AP Statistics: Lecture Notes

These are the notes of my AP Statistics lectures, organised by the units of the official AP course. The examples are the ones we worked through on the board. Every unit ends with three practice sets, each with a 15-question multiple-choice quiz that checks your answers and AP-style free-response questions with full solutions. Try each question on paper before opening the solution.

Unit 1: Exploring One-Variable Data

The Big Picture

The course has three big parts:

1. Exploratory data analysis (Units 1–2): looking at data, drawing it and summarising it with numbers.
2. Probability (Units 4–5): the mathematics of chance, which tells us how much samples vary.
3. Inference (Units 6–9): using a sample to draw conclusions about a whole population.

In the same way, statistics itself splits into two kinds:

• Descriptive statistics describes the data we actually have.
• Inferential statistics uses the data we have (a sample) to say something about data we do not have (the population).

Population, Sample, Parameter, Statistic

• The population is the whole group we are interested in. A number that describes the population is a parameter.
• A sample is the part of the population we actually observe. A number calculated from the sample is a statistic.

An easy way to remember: population goes with parameter, sample goes with statistic. We usually do not know the parameter; we estimate it with a statistic.

Notation follows the same split: statistics get Latin letters, parameters get Greek letters.

Types of Variables

• A categorical variable puts each individual into a group: eye colour, favourite subject, city of birth. Even if the categories are written with numbers (a postcode, a jersey number), they are categorical if adding or averaging them makes no sense.
• A quantitative variable is a number that we can do arithmetic with: height, exam score, number of siblings.

Quantitative variables are of two kinds:

• Discrete: the possible values can be listed, usually counts. The number of heads in 20 coin flips can only be $0,1,2,\dots ,20$.
• Continuous: any value in an interval is possible, usually measurements. A height can be $172$, $172.4$, $172.43$ cm, …

Describing a Distribution

When you are asked to describe a distribution of quantitative data, always comment on four things, in context (name the variable and the units):

1. Shape: symmetric or skewed, how many peaks, any gaps or clusters.
2. Center: a typical value (mean or median).
3. Spread: how much the values vary.
4. Unusual features: outliers, gaps, clusters.

The common shapes:

• Mound-shaped (bell-shaped): one peak in the middle, falling off on both sides.
• Roughly symmetric: the left half is close to a mirror image of the right half.
• Skewed left (negatively skewed): the long tail is on the left, most values are high.
• Skewed right (positively skewed): the long tail is on the right, most values are low.
• Bimodal: two clear peaks; often a sign that two different groups are mixed together.
• Uniform: all bars roughly the same height, no peak.

The direction of skew is the direction of the tail, not of the peak.

Dotplots

Draw a number line and put one dot above it for every value; repeated values are stacked on top of each other. Dotplots are quick and show every single value, so they are best for small datasets.

Stemplots (Stem-and-Leaf Plots)

Every value is split into a stem (the leading digits) and a leaf (the last digit). The stems are written once in a column, and the leaves are written next to their stem, in order.

Always give a key. Without it, nobody knows what the numbers mean:

• Split stems: when there are too few stems, each stem is used twice: once for leaves $0$–$4$ and once for leaves $5$–$9$. This stretches the plot so the shape becomes visible.
• Back-to-back stemplot: two datasets share one column of stems; one set of leaves goes to the left, the other to the right. It is a quick way to compare two groups. One key works for both sides (for example $1\mid 5$ means $15$).

A stemplot is a histogram turned on its side that still keeps the actual values.

Bar Charts (Categorical Data)

The categories go on the horizontal axis; the height of each bar is the frequency (count) or the relative frequency (proportion).

Example. 25 students choose their favourite subject, and 5 of them choose mathematics. The frequency of "mathematics" is $5$ and the relative frequency is

$$\frac{5}{25}=0.20=20\mathrm{\%}.$$

Bars in a bar chart are drawn with gaps between them, because the categories are separate and their order usually does not matter.

A frequency table lists each category with its count; a relative frequency table lists the proportions, which add up to 1 (up to rounding). A pie chart shows the same proportions as slices of a circle. It only works when the categories are parts of one whole, and it makes small differences hard to see, so bar charts are usually the better choice.

For categorical data, "shape" means little: the categories can be put in any order. Describe which categories are most and least common instead.

Histograms (Quantitative Data)

Split the number line into intervals of equal width (bins or classes) and draw a bar over each one whose height is the number (or proportion) of values that fall in it. A value on the boundary belongs to exactly one bin, so bins are written like $20\le x<30$. There are no gaps between the bars, because the number line is continuous.

