about the geometric location of all pairs \((x, \hat y)\). - Correlation reaches an extreme value when points lie exactly on a nonhorizontal straight line. - Distinguish fitted values from the original observed responses.

Solution

1. Every predicted value has the form \(\hat y=a+bx\) with the same negative slope \(b\). 2. Therefore, the ordered pairs \((x, \hat y)\) lie exactly on a decreasing straight line. 3. For nonconstant \(x\)-values, an exact decreasing linear relationship has correlation \(-1\). 4. The observed responses do not all lie on the fitted line, so their correlation with \(x\) is only \(-0.62\).

Answer

The correlation between \(x\) and \(\hat y\) is \(-1\). The fitted values lie exactly on a decreasing straight line, whereas the observed responses include deviations from that line.
Correlation 59
Ridgeway Instruments records the following paired values for a calibration setting \(x\) and an instrument response \(y\): <table> <thead><tr><th>\(x\)</th><th>Recorded \(y\)</th></tr></thead> <tbody> <tr><td>\(1\)</td><td>\(2\)</td></tr> <tr><td>\(2\)</td><td>\(4\)</td></tr> <tr><td>\(3\)</td><td>\(6\)</td></tr> <tr><td>\(4\)</td><td>\(8\)</td></tr> <tr><td>\(5\)</td><td>\(50\)</td></tr> </tbody> </table> Use technology to compute the correlation. A data check then shows that the last response was entered incorrectly and should be \(10\), not \(50\). Recompute the correlation after the correction and explain what the change illustrates. Round the first correlation to three decimals.

Hints

- Compute the correlation using all recorded pairs first. - Then change only the verified data-entry error and recompute. - Compare how closely each version of the data follows a straight line.

Solution

1. With the recorded value \(50\), the correlation is \(r\approx0.781\). 2. After replacing \(50\) with \(10\), the points are \((1, 2)\), \((2, 4)\), \((3, 6)\), \((4, 8)\), and \((5, 10)\), which lie exactly on an increasing straight line. 3. The corrected correlation is therefore \(r=1\). 4. The large change shows that correlation can be sensitive to an unusual or erroneous observation, especially in a small data set.

Answer

Before correction, \(r\approx0.781\). After correction, \(r=1\). The change illustrates that one unusual data value can substantially affect correlation in a small sample.
Correlation 60
For the three paired observations \((1, 1), (2, 3), (3, 2)\), compute the sample correlation \(r\).

Hints

- Find the mean of each variable before comparing paired deviations. - Use the centered products to measure whether the variables tend to vary together. - Standardize the covariance by the two sample standard deviations.

Solution

1. The means are \(\bar x=2\) and \(\bar y=2\). 2. The centered values are \((-1, 0, 1)\) for \(x\) and \((-1, 1, 0)\) for \(y\). 3. The sample standard deviations are \(s_x=1\) and \(s_y=1\), and the sample covariance is \(\frac{(-1)\cdot(-1)+0\cdot1+1\cdot0}{2}=0.5\). 4. Therefore, \(r=\frac{0.5}{1\cdot1}=0.50\).

Answer

The sample correlation is \(r=0.50\).
Correlation 61
A soccer coach compares practice sessions with penalty shots scored by four players. Find the correlation \(r\) from the plotted pairs. Explain why correlation has no units. Round to two decimal places.
Figure for problem 563072

Hints

- Read all four coordinates and find the mean of each variable. - For each dot, use its deviations from the two means in the correlation formula. - Compare the units in the cross-products and the two squared-deviation sums.

Solution

1. The points are \((1,2)\), \((2,1)\), \((3,4)\), and \((4,3)\). Both sample means are \(2.5\). 2. The sum of products of paired deviations is \(\sum(x-\bar x)(y-\bar y)=3\). Each sum of squared deviations is \(5\). 3. Thus \(r=\frac{3}{\sqrt{5\cdot5}}=0.60\). Numerator and denominator have the same combined measurement units, which cancel.

Answer

\(r=0.60\), with no units because the measurement units cancel.
Correlation 62
A sample has correlation \(r=0.72\) between two quantitative variables. A new observation is added whose coordinates are exactly the original sample means \((\bar x, \bar y)\). What happens to the correlation? Explain without recomputing it from all observations.

Hints

- Think about the deviations of the new point from the existing means. - Ask whether the new point changes either center. - Consider how a point with zero centered deviations contributes to a linear-association measure.

Solution

1. Adding \((\bar x, \bar y)\) does not change either sample mean because the new point is located at the existing mean of each variable. 2. Relative to those means, the new point has deviations \(0\) and \(0\), so it adds nothing to the centered cross-product sum or to either centered sum of squares. 3. In the correlation formula, the centered cross-product sum and both centered sums of squares are therefore unchanged, so the correlation remains \(0.72\).

Answer

The correlation remains \(r=0.72\).
Correlation 63
Northstar Controls tests a tuning setting \(x\) and a response score \(y\) on two different machines. <table> <thead><tr><th>Machine</th><th>\(x\)</th><th>\(y\)</th></tr></thead> <tbody> <tr><td>A</td><td>\(1\)</td><td>\(8\)</td></tr> <tr><td>A</td><td>\(2\)</td><td>\(9\)</td></tr> <tr><td>A</td><td>\(3\)</td><td>\(10\)</td></tr> <tr><td>B</td><td>\(7\)</td><td>\(1\)</td></tr> <tr><td>B</td><td>\(8\)</td><td>\(2\)</td></tr> <tr><td>B</td><td>\(9\)</td><td>\(3\)</td></tr> </tbody> </table> Use technology to find the correlation for each machine separately and for all six observations pooled together. Round the pooled correlation to three decimals. Explain why the pooled result does not have to resemble either within-machine correlation.

Hints

- Compute the linear association within each group before combining the observations. - Compare where the two groups sit relative to one another in the coordinate plane. - The correlation of pooled data is not determined by averaging the separate correlations.

Solution

1. For Machine A, the three points lie exactly on \(y=x+7\), so \(r_A=1\). 2. For Machine B, the three points lie exactly on \(y=x-6\), so \(r_B=1\). 3. Pooling all six observations gives \(r\approx-0.880\). 4. The machine groups occupy different regions of the scatterplot. Machine B has larger \(x\)-values but much smaller \(y\)-values than Machine A, so the between-group pattern dominates the pooled correlation even though each within-group relationship is perfectly positive.

Answer

Machine A: \(r=1\). Machine B: \(r=1\). Pooled data: \(r\approx-0.880\). Correlation depends on the full pattern of points. Pooling groups with different locations can produce a very different overall linear association from the association within each group.
Correlation 64
A data set contains the three points \((0, 0), (1, 2), (2, 1)\), whose correlation is \(0.50\). The point \((0, 0)\) is then duplicated once, while the other two points remain single observations. Does the correlation stay the same? Compute the new correlation.

Hints

- Contrast duplicating one observation with duplicating every observation equally. - Recompute the means because the duplicated point changes the center of the data. - Use the centered paired deviations to find the new correlation.

Solution

1. Duplicating only one point changes that point's weight in the paired data, so correlation need not remain unchanged. 2. For the four points, \(\bar x=\bar y=0.75\). 3. The centered cross-product sum is \(1.75\), and each centered square sum is \(2.75\). 4. Therefore, \(r=\frac{1.75}{\sqrt{2.75\cdot2.75}}=\frac{1.75}{2.75}\approx0.636\).

Answer

No. The new correlation is approximately \(r=0.636\). Duplicating one observation changes its weight and can change correlation.

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