Unit 22 · Statistics in Algebra I

A teacher asks 12 students how many siblings they have and writes down the answers: 0, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 6. A list like this is hard to read. A dot plot fixes that. Draw a number line from 0 to 6. For each answer, put one dot above that number, stacking dots when a number repeats. You get 1 dot above 0, 3 dots above 1, 4 dots above 2, 2 dots above 3, 1 dot above 4, none above 5, and 1 dot above 6. The 12 dots make a picture of the whole distribution, which is the full set of values and how often each one occurs.

Once the dots are stacked, describe the shape. Ask three questions. Where is the tallest stack? Here it is at 2, so 2 siblings is the most common answer. Is the picture balanced? No. Most dots crowd the left side, and a thin tail stretches to the right toward the lone 6. A distribution shaped like this is skewed right, because the long tail points right. If the tail pointed left, it would be skewed left. A distribution with two matching halves, like a hill with the same slope on each side, is symmetric.

Students often mix up which way a skew points. The skew is named for the tail, not the peak. Picture a slide: the pile of dots is the platform, and the tail is the slide going down. In the sibling data, the slide goes to the right, so it is skewed right. Another common slip is forgetting a value that appears zero times. There are no students with 5 siblings, so 5 gets no dot, but 5 still stays on the number line. Leaving it out would squeeze the 6 next to the 4 and hide the gap.

A dot plot works for 12 answers. It falls apart for 300 commute times measured to the minute, because almost every value is different and the dots never stack. A histogram solves this by grouping. Suppose 30 students report their morning commute in minutes. Split the number line into equal-width bins: 0 to 9 minutes, 10 to 19, 20 to 29, 30 to 39, and 40 to 49. Count how many commutes land in each bin. Say the counts are 4, 9, 10, 5 and 2. Those five counts add to 30, which is a quick check that no student was lost or counted twice.

Now draw. The horizontal axis is minutes, split at 0, 10, 20, 30, 40 and 50. Above each bin draw a bar whose height is the frequency, the count for that bin. The bars touch each other, with no gaps, because the bins cover the whole number line without skipping any minutes. That is the visual difference between a histogram and a bar graph of categories like favorite pizza topping, where the bars are separated because the categories are not numbers on a line.

Reading a histogram means adding bins. How many students commute less than 20 minutes? Add the first two bins: 4 + 9 = 13 students. What share commute 30 minutes or more? Add the last two bins: 5 + 2 = 7, and 7/30 is about 0.233, or 23%. The most common bin is 20 to 29 minutes with 10 students, and the shape has a slight tail to the right, since the long commutes trail off.

The width of the bins changes the picture. Bins of width 5 would show more detail but might look jagged. Bins of width 25 would smooth everything into two fat bars and hide the peak. A frequent mistake is making bins that overlap, like 0 to 10 and 10 to 20. A commute of exactly 10 minutes would then belong to both bins. Use bins like 0 to 9 and 10 to 19, or agree on a rule such as ’the left edge is included and the right edge is not.’

Eleven students take a test and score 62, 68, 70, 74, 75, 78, 80, 84, 88, 90 and 95, already in order. The median is the middle value. With 11 scores, the middle is the 6th one, which is 78. Five scores sit below it and five above. Now split each half again. The lower half is 62, 68, 70, 74, 75, and its middle is 70. That is the first quartile, written Q1. The upper half is 80, 84, 88, 90, 95, and its middle is 88. That is the third quartile, Q3. Together with the minimum 62 and the maximum 95, these make the five-number summary: 62, 70, 78, 88, 95.

A box plot draws exactly those five numbers above a number line. Draw a box from Q1 = 70 to Q3 = 88. Draw a line inside the box at the median, 78. From each end of the box draw a whisker, a thin line out to the minimum 62 on the left and the maximum 95 on the right. Roughly one quarter of the scores lie in each of the four pieces: below the box, in the left part of the box, in the right part, and beyond the box. The box itself holds the middle half of the class.

When the count is even, the median is the average of the two middle values. Take 3, 5, 6, 8, 10, 12, 15, 20. The two middle values are 8 and 10, so the median is (8 + 10) ÷ 2 = 9. The lower half is 3, 5, 6, 8 and Q1 is (5 + 6) ÷ 2 = 5.5. The upper half is 10, 12, 15, 20 and Q3 is (12 + 15) ÷ 2 = 13.5. A common mistake is finding the quartiles from the unsorted list. Always sort first. Another is including the median itself in both halves when the count is odd; in this course, leave the middle value out of both halves.

Five people work at a small café in Pilsen. Four of them earn $12, $13, $13 and $14 an hour. The manager earns $48 an hour. The mean is the total shared equally: add the wages, 12 + 13 + 13 + 14 + 48 = 100, and divide by the 5 workers to get $20 an hour. The median is the middle wage when the list is sorted: 12, 13, 13, 14, 48, so the median is $13. If a job ad said ’our workers average $20 an hour,’ it would be telling the truth and still misleading four of the five people who applied.

The mean is sensitive to every value. Raise the manager’s wage from $48 to $98 and the total becomes 150, so the mean jumps to $30. The median stays $13, because the middle of the sorted list has not moved. A measure that barely changes when one extreme value changes is called resistant. The median is resistant. The mean is not. That is why news reports about house prices or incomes usually give the median: a few mansions or billionaires would drag the mean far above what a typical family sees.

Neither measure is wrong. They answer different questions. The mean answers this: if the total were shared equally, how much would each get? That matters for a budget. The café’s payroll is 5 × 20 = $100 an hour no matter how the wages are split. The median answers a different question: what does a typical member of this group look like? For a symmetric distribution the two are close together. For a skewed distribution they pull apart. The mean gets dragged toward the tail. Here is a quick check on any data set. If the mean is far above the median, expect a right tail of large values.

One week in June, a student records the daily high temperature outside her apartment in Chicago: 71, 75, 68, 80, 74, 73 and 90 degrees Fahrenheit. Sort them: 68, 71, 73, 74, 75, 80, 90. The range is the biggest value minus the smallest: 90 − 68 = 22 degrees. The range is quick and easy, but it depends entirely on the two most extreme days. That single 90-degree Saturday sets the whole range by itself.

The interquartile range, or IQR, ignores the extremes. Find the median first. With 7 values the middle is the 4th, 74. The lower half is 68, 71, 73, so Q1 = 71. The upper half is 75, 80, 90, so Q3 = 80. The IQR is Q3 − Q1 = 80 − 71 = 9 degrees. That is the width of the box in a box plot, the stretch of the middle half of the week. Notice that the 90 does not appear anywhere in that subtraction.

Now suppose the Saturday reading was a typo and it should have been 120, a number that would break Chicago’s all-time record. The range becomes 120 − 68 = 52, more than doubling. The IQR is still 80 − 71 = 9, because Q3 is still the middle of the upper half, and 120 is still just the biggest value in that half. Like the median, the IQR is resistant. Range and IQR are both measures of spread: how far apart the values are. The range is honest about the extremes; the IQR is honest about the crowd in the middle.

The IQR uses only two numbers from the data. The standard deviation uses all of them. Start with five quiz scores: 6, 7, 8, 9, 10. The mean is 40 ÷ 5 = 8. Now find each score’s deviation, its distance from the mean with a sign: 6 − 8 = −2, 7 − 8 = −1, 8 − 8 = 0, 9 − 8 = 1, and 10 − 8 = 2. The deviations always add to zero, −2 − 1 + 0 + 1 + 2 = 0, so averaging them tells you nothing. That is the reason for the next step.

Square each deviation to erase the signs: 4, 1, 0, 1, 4. Add them: 4 + 1 + 0 + 1 + 4 = 10. Divide by the number of values, 5, to get 2. That number is the variance. Its units are squared points, which is awkward, so take the square root: √2 ≈ 1.41. The standard deviation of these scores is about 1.41 points. Read it this way: a typical quiz score in this set sits about 1.4 points from the mean of 8.

Compare a second set with the same mean: 8, 8, 8, 8, 8. Every deviation is 0, so the variance is 0 and the standard deviation is 0. A third set, 4, 6, 8, 10, 12, has deviations −4, −2, 0, 2, 4, squares 16, 4, 0, 4, 16 that add to 40, variance 40 ÷ 5 = 8, and standard deviation √8 ≈ 2.83. Same mean, twice the spread, and the standard deviation doubles from 1.41 to 2.83. Larger standard deviation means the values wander farther from the mean.

Two warnings. First, many calculators offer two versions, one dividing by the number of values and one dividing by one less than that. The second version gives a slightly bigger answer, √(10 ÷ 4) ≈ 1.58 for the quiz scores, and is used when the data are a sample from a bigger group. In this course, divide by the number of values unless you are told otherwise. Second, do not skip the square root. A variance of 2 and a standard deviation of 1.41 are different numbers, and only the second is in the same units as the data.

Eight friends count the text messages they sent in one hour: 12, 14, 15, 15, 16, 17, 18 and 40. Seven of the numbers are between 12 and 18. The 40 is far from the rest. A value like that is an outlier. Sometimes it is a mistake, like a typo. Sometimes it is real: one friend was planning a party. Either way, you need a rule for deciding when a value is far enough away to count, and you need to know what it does to your summaries.

The standard rule uses the IQR. The median of the eight values is (15 + 16) ÷ 2 = 15.5. The lower half, 12, 14, 15, 15, has median (14 + 15) ÷ 2 = 14.5, so Q1 = 14.5. The upper half, 16, 17, 18, 40, has median (17 + 18) ÷ 2 = 17.5, so Q3 = 17.5. The IQR is 17.5 − 14.5 = 3. Multiply it by 1.5 to get 4.5. Any value more than 4.5 below Q1 or more than 4.5 above Q3 is an outlier. The fences are 14.5 − 4.5 = 10 and 17.5 + 4.5 = 22. The 40 is above 22, so it is an outlier. The 12 is not below 10, so it is fine.

Now see the pull. With the 40, the total is 147 and the mean is 147 ÷ 8 ≈ 18.4. Without it, the total is 107 and the mean is 107 ÷ 7 ≈ 15.3. One value moved the mean by about 3 messages. The median moves from 15.5 to 15, half a message. The range collapses from 40 − 12 = 28 to 18 − 12 = 6. Without the 40, Q1 is 14 and Q3 is 17, so the IQR is still 3, unchanged. This is the whole reason the median and IQR are called resistant. The mean and range are not.

Do not just delete an outlier because it is inconvenient. First ask whether it is an error. A height of 5.9 meters is a typo for 1.59 meters; fix it. A student who really did send 40 texts belongs in the data, and the honest move is to report the summary both ways and say why. On a box plot, outliers are drawn as separate dots beyond the whiskers, and the whisker stops at the last value inside the fence, here at 18.

Back to Coach Alvarez. Dana’s runs allowed in eight starts: 3, 3, 2, 4, 3, 3, 4, 2. Marcus’s: 0, 1, 0, 7, 0, 8, 1, 7. Both totals are 24, so both means are 3. A one-number comparison ends there in a tie. A three-part comparison, shape, center and spread, does not. Sort Dana: 2, 2, 3, 3, 3, 3, 4, 4. Sort Marcus: 0, 0, 0, 1, 1, 7, 7, 8. Dana’s dot plot is a single symmetric hill. Marcus’s has two clumps with a gap in the middle, one at 0 and 1 and one at 7 and 8.

Center: Dana’s median is (3 + 3) ÷ 2 = 3. Marcus’s median is (1 + 1) ÷ 2 = 1. Spread: Dana’s range is 4 − 2 = 2 and Marcus’s is 8 − 0 = 8. For the IQR, Dana’s Q1 is the middle of 2, 2, 3, 3, which is 2.5, and Q3 is the middle of 3, 3, 4, 4, which is 3.5, so the IQR is 1. Marcus’s Q1 is the middle of 0, 0, 0, 1, which is 0, and Q3 is the middle of 1, 7, 7, 8, which is 7, so the IQR is 7. Dana is far more consistent: her spread is a fraction of his by every measure.

Write the comparison in sentences, with the numbers. ’Both pitchers allow a mean of 3 runs per start, but their distributions differ. Dana’s runs are symmetric and tightly clustered, with an IQR of 1. Marcus’s are split into two clumps, with a median of only 1 but an IQR of 7.’ When two groups are drawn as box plots on the same number line, look for overlap: if one box sits entirely to the right of the other, the groups differ clearly. If the boxes overlap heavily, the difference between the groups is smaller than the variability inside each group.

One trap is comparing groups of different sizes by raw counts. Suppose 40 students in one class and 20 in another take a survey. Then 10 ’yes’ answers mean 25% in the first class and 50% in the second. When the groups are not the same size, compare percents, medians or means, never raw counts. And always use the same scale on both plots. A box plot drawn from 0 to 10 next to one drawn from 0 to 100 makes the second group look more crowded than it is.

Every data set can be summarized two ways. The mean and standard deviation form one pair: both use every value, and both get pulled by extremes. The median and IQR form the other pair: both depend on position in the sorted list, and both resist extremes. Choosing between the pairs is a judgment about the shape of the data, so look at a dot plot, histogram or box plot before you pick.

Take the café wages: 12, 13, 13, 14, 48. The distribution is strongly skewed right, with the manager’s 48 far out in the tail. The mean of 20 describes nobody. The median of 13 and the IQR describe the four ordinary workers well. For skewed data or data with outliers, report the median and IQR. Now take Dana’s runs: 2, 2, 3, 3, 3, 3, 4, 4. Symmetric, no outliers, a tidy hill. Here the mean of 3 and the standard deviation tell the story completely, and they use all eight games instead of just positions in a list. For symmetric data without outliers, report the mean and standard deviation.

Compute Dana’s standard deviation to see the pair in action. The mean is 3. Deviations are −1, −1, 0, 0, 0, 0, 1, 1. Squares: 1, 1, 0, 0, 0, 0, 1, 1, summing to 4. Divide by 8 to get 0.5, and take the square root: √0.5 ≈ 0.71. A typical Dana start is about 0.7 runs from her mean. For Marcus, deviations from 3 are −3, −3, −3, −2, −2, 4, 4, 5, squares 9, 9, 9, 4, 4, 16, 16, 25 summing to 92, variance 92 ÷ 8 = 11.5, standard deviation √11.5 ≈ 3.39. About five times Dana’s. But notice Marcus’s two-clump shape: no single center describes him well, so any summary should come with the picture.

The most common mistake is reporting a mean because it is familiar, without checking the shape. The second is reporting a spread measure from the wrong pair, like a median with a standard deviation. Keep the pairs together: mean with standard deviation, median with IQR. And when in doubt, report both pairs and show the plot. A reader can always ignore an extra number; a reader cannot recover the one you left out.

A school surveys 100 students with two questions: what grade are you in, and how do you get to school? Every student is in grade 9 or grade 10, and every student either rides the bus or walks. Two questions with two answers each make four combinations, and a two-way table has one cell for each. Say 35 ninth graders ride the bus and 25 walk, while 20 tenth graders ride the bus and 20 walk. Rows are grades, columns are how they travel. The four cells hold 35, 25, 20 and 20, which add to 100.

The numbers inside the four cells are joint frequencies. Each one counts students who fit two categories at the same time, like ’grade 9 and bus.’ Now add across each row and down each column. Row totals: grade 9 has 35 + 25 = 60 students, and grade 10 has 20 + 20 = 40. Column totals: bus riders number 35 + 20 = 55, and walkers number 25 + 20 = 45. These totals are written in the margins of the table. They are the marginal frequencies. Each margin counts one category by itself, ignoring the other question. Both sets of margins add to the same grand total: 60 + 40 = 100 and 55 + 45 = 100. That is a built-in check on your arithmetic.

Reading a two-way table is a matter of finding the right cell. How many tenth graders walk? Go to the grade 10 row and the walk column: 20. How many students ride the bus in total? Read the bus column margin: 55. How many students are in grade 9? Read the grade 9 row margin: 60. A common mistake is answering a marginal question with a joint cell, like saying 35 ninth graders when the question asked for all ninth graders. Read the question twice: does it name one category or two?

The survey table says 35 ninth graders and 20 tenth graders ride the bus. Does that mean ninth graders like the bus more? Not yet. There are 60 ninth graders and only 40 tenth graders, so of course the ninth grade has more of everything. To compare fairly, turn counts into fractions of the right group. A relative frequency is a cell count divided by some total. Divide by the grand total, 100, and you get the joint relative frequency: 35 ÷ 100 = 0.35, so 35% of all students are ninth graders who ride the bus.

The more useful move is to divide by a row or column total. That gives a conditional relative frequency, because you are asking a question on the condition that someone belongs to a certain group. Among ninth graders, what fraction ride the bus? Divide the cell by the row total: 35 ÷ 60 ≈ 0.583, about 58%. Among tenth graders: 20 ÷ 40 = 0.50, exactly 50%. Now the comparison is fair. Ninth graders are somewhat more likely to ride the bus than tenth graders, 58% versus 50%.

You can condition on the columns instead. Among bus riders, what fraction are in grade 9? Divide by the column total: 35 ÷ 55 ≈ 0.636, about 64%. Among walkers: 25 ÷ 45 ≈ 0.556, about 56%. Notice that ’the fraction of ninth graders who ride the bus’ (58%) and ’the fraction of bus riders who are ninth graders’ (64%) are different questions with different answers. The most common mistake in this whole topic is dividing by the wrong total. Say the condition out loud: ’of the ninth graders’ means divide by 60; ’of the bus riders’ means divide by 55.

When the conditional relative frequencies differ from group to group, the two categories have an association: knowing a student’s grade tells you something about how they travel. If every row gave the same percents, the categories would be independent of each other. Here the gap is 58% versus 50%, a modest association. Whether a gap that size matters is a question for later courses; for now, the skill is computing the right fraction and stating clearly which group it describes.

A poster shows two bars: last year’s test average, 50, and this year’s, 52. The bar for 52 is twice as tall as the bar for 50. How? The vertical axis does not start at 0. It starts at 48. Measured from 48, the first bar is 2 units tall and the second is 4 units tall, so it looks like a doubling. The real change is 52 ÷ 50 = 1.04, a 4% rise. A truncated axis, one that cuts off the bottom, is the most common trick in graphs. Always find the number where the axis begins before you compare bar heights.

The second thing to check is who was asked. A gym hands out a survey about exercise habits to its own members and reports that 90% of people exercise three times a week. The sample, the group that was actually measured, is gym members, and gym members are not typical of everyone. The sample is biased: it leans in one direction before a single question is asked. A fair sample gives every person in the group you care about a chance to be picked. Small samples are a related problem: a 3-out-of-4 result from four people is not a 75% result for a school of 1,600.

The third check is the summary itself. A company with 40 workers earning $40,000 and one owner earning $2,000,000 can honestly say its mean salary is over $87,000: the total is 1,600,000 + 2,000,000 = 3,600,000, divided by 41 is about $87,800. The median is $40,000. Both numbers are true. Only one of them describes a typical worker. When you see an average, ask which kind, and ask what the shape of the data looks like. When you see a percent, ask what the total was. When you see a graph, find zero on the axis.

None of this makes statistics untrustworthy. It makes you a careful reader. The same tools that can mislead, axes, samples and summaries, are the tools that tell the truth when used well. A histogram with an honest axis, a sample chosen fairly and a summary that fits the shape of the data is the strongest kind of argument there is. Your job as a reader is to check that all three are in place, and your job as a writer is to make sure they are.

Five students record how many hours they studied for a quiz and the score they got: 1 hour and 60 points, 2 hours and 65, 3 hours and 75, 4 hours and 80, 5 hours and 90. Each student gives one pair of numbers, and each pair becomes one dot. A scatter plot draws those dots on a grid: hours on the horizontal axis, score on the vertical axis. The student who studied 3 hours and scored 75 is the dot at (3, 75). Five students, five dots.

Which variable goes on which axis is not random. The explanatory variable is the one you think does the explaining or comes first, and it goes on the horizontal axis. Hours of studying come before the quiz, so hours are explanatory. The response variable is the outcome you want to predict, and it goes on the vertical axis. The score responds to the studying. In an experiment the explanatory variable is the one you control; in a survey it is the one you would use to predict the other.

Now look at the drift. As hours go up, scores go up: the dots climb from lower left to upper right. That is a positive association. If the dots fell from upper left to lower right, like hours of TV against quiz score, that would be a negative association. If the dots formed a shapeless cloud with no drift, there would be no association. Also note the form, whether the dots follow a straight line or a curve, and the strength, whether they hug the pattern tightly or scatter widely. These five dots follow a nearly straight line, tightly, going up.

A common mistake is reading a scatter plot as a time graph and connecting the dots in order. Do not connect them. Each dot is a separate student, not a step in a sequence, and the pattern is in the cloud as a whole. Another is describing a single dot as ’high’ without saying which axis: the student at (5, 90) is high in hours and high in score, and both facts matter.

The five study dots, (1, 60), (2, 65), (3, 75), (4, 80) and (5, 90), nearly form a line. A line of fit is a straight line drawn through the cloud so that the dots are balanced around it, about as many above as below and none very far away. To find an equation by hand, pick two points that seem to sit on the trend, not necessarily data points. Here the first and last dots look right. Use (1, 60) and (5, 90).

The slope is rise over run: (90 − 60) ÷ (5 − 1) = 30 ÷ 4 = 7.5. Each extra hour of studying goes with 7.5 more points. Now find the y-intercept using one point. If y = 7.5x + b and the point (1, 60) is on the line, then 60 = 7.5 × 1 + b, so b = 52.5. The line is y = 7.5x + 52.5. Check it with a point you did not use: at x = 3, y = 7.5 × 3 + 52.5 = 22.5 + 52.5 = 75, which is exactly the middle dot. At x = 4, y = 30 + 52.5 = 82.5, close to the actual 80.

A calculator or spreadsheet finds the least-squares line. That is the one line that makes the total of the squared vertical gaps as small as possible. For these data it gives y = 7.5x + 51.5, just 1 point below the hand-drawn line. Two people drawing by eye will get slightly different lines. That is fine, as long as each line runs through the middle of the cloud. The least-squares line is the standard answer everyone can agree on. This chapter uses it from here on.

The most common error is picking the two points badly, such as the highest dot and the lowest dot even when they are off the trend, or forcing the line through the origin. The line does not have to pass through (0, 0); it has to pass through the middle of the data. Another slip is mixing up rise and run. Slope is change in y divided by change in x, vertical over horizontal, always.

A used-car website fits a line to the ages and prices of a certain model. With age a in years and value v in thousands of dollars, the line is v = 24 − 2.5a. Numbers in an equation are only useful once you can say them in words. The slope is −2.5. In context: for each additional year of age, the predicted value drops by 2.5 thousand dollars, that is, $2,500 per year. The intercept is 24. In context: a car of age 0, brand new, is predicted to be worth $24,000.

Use the line to predict. A 4-year-old car: v = 24 − 2.5 × 4 = 24 − 10 = 14, so about $14,000. A 6-year-old car: v = 24 − 15 = 9, about $9,000. Notice the word ’predicted’ or ’about’ in every sentence. The line describes the trend across many cars; any one car can sit above or below it because of mileage, accidents or paint color. A line of fit predicts the typical value, not the exact value.

Always attach units to slope and intercept. ’The slope is −2.5’ is incomplete. ’Value drops by $2,500 per year’ is a sentence a car buyer can use. The slope’s units are always the response units per one explanatory unit: dollars per year, points per hour, centimeters per month. The intercept has the response’s units alone.

Sometimes the intercept has no real meaning. A line predicting a child’s height from age might have an intercept of 50 cm, which is sensible. But a line predicting shoe size from height might have an intercept of −20. Nobody has a negative shoe size. The intercept is just where the line crosses the axis. It only describes reality if x = 0 is a situation that actually occurs in the data. A common mistake is to explain a slope backwards, as ’years per dollar.’ Read it as y per x. The response changes this much when the explanatory variable rises by one.

A line of fit rarely passes through every dot, so measure the misses. Use the least-squares line for the study data, y = 7.5x + 51.5. For the student at (2, 65), the line predicts y = 7.5 × 2 + 51.5 = 15 + 51.5 = 66.5. The student actually scored 65. The residual is actual minus predicted: 65 − 66.5 = −1.5. The dot sits 1.5 points below the line. Negative residual, dot below.

Do all five. At x = 1 the predicted value is 7.5 + 51.5 = 59 and the actual is 60, so the residual is 60 − 59 = 1. At x = 3 the prediction is 22.5 + 51.5 = 74 and the actual is 75, residual 1. At x = 4 the prediction is 30 + 51.5 = 81.5 and the actual is 80, residual −1.5. At x = 5 the prediction is 37.5 + 51.5 = 89 and the actual is 90, residual 1. The residuals are 1, −1.5, 1, −1.5, 1. Add them: 1 − 1.5 + 1 − 1.5 + 1 = 0. That is not a coincidence. The least-squares line always makes the residuals sum to zero; the dots above balance the dots below exactly.

Residuals are how the least-squares line earns its name. Square each residual: 1, 2.25, 1, 2.25, 1, which add to 7.5. Any other line through these dots gives a bigger total. Try the hand-drawn line y = 7.5x + 52.5 from the last lesson: its residuals are 0, −2.5, 0, −2.5, 0, and their squares add to 12.5, larger than 7.5. Smaller total means better fit, and the least-squares line is the champion by definition.

Keep the order straight: actual minus predicted, never the reverse. Students who flip it get the right size and the wrong sign, and then describe a dot as above the line when it is below. A second slip is reading the residual off the horizontal axis. A residual is a vertical distance, measured straight up or down from the dot to the line, because it is a miss in the predicted response, not in the explanatory variable.

A residual plot draws the residuals themselves: x on the horizontal axis and each residual on the vertical axis, with a horizontal line at 0 marking the fit line. For the study data the residuals are 1, −1.5, 1, −1.5, 1. They bounce above and below zero with no trend, all within 1.5 points. That is what a good fit looks like. If a line is the right model, the leftover misses should look like random noise, small and patternless.

Now try data that grow faster and faster: (1, 2), (2, 5), (3, 10), (4, 17), (5, 26). The least-squares line is y = 6x − 6. Its predictions are 0, 6, 12, 18 and 24. The residuals are 2 − 0 = 2, 5 − 6 = −1, 10 − 12 = −2, 17 − 18 = −1, and 26 − 24 = 2. In order they read 2, −1, −2, −1, 2: positive at both ends, negative in the middle. Plotted, they form a U. The line is too high in the middle and too low at both ends, because the data bend upward and a straight line cannot bend.

A U or an upside-down U in the residual plot means a curve would fit better than a line. A fan shape is different: residuals are tiny on the left and huge on the right. It means the line’s predictions get less reliable as x grows. Either pattern is a warning. These patterns are easy to miss on the scatter plot, where the dots may look close to the line. They are impossible to miss once the line is flattened out to zero. That is the point of the residual plot.

Do not judge a fit by the size of the residuals alone. The U-shaped residuals above are only 2 units off on data that reach 26, which sounds fine, yet the pattern says the linear model is wrong and will be badly wrong beyond x = 5. Small residuals with a pattern are a worse sign than larger residuals without one. Also remember that a residual plot uses the same x-values as the scatter plot; only the vertical axis changes.

Words like ’strong’ and ’weak’ are vague. The correlation coefficient, written r, replaces them with a number. It is always between −1 and 1. The sign gives the direction: positive r means the dots drift upward, negative r means they drift downward. The size gives the strength: r near 1 or −1 means the dots hug a straight line tightly, and r near 0 means the dots scatter with little straight-line pattern. If every dot sits exactly on an upward line, r = 1. On a downward line, r = −1.

A calculator computes r from the data. The formula multiplies how far each x is from its mean by how far the matching y is from its mean. It adds those products and scales the total so the result lands between −1 and 1. For the study data, (1, 60) through (5, 90), r ≈ 0.99. That is a very strong positive correlation, and it matches the picture of five dots almost on a line. Here are rough guides for reading r. Above 0.8 in size is strong. From 0.5 to 0.8 is moderate. Below 0.5 is weak. The cutoffs shift by field, though. A medical study may call 0.3 meaningful; a physics lab would call it noise.

Two facts make r easy to work with. First, r has no units. Change hours to minutes or points to percents and r does not move, because it measures pattern, not scale. Second, r is the same whichever variable you put on which axis, since it treats x and y symmetrically. The slope of the line of fit has neither property; it changes with the units and flips when you swap the axes.

Students often confuse the strength of r with the steepness of the line. A slope of 0.001 can have r = 1 if the dots sit exactly on that gentle line, and a steep slope of 50 can have r = 0.2 if the dots scatter widely around it. Steepness is slope. Tightness is r. Another mistake is reading r = −0.9 as weaker than r = 0.6 because it is ’negative.’ Compare sizes: 0.9 is larger than 0.6, so the negative relationship is the stronger one.

Take five dots that sit perfectly on a U: (−2, 4), (−1, 1), (0, 0), (1, 1) and (2, 4). Every y equals x squared, so the relationship is as exact as a relationship can be. Compute r and you get exactly 0. Why? The left half of the U falls and the right half rises, and the two halves cancel in the formula. A student who looks only at r would say ’no relationship’ and be completely wrong. The value r = 0 means no straight-line relationship, nothing more. Always look at the scatter plot before you trust r.

Now the reverse trap. The curving data from the residual lesson, (1, 2), (2, 5), (3, 10), (4, 17), (5, 26), give r ≈ 0.98. That sounds like a perfect line. But the residual plot was a clear U, and the line predicts 6 × 10 − 6 = 54 at x = 10 when the curve y = x^2 + 1 gives 101. A high r does not prove that a line is the right model; it only says the dots rise together. The residual plot catches what r misses.

A single outlier can also swing r. Picture nine dots forming a shapeless blob with r near 0, and then add one dot far away in the upper right. The formula counts that dot’s huge distances from both means, and r can jump to 0.7 or more, from a single point. The opposite happens too: one dot far off a tight line can drag r from 0.95 down to 0.5. When r changes a lot after removing one point, report both values and say so.

Finally, r says nothing about how much y changes per unit of x. That is the slope’s job. And r says nothing about why the variables move together, which is the next lesson. The correlation coefficient is one summary of one feature of a scatter plot. It is a good summary of that one feature. Everything else, form, outliers, clusters and cause, needs the picture and your judgment.

In the chapter story, ice cream sales and water rescues had a correlation near 0.93, and a council member concluded that ice cream causes drowning. His graph was right and his conclusion was wrong. Heat, the lurking variable, drives both: hot days send people to the ice cream stand and into the lake. A lurking variable is one that was left out of the plot but affects both variables in it. Whenever two things rise and fall together, ask what else changes at the same time.

Here is another. Among elementary school children, shoe size and reading level are strongly correlated. Bigger feet, better readers. Does buying larger shoes improve reading? Of course not; the lurking variable is age. Older children have bigger feet and have had more years of reading lessons. The correlation is real and could even be used to predict: tell me a child’s shoe size and I can guess their reading level better than chance. Prediction works. Causation, one thing making another happen, was never shown.

Correlation can also run backwards. Cities with more police officers tend to have more reported crime. Do officers cause crime? More likely, crime leads cities to hire officers. The direction of cause is not in the scatter plot; both variables are just numbers on axes. And sometimes a correlation is pure coincidence: with thousands of variables being tracked, some pairs will move together by chance, like the number of movies an actor makes and the price of a vegetable.

How do scientists ever show cause? With an experiment: they change the explanatory variable on purpose, assign subjects to groups at random so lurking variables spread evenly, and then compare the responses. A survey or a record book, where nobody controlled anything, is observational, and it can show correlation only. Every headline that says ’X linked to Y’ is reporting a correlation. The careful reader asks: was this an experiment, or could a lurking variable explain it, or could the cause run the other way?

A biology class counts bacteria in a dish every hour: 100 at hour 0, then 200, 400, 800 and 1,600. Try a line. The changes from hour to hour are +100, +200, +400 and +800. A line needs the same change every step, and these changes double. A line fit to these points would be too high at hours 1 and 3 and far too low at hour 4, and the residual plot would show a U. The line fails because the pattern is not additive. It is multiplicative.

Look at ratios instead of differences. 200 ÷ 100 = 2, 400 ÷ 200 = 2, 800 ÷ 400 = 2, 1,600 ÷ 800 = 2. Every hour the count multiplies by 2. A pattern with a constant ratio is exponential. Its model is y = a × b^x, where a is the starting value and b is the growth factor, the constant ratio. Here y = 100 × 2^x. Check: at x = 3, 100 × 2^3 = 100 × 8 = 800. To predict hour 6, compute 100 × 2^6 = 100 × 64 = 6,400.

Real data are never this clean, so the test is whether the ratios are roughly constant. Say a savings account shows $1,000, $1,050, $1,102, $1,158 over four years. Ratios: 1.05, 1.0495, 1.0508. Close enough to call it exponential with growth factor about 1.05, or 5% per year. The differences, 50, 52 and 56, creep upward, which is the fingerprint of exponential growth: each year’s increase is a little bigger than the last because it is a percent of a bigger amount.

The test for a line is constant differences; the test for an exponential is constant ratios. Students often check only differences, see them change, and give up. Check ratios next. A second common mistake is writing the growth factor as the percent: 5% growth is a factor of 1.05, not 0.05 or 5. And a ratio less than 1, like 0.9, still gives an exponential model, one that decays, such as a medication leaving the bloodstream by 10% each hour.

A ball is dropped from a tall building, and a camera records how far it has fallen after each second, in feet: 0, 16, 64, 144, 256. First differences: 16, 48, 80, 112. Not constant, so not a line. Ratios: 64 ÷ 16 = 4, 144 ÷ 64 = 2.25, 256 ÷ 144 ≈ 1.78. Not constant either, so not exponential. Now take differences of the differences: 48 − 16 = 32, 80 − 48 = 32, 112 − 80 = 32. The second differences are constant. That is the fingerprint of a quadratic model, an equation with an x^2 term.

Here the model is d = 16t^2, distance in feet after t seconds. Check: at t = 3, 16 × 9 = 144. Predict t = 5: 16 × 25 = 400 feet. A simpler example: y-values 0, 5, 20, 45, 80 for x = 0 through 4 have first differences 5, 15, 25, 35 and second differences 10, 10, 10. The model is y = 5x^2. At x = 6 it predicts 5 × 36 = 180.

Quadratic patterns show up whenever something speeds up steadily or turns around. A falling object is one example. So is the height of a thrown basketball on its way up and back down. So is the area of a square as its side grows. A calculator can fit a quadratic to messy data just as it fits a line. The same tools apply: look at the residual plot. If a line leaves a U in the residuals and a quadratic leaves patternless noise, the quadratic is the better model.

Keep the three tests in one place. Constant first differences: linear, y = mx + b. Constant ratios: exponential, y = a × b^x. Constant second differences: quadratic, with an x^2 term. Always check the tests in that order, and only on equally spaced x-values; differences between x = 1 and x = 2 and between x = 2 and x = 5 are not comparable. A common slip is to compute second differences from the original values instead of from the first differences.

The study-time line, y = 7.5x + 51.5, was built from students who studied 1 to 5 hours. Ask it about 3.5 hours: y = 7.5 × 3.5 + 51.5 = 26.25 + 51.5 = 77.75, about 78 points. That is interpolation, a prediction between data points you actually have. Nobody in the data studied exactly 3.5 hours, but people studied 3 and 4, and the line between them is a reasonable guess.

Now ask about 10 hours: y = 7.5 × 10 + 51.5 = 75 + 51.5 = 126.5. A score of 126.5 on a 100-point quiz is impossible. The line did nothing wrong; it kept going straight, as lines do. The mistake was asking it about a place with no data. That is extrapolation, predicting beyond the range that was measured. Real relationships bend, level off or stop, and the line cannot know that. Ask about 0 hours and the line says 51.5, which might be reasonable, or might not; no student in the data studied 0 hours either.

Extrapolation is not forbidden, but it needs a warning label, and the farther you go, the louder the label. Predicting 5.5 hours from data that reach 5 is a small stretch. Predicting 20 hours is a leap into the dark. Exponential models extrapolate especially badly: the bacteria model y = 100 × 2^x predicts 100 × 2^24, over 1.6 billion, after one day, and the dish would have run out of food long before. Every model has a range where it is trustworthy, and that range is roughly the range of the data that built it.

When you report a prediction, state the range of the data next to it. For example: ’Based on students who studied 1 to 5 hours.’ That one clause tells the reader whether your prediction is interpolation or extrapolation. A frequent mistake is reading a prediction off a line that has been extended on the graph to fill the page. The drawn line may extend to the edges. The data do not. Only the data part deserves trust.