• The bin width changes the picture: with width 10 you see the rough shape, with width 5 you see more detail (and more noise).
• A histogram loses the individual values: you only know how many values fall in each bin.

Example (class boundaries). A histogram has bars centred at $15,20,25,\dots ,50$. Consecutive midpoints are $5$ apart, so every bar is $5$ wide and the bars reach $2.5$ on either side of their midpoint. The boundaries are

$$12.5,17.5,22.5,\dots ,52.5.$$

If the first three bars hold $2,3$ and $5$ of the $30$ values, the proportion of values below $27.5$ is

$$\frac{2+3+5}{30}=\frac{10}{30}\approx 0.33.$$

Example (normal-looking data). The heights of 100 adult women, drawn with bins such as $59\le x<61$, give a roughly symmetric, mound-shaped histogram centred near 66 inches. Data like this is called approximately normal; the normal curve comes later in this unit.

Measures of Center

The mean (average, ortalama) adds all the values and divides by how many there are:

$$\overline{x}=\frac{\sum x_i}{n}=\frac{x_1+x_2+\cdots +x_n}{n}.$$

Example. Quiz scores $6,8,9,10,12$: $\overline{x}={\displaystyle \frac{45}{5}}=9$.

The median is the middle value of the sorted data: at least half of the values are at or below it and at least half are at or above it.

1. Sort the data first.
2. With $n$ values, the median sits at position ${\displaystyle \frac{n+1}{2}}$.
3. If that position is a whole number, the median is that value; if it ends in $.5$, the median is the mean of the two values on either side.

Example. For $10,20,30,40$ we have $n=4$, so the position is ${\displaystyle \frac{4+1}{2}}=2.5$: halfway between the 2nd and 3rd values, and the median is ${\displaystyle \frac{20+30}{2}}=25$.

${\displaystyle \frac{n+1}{2}}$ is the position (index) of the median, not the median itself.

Resistance

A statistic is resistant if a few extreme values cannot change it much.

Example. For $10,20,30$ the median is $20$ and the mean is $20$. Change the last value to $300$:

$$10,20,300\Rightarrow \text{median}=20,\overline{x}=\frac{330}{3}=110.$$

The median did not move at all; the mean jumped from 20 to 110. The median is resistant, the mean is not.

This tells us which one to use:

• Symmetric data: mean and median are close; either works.
• Skewed data or data with outliers: the median describes the center better.

Mean versus median in a skewed distribution. Look at the tail: the mean is pulled toward the tail.

• Skewed right: mean $>$ median.
• Skewed left: mean $<$ median.

Estimating the Median from Grouped Data

If you only have a frequency table or a histogram, you cannot find the median exactly, but you can find which interval it is in. Add a cumulative frequency column (the running total).

Example. The quiz scores of 60 students:

The median is at position ${\displaystyle \frac{60+1}{2}}=30.5$, between the 30th and 31st values. The cumulative frequency is 28 up to 30, and reaches 46 at 40, so the 30th and 31st values are both in $30\le x<40$. The median is somewhere in that interval.

The same idea locates the quartiles. The first quartile is the median of the lower 30 values, at position $15.5$; the cumulative frequency passes 15 in the third row, so $Q_1$ is in $20\le x<30$.

The mode is the most frequent value (or, in a histogram, the tallest bar). It is the only measure of center that also works for categorical data.

Measures of Spread

Two classes can have the same mean and still be very different: in one everybody scored close to 70, in the other the scores run from 30 to 100. A measure of spread (variability) tells these apart.

• Range $=max-min$. Easy, but it depends on only two values, the most extreme ones, so it is not resistant.
• Interquartile range $\text{IQR}=Q_3-Q_1$: the range of the middle half of the data. It is resistant.
• Standard deviation $s_x$: roughly the typical distance of the values from the mean.

The standard deviation. For each value find its deviation from the mean, $x_i-\overline{x}$. The deviations always add up to 0 (positive and negative ones cancel), so we square them, average them (dividing by $n-1$) and take the square root:

$$s_x^2=\frac{\sum (x_i-\overline{x}{)}^2}{n-1},s_x=\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{n-1}}.$$

$s_x^2$ is the variance. For a whole population we divide by $N$ instead and write $\sigma$.

Example. Quiz scores $6,8,9,10,12$ with $\overline{x}=9$:

$$s_x^2=\frac{20}{5-1}=5,s_x=\sqrt{5}\approx 2.24.$$

Interpretation sentence (learn it): "The quiz scores typically vary by about 2.24 points from the mean of 9 points."

Facts about $s_x$:

• $s_x\ge 0$, and $s_x=0$ only when all values are equal.
• It has the same units as the data (the variance has squared units).
• Like the mean, it is not resistant: one extreme value can make it much larger.
• On the exam use $s_x$ (sample) unless the data is a whole population.

1-Var Stats
 x̄=9
 Σx=45
 Σx²=425
 Sx=2.236067977
 σx=2
 n=5
 minX=6
 Q1=7
 Med=9
 Q3=11
 maxX=12

Which pair to report? Mean with standard deviation for roughly symmetric data without outliers; median with IQR for skewed data or data with outliers.

Quartiles, the Five-Number Summary and Boxplots

The quartiles split the sorted data into four parts with (about) a quarter of the values each:

• $Q_1$ is the median of the lower half of the data,
• the median is $Q_2$,
• $Q_3$ is the median of the upper half.

When $n$ is odd, the median itself is left out of both halves. The five-number summary is

$$\text{min},Q_1,\text{median},Q_3,\text{max}.$$

Outliers: the $1.5\cdot \text{IQR}$ rule. A value is an outlier if it lies more than $1.5\cdot \text{IQR}$ beyond the quartiles, that is, outside the fences

$$Q_1-1.5\cdot \text{IQR}\text{and}{Q}_3+1.5\cdot \text{IQR}.$$

For roughly normal data a second rule is also used: a value more than 2 standard deviations from the mean is unusual. The two rules can disagree; say which one you use.

A boxplot draws the five-number summary: a box from $Q_1$ to $Q_3$ with a line at the median, and whiskers out to the smallest and largest values that are not outliers. Outliers are drawn as separate points. (This is a modified boxplot, the one the AP exam expects.)

Example. The numbers of books read last year by 15 students:

$$2,5,7,8,9,10,11,12,12,13,14,15,16,18,35.$$

$n=15$, so the median is the 8th value, 12. The lower half is $2,5,7,8,9,10,11$, so $Q_1=8$; the upper half is $12,13,14,15,16,18,35$, so $Q_3=15$. Then $\text{IQR}=7$, $1.5\cdot 7=10.5$, and the fences are $8-10.5=-2.5$ and $15+10.5=25.5$. The value 35 is above $25.5$, so it is an outlier; the upper whisker stops at 18.

What a boxplot shows and hides:

• It shows center (median), spread (IQR and range), skewness (a longer whisker or a longer half of the box points to the skew) and outliers.
• It does not show gaps, clusters or the number of peaks: a bimodal distribution can have a perfectly ordinary-looking boxplot. It also does not show how many values there are.
• Each of the four parts holds about 25% of the values, however long or short it is drawn. A long part means the values there are spread out, not that there are more of them.

Percentiles and Cumulative Relative Frequency Plots

The $p$th percentile is the value with $p\mathrm{\%}$ of the data at or below it. So $Q_1$ is the 25th percentile, the median the 50th and $Q_3$ the 75th. A student "in the 90th percentile" scored as well as or better than 90% of the students, which says nothing about getting 90% of the points.

A cumulative relative frequency plot (ogive) shows, for each value, the proportion of data at or below it. It always rises from 0 to 1. Read it in two directions:

• from a value up to the curve and across: the percentile of that value;
• from a proportion across to the curve and down: the value at that percentile (0.5 gives the median, 0.25 and 0.75 the quartiles).

Where the curve is steep, many values are packed into a short interval; where it is flat, there are few values.

Adding and Multiplying: How Summary Statistics Change

If we add the same constant $a$ to every value, or multiply every value by a positive constant $b$ (changing units, adding a bonus):

Adding a constant slides the whole distribution along the number line without stretching it, so the spread stays the same.

Example. Daily temperatures have mean ${20}^{\circ }$C and standard deviation $5^{\circ }$C. In Fahrenheit, $F=32+1.8C$, so the mean is $32+1.8\cdot 20={68}^{\circ }$F and the standard deviation is $1.8\cdot 5=9^{\circ }$F (the $+32$ does not affect it).

$z$-scores do not change when the units change: converting heights from inches to centimetres changes every value, the mean and the standard deviation by the same factor, so every $z$-score stays the same.

Comparing Distributions

To compare two or more groups, draw them on the same scale (side-by-side boxplots, back-to-back stemplots, dotplots stacked over one axis) and compare shape, center, spread and outliers, using comparative words: "the median of A is greater than the median of B", not just "A's median is 75 and B's is 70".

Example (a model answer). "The median score of class A (75) is greater than the median score of class B (70). The scores of class B are more variable: its IQR ($86-60=26$ points) is twice that of class A ($81-68=13$ points), and its range (53) is larger than A's (42). Class A is roughly symmetric, while class B is skewed right (the upper half of its box, from 70 to 86, is longer than the lower half). Neither class has outliers."

The Normal Distribution

Many quantitative variables (heights, test scores, measurement errors) have a symmetric, bell-shaped distribution. We model them with a normal curve. A normal distribution is fixed by two numbers, its mean $\mu$ and its standard deviation $\sigma$, and we write $X\sim N(\mu ,\sigma )$.

• The curve is symmetric about $\mu$, so mean = median = mode.
• The total area under the curve is 1, and the area over an interval is the proportion (probability) of values in that interval.
• The tails go on forever, but very little area is more than 3 standard deviations from the mean.

The 68–95–99.7 rule. In a normal distribution about 68% of the values are within one standard deviation of the mean, about 95% within two and about 99.7% within three. These are approximations: the exact value for two standard deviations is $0.9545$.

$z$-scores. The $z$-score of a value $x$ says how many standard deviations it lies from the mean:

$$z=\frac{x-\mu }{\sigma }.$$

Standardizing turns any normal distribution into the standard normal distribution $N(0,1)$, so a single table (or one calculator function) is enough for all of them.

Example. In the standard normal distribution, $P(z<1.5)=0.9332$ (from the table of standard normal probabilities, which gives the area to the left of $z$). Since $z$ is continuous, $P(z\le 1.5)$ is the same number.

Finding an area. On the TI-83/84, $\text{normalcdf}(\text{lower},\text{upper},\mu ,\sigma )$ gives the area between two values. For "no lower bound" use a very large negative number, such as $-1000$ or $-{10}^{99}$.

normalcdf(-1E99,65,68,3)
           .1586552596
normalcdf(65,70,68,3)
           .5888522522
normalcdf(70,1E99,68,3)
           .2524925375

Example (heights). The heights $X$ of the students of a college are approximately $N(68,3)$, in inches.

• $P(X<65)$: $z={\displaystyle \frac{65-68}{3}}=-1$, so $P(X<65)=P(z<-1)=0.1587$. On the calculator: $\text{normalcdf}(-1000,65,68,3)$.
• $P(X>65)=1-0.1587=0.8413$.
• $P(65<X<70)$: the $z$-scores are $-1$ and ${\displaystyle \frac{70-68}{3}}\approx 0.67$, so the area is $0.7486-0.1587=0.5899$. The calculator gives $\text{normalcdf}(65,70,68,3)=0.5889$; the small difference comes from rounding $z$ to $0.67$ for the table.
• $P(70<X<75)$: the area to the left of 75 ($z\approx 2.33$) minus the area to the left of 70, $0.9901-0.7486=0.2415$.

Finding a value from an area. Sometimes the area is given and the value is unknown. Then read the table from the inside out, or use $\text{invNorm}(\text{area to the left},\mu ,\sigma )$, and solve the $z$-score equation for $x$.

Example (top 15%). SAT scores are approximately $N(500,100)$. What is the lowest score that is still in the top 15%? The top 15% means 85% of the scores are to the left, and the $z$-score with area $0.85$ to its left is $z\approx 1.04$. So

$$\frac{x-500}{100}=1.04⟹x=500+1.04\cdot 100=604.$$

On the calculator, $\text{invNorm}(0.85,500,100)\approx 604$. Even when you use the calculator, write down the $z$-score equation: the AP rubrics expect it.

invNorm(.85,500,100,LEFT)
           603.643338
invNorm(.15,500,100,RIGHT)
           603.643338
invNorm(.90,500,100,CENTER)
     {335.5146373 664.4853627}

Example (unknown mean). A normal variable has $\sigma =12$ and $P(X>50)=0.90$. Then 10% of the area is to the left of 50, so $z=-1.28$ and

$$-1.28=\frac{50-\mu }{12}⟹\mu =50+1.28\cdot 12\approx 65.4.$$

Comparing with $z$-scores. $z$-scores let us compare values measured on different scales. Ada scored 680 on a test with mean 600 and standard deviation 50; Ben scored 29 on a test with mean 24 and standard deviation 4. Then

$$z_{\text{Ada}}=\frac{680-600}{50}=1.6,z_{\text{Ben}}=\frac{29-24}{4}=1.25,$$

so Ada did better relative to the other test takers: she was 1.6 standard deviations above the mean, Ben only 1.25. (This comparison does not need normality; turning a $z$-score into a percentile does.)

Is the data approximately normal? Before using normal calculations on real data, check that the data is roughly normal:

• a histogram or dotplot is unimodal and roughly symmetric, without outliers;
• the 68–95–99.7 rule roughly holds for the data;
• a normal probability plot (data against their expected $z$-scores, on the calculator the last STAT PLOT type) is close to a straight line. A curved plot means skewness; isolated points at the ends are outliers.

Unit 1 Practice 1

Fifteen multiple-choice questions in the style of the AP exam, then free-response questions with full solutions. The questions are adapted (new numbers and contexts) from Barron's AP Q&A Statistics and AP Statistics Premium, 5 Steps to a 5: AP Statistics and released AP exams; each free-response question names the question it is modelled on.

Multiple-Choice Questions

Free-Response Questions

Write complete answers: show the method, give the numbers, and finish with a sentence in context. On the AP exam each question is scored E (essentially correct), P (partially correct) or I (incorrect) part by part.

FRQ 1 (outlier rules). Modelled on AP Statistics 2021, FRQ 1. A library recorded how many days each of a random sample of 40 borrowed books stayed out before being returned.

(a) Find the five-number summary of the number of days.

(b) (i) Using the $1.5\cdot \text{IQR}$ rule (method A), identify any outliers. (ii) The mean is 9.58 days and the standard deviation is 2.98 days. Using the rule "more than 2 standard deviations from the mean" (method B), identify any outliers.

(c) Explain why, for a distribution that is strongly skewed to the right, method A can identify more outliers than method B.

FRQ 2 (comparing distributions). Modelled on AP Statistics 2015, FRQ 1. Two large companies, A and B, both hired graduates as junior engineers five years ago at the same starting salary of 40 thousand euros. For 30 such engineers still working at each company, the current yearly salaries (thousands of euros) are summarised below.

(a) Write a few sentences comparing the distributions of salaries at the two companies.

(b) You are offered a junior job at both companies for 40 thousand euros. (i) Using the summaries, give one reason to choose company A. (ii) Give one reason to choose company B.

FRQ 3 (what a boxplot hides). Modelled on AP Statistics 2019, FRQ 1. The floor areas (m²) of the 20 apartments in a building are shown in the stemplot.

11 | 8
12 | 6
13 | 1 7
14 | 0 4 9
15 | 2 8
16 |
17 |
18 |
19 |
20 |
21 |
22 |
23 |
24 |
25 |
26 | 2
27 | 0 6
28 | 1 5 8
29 | 2 6
30 | 1 5
31 | 2
Key: 14 | 0 means 140 m²

(a) Describe the distribution of floor area.

(b) Find the five-number summary, decide whether there are outliers, and describe the boxplot.

(c) What feature of the distribution is clear from the stemplot but would be invisible in the boxplot?

FRQ 4 (normal calculations). Modelled on the free-response problems of Barron's AP Q&A Statistics, Chapter 5. The lengths of adult trout in a lake are approximately normal with mean 32 cm and standard deviation 4 cm.

(a) What proportion of adult trout are longer than 38 cm?

(b) Fishing rules say that the shortest 10% of trout must be released. Below what length must a trout be released?

(c) In a second lake, adult perch have lengths that are approximately normal with mean 48 cm and standard deviation 5 cm. Which is more unusually long for its species: a 38 cm trout or a 55 cm perch? Justify.

Unit 1 Practice 2

A second set of fifteen multiple-choice questions and three free-response questions on Unit 1.

Multiple-Choice Questions

Free-Response Questions

FRQ 5 (reading grouped data). Modelled on AP Statistics 2018, FRQ 5. The numbers of years that teachers at two large schools have been teaching are summarised below (40 teachers at each school).

(Each interval includes its left end, for example $5\le x<10$.)

(a) Describe the shape of each distribution.

(b) In which interval is the median of each school? Which school has the greater median?

(c) For school A, is the mean greater than, less than or about equal to the median? Explain.

FRQ 6 (changing units). Modelled on the free-response problems of Barron's AP Q&A Statistics, Chapter 5. A weather station recorded the daily maximum temperature for 60 days. In degrees Celsius: mean 18, median 17, standard deviation 4, IQR 5, minimum 9.

(a) Find the mean, median, standard deviation, IQR and minimum in degrees Fahrenheit, using $F=32+1.8C$.

(b) Find the $z$-score of a day with a maximum of 26 °C. What is the $z$-score of the same day in Fahrenheit?

(c) The station later found that its thermometer read 1.5 °C too low every day. Which of the summaries in degrees Celsius change after the correction, and how?

FRQ 7 (normal model for a filling machine). Modelled on AP Statistics 2022, FRQ 3. A machine fills cereal boxes labelled 500 g. The amount of cereal is approximately normal with mean 510 g and standard deviation 6 g.

(a) What proportion of boxes contain less than the labelled 500 g?

(b) The company wants only 1% of boxes under 500 g. If the standard deviation stays 6 g, what mean should the machine be set to?

(c) Instead, the company keeps the mean at 510 g and improves the machine. What standard deviation would give only 1% of boxes under 500 g?

Unit 1 Practice 3

A third set of fifteen multiple-choice questions and three free-response questions on Unit 1.

Multiple-Choice Questions

Free-Response Questions

FRQ 8 (a stemplot). Modelled on AP Statistics 2013, FRQ 1(a). The noise levels (decibels) in 20 classrooms during a lesson are shown in the stemplot.

 5 | 2 4 7
 6 | 0 1 3 5 5 8
 7 | 0 2 2 3 6 9
 8 | 1 4 7
 9 | 0
10 |
11 | 3
Key: 6 | 3 means 63 dB

(a) Noise above 85 dB is considered harmful. What proportion of the classrooms had harmful noise levels?

(b) Find the five-number summary and determine whether there are outliers.

(c) The mean noise level is 72.1 dB. Explain why the mean is greater than the median.

FRQ 9 (choosing between two services). Modelled on the comparison problems in 5 Steps to a 5, Chapter 5. The delivery times (minutes) of 25 orders from each of two pizza services were recorded.

(a) Are there outliers in either data set? Justify.

(b) Compare the two distributions.

(c) You need a pizza delivered within 35 minutes for a party. Which service would you choose? Use the summaries to justify.

FRQ 10 (normal model and relative position). Modelled on AP Statistics 2017, FRQ 3(a). The diameters of apples from orchard A are approximately normal with mean 75 mm and standard deviation 4 mm; those from orchard B are approximately normal with mean 70 mm and standard deviation 5 mm.

(a) What proportion of orchard A's apples have a diameter over 80 mm?

(b) Find the 95th percentile of the diameters of orchard A's apples.

(c) An apple has a diameter of 78 mm. Would this apple be more unusual if it came from orchard A or from orchard B? Justify.

Unit 2: Exploring Two-Variable Data

Two Categorical Variables

When both variables are categorical, the data is summarised in a two-way table (contingency table). The table below gives the country of origin and blood type of the 500 people at a conference; we will use it again in Unit 4.

• The numbers in the Total row and column are the marginal frequencies; the numbers inside the table are the joint frequencies.
• Dividing by the grand total (500) gives relative frequencies.
• The marginal distribution of one variable uses only the totals: for blood type, $\frac{225}{500}=45\mathrm{\%}$ O, $34\mathrm{\%}$ A, $16\mathrm{\%}$ B and $5\mathrm{\%}$ AB.
• A conditional distribution looks at one variable within one group of the other: we divide by the row (or column) total, not by 500.

Conditional Distributions and Segmented Bar Graphs

To compare a categorical variable across groups we use conditional relative frequencies: we treat each group on its own. In Country 1, 150 of the 250 people have type O, a conditional relative frequency of ${\displaystyle \frac{150}{250}}=60\mathrm{\%}$. Doing this for every country gives the conditional distribution of blood type by country:

Comparing raw counts would be misleading here: Country 1 has more people of every type simply because it is the largest group. Always compare percentages within groups.

• A side-by-side bar graph draws, for each country, one bar per blood type (heights are the conditional percentages).
• A segmented bar graph draws one bar of the same length for each group, divided into segments by these percentages.
• A mosaic plot does the same, but makes the width of each bar proportional to the size of the group. So it also shows how many people are in each group, and the area of a rectangle is proportional to the number of people in that cell.

Association. Two categorical variables are associated if the conditional distributions of one variable differ from group to group of the other. Here they clearly differ: type O is 60% of Country 1 but under a third of Countries 2 and 3, and type B is far more common in Country 3. So blood type and country are associated; knowing a person's country changes what we expect their blood type to be. If there were no association, all the bars would be split the same way. (In Unit 4 the same idea appears as independence of events.) The mosaic plot also shows that the number of type A people is about the same in Countries 1 and 2 (73 and 75), since those two rectangles have about the same area.

An association in a two-way table, like a correlation, does not show cause and effect.

Scatterplots

In Unit 1 we looked at one variable at a time. Now every individual gives us two quantitative values, a pair $(x,y)$, and we want to know how they are related. The graph for this is a scatterplot: one point per individual.

• The explanatory variable $x$ (the one we think explains or predicts) goes on the horizontal axis.
• The response variable $y$ (the one we measure as the outcome) goes on the vertical axis.

Describe a scatterplot with direction, form and strength (plus unusual points), in context:

• Positive association: above-average $x$ goes with above-average $y$; the cloud rises from left to right.
• Negative association: above-average $x$ goes with below-average $y$; the cloud falls from left to right.
• Form: linear, or curved (for example U-shaped).
• Strength: how tightly the points follow the form.

On the exam a quick hand sketch of the scatterplot is enough, but the axes must be labelled and scaled.

The Correlation Coefficient $r$

The correlation $r$ measures the direction and the strength of a linear relationship:

$$r=\frac{1}{n-1}\sum \left(\frac{x_i-\overline{x}}{s_x}\right)\left(\frac{y_i-\overline{y}}{s_y}\right)=\frac{1}{n-1}\sum z_{x_i}{z}_{y_i}.$$

So $r$ is (almost) the average of the products of the $z$-scores. When a point is above average in both $x$ and $y$, or below average in both, the product is positive; this is why positive association gives a positive $r$. You will not compute it by hand; the formula is on the formula sheet and the calculator gives $r$.

Properties of $r$:

• $-1\le r\le 1$. At $r=\pm 1$ the points lie exactly on a line.
• The sign gives the direction: $r>0$ positive, $r<0$ negative.
• A rough guide to strength:

 -1        -0.8        -0.5         0         0.5        0.8         1
  |  strong  | moderate  |   weak   |   weak   | moderate  |  strong  |

• $r=0$ means no linear relationship. There may still be a strong curved one: points on a U-shape can have $r$ close to 0.
• Swapping $x$ and $y$ does not change $r$.
• Changing units (hours to minutes, inches to cm) does not change $r$.
• $r$ is not resistant. One point far away from the rest can change $r$ a lot, even turn a negative association into a positive-looking one. Always look at the scatterplot, not just at $r$.

Correlation is not slope. A steep line with points widely scattered around it has a large slope but a small $r$; a nearly flat line with points tightly on it has a small slope but $r$ close to 1. $r$ measures how close the points are to a line, not how steep the line is.

Correlation Does Not Imply Causation

A strong correlation between two variables does not show that one causes the other. Often a third variable (a lurking or confounding variable) drives both. Ice cream sales and drownings rise together, but hot weather causes both. Only a well-designed experiment (Unit 3) can show cause and effect.

The Least-Squares Regression Line (LSRL)

If the scatterplot looks linear, we fit a line to predict $y$ from $x$. Here, unlike for $r$, it matters which variable is $x$ and which is $y$.

$$\hat{y}=a+bx$$

• $\hat{y}$ ("y-hat") is the predicted value of $y$.
• $b$ is the slope, $a$ is the $y$-intercept (the predicted $y$ when $x=0$).

For every point, the vertical distance from the point to the line is its residual:

$$\text{residual}=y-\hat{y}=\text{actual}-\text{predicted}.$$

For the regression line the residuals always add up to zero, $\sum (y-\hat{y})=0$, so their sum cannot tell us which line is best. Instead we square them: the least-squares regression line (the "line of best fit") is the line that makes the sum of squared residuals $\sum (y-\hat{y}{)}^2$ as small as possible.

Formulas.

$$b=r\frac{s_y}{s_x},a=\overline{y}-b\overline{x}.$$

The second formula says that the LSRL always passes through the point $(\overline{x},\overline{y})$. The first says that the slope and $r$ always have the same sign.

Example. For hours studied ($x$) and test score ($y$): $\overline{x}=3$, $\overline{y}=80$, $s_x=2$, $s_y=10$, $r=0.8$. Then

$$b=0.8\cdot \frac{10}{2}=4,a=80-4\cdot 3=68,\widehat{\text{score}}=68+4(\text{hours}).$$

• Slope in context: for each additional hour of study, the predicted score increases by 4 points.
• Intercept in context: a student who does not study at all is predicted to score 68.
• Prediction: a student who studies 2.5 hours is predicted to score $68+4(2.5)=78$. This is a prediction, not a guarantee.

Always define the variables in context. "$\hat{y}=68+4x$" with no explanation of $x$ and $y$ does not get full credit.

Reading computer output. Regression software prints a table like this:

Predictor      Coef    SE Coef      T       P
Constant     68.000      ...       ...     ...
Hours         4.000      ...       ...     ...

S = ...      R-Sq = 64.0%

The number next to Constant is the intercept $a$, the number next to the variable name is the slope $b$, and R-Sq is $r^2$. The other columns belong to inference (Unit 9).

LinReg
 y=a+bx
 a=68
 b=4
 r²=.64
 r=.8

Residuals and Residual Plots

• A positive residual means the point is above the line: the actual value is higher than predicted (the line underestimated).
• A negative residual means the point is below the line: the line overestimated.

Example. With $\widehat{\text{score}}=68+4(\text{hours})$, a student who studied 4 hours and scored 82 has predicted score $68+16=84$ and residual $82-84=-2$. The line overestimated this student by 2 points.

A residual plot puts the residuals on the vertical axis against $x$ (or against the predicted values $\hat{y}$). It is the main tool for checking whether a line is a good model. A line is appropriate when the residuals are

1. randomly scattered with no pattern, and
2. spread around 0.

Patterns that warn you:

• a curve (U-shape or upside-down U): the relationship is not linear;
• a fan that opens out: the predictions get less reliable as $x$ grows;
• residuals all positive at both ends and negative in the middle (or the reverse): again, a curve.

Example. Population growth is usually not linear but exponential. A line fitted to it misses at both ends, and the residual plot shows a clear U-shape.

Interpolation and Extrapolation

• Interpolation: predicting for an $x$-value inside the range of the data. Fairly safe if the line fits well.
• Extrapolation: predicting for an $x$-value outside that range. Dangerous: nothing guarantees the linear pattern continues. A line that predicts children's height from age between 18 and 29 months will give absurd heights if you plug in an age of 12 years.

The Coefficient of Determination $r^2$

Without any $x$ information, the best prediction for $y$ is simply $\overline{y}$. Compare the total error of the two ways of predicting:

$$\text{SST}=\sum (y-\overline{y}{)}^2\text{(error using the mean)},\text{SSE}=\sum (y-\hat{y}{)}^2\text{(error using the line)}.$$

$$r^2=\frac{\text{SST}-\text{SSE}}{\text{SST}}$$

is the proportion of the variation in $y$ that is explained by the linear relationship with $x$. It is exactly the square of the correlation $r$.

Interpretation sentence (learn it): "$r^2=0.64$ means that 64% of the variation in test score is explained by the linear relationship with hours studied."

Going back from $r^2$ to $r$: $r=\pm \sqrt{r^2}$, and the sign is the sign of the slope. If $r^2=0.64$ and the slope is positive, $r=0.8$; if the slope is negative, $r=-0.8$.

The Standard Deviation of the Residuals $s$

$r^2$ tells us what fraction of the variation the line explains; $s$ tells us how big the prediction errors typically are, in the units of $y$:

$$s=\sqrt{\frac{\sum (y_i-{\hat{y}}_i{)}^2}{n-2}}=\sqrt{\frac{\text{SSE}}{n-2}}.$$

In computer output it is the number labelled S.

Interpretation sentence (learn it): "$s=6.2$ means that the actual test scores typically differ from the scores predicted by the line by about 6.2 points."

A good model has a small $s$ compared with the size of the $y$-values, a high $r^2$ and a residual plot without a pattern. Use all three; none of them alone proves that a line is appropriate.

Example (reading output). Regression of the time (seconds) to finish checkout on the number of customers already in line gives

Predictor        Coef    SE Coef      T       P
Constant        70.20     ...        ...     ...
Customers       165.80    ...        ...     ...

S = 185.0    R-Sq = 71.4%

• Slope: for each additional customer in line, the predicted checkout time increases by about 166 seconds.
• Intercept: a customer who arrives at an empty line is predicted to need about 70 seconds.
• $r^2$: 71.4% of the variation in checkout time is explained by the linear relationship with the number of customers in line.
• $s$: actual checkout times typically differ from the predicted times by about 185 seconds.
• $r=+\sqrt{0.714}\approx 0.845$ (positive, because the slope is positive).

Slope in standard deviations. Since $b=r{s}_y/s_x$, an increase of one standard deviation in $x$ predicts an increase of only $r$ standard deviations in $y$. With $|r|<1$, predictions are pulled toward the mean of $y$: this is regression to the mean. Very tall parents tend to have tall children, but on average not as tall as themselves.

Regressing $x$ on $y$ is a different line. The LSRL of $y$ on $x$ minimises vertical distances. To predict $x$ from $y$ we must compute a new regression (with $x$ and $y$ swapped); solving $\hat{y}=a+bx$ for $x$ does not give it, except when $r=\pm 1$.

Outliers and Influential Points

• An outlier in regression is a point with a large residual: far from the line, outside the overall pattern.
• An influential point is one that changes the regression a lot (slope, intercept, $r$) when you remove it. Points that are far out in the $x$-direction (high leverage) are often influential: they pull the line toward themselves, so they may even have a small residual.

To decide, imagine removing the point: if the line or $r$ changes a lot, the point is influential.

Transformations to Achieve Linearity

If the scatterplot or the residual plot shows a curve, a line is not a suitable model for the original data. Sometimes we can transform a variable so that the relationship becomes linear, fit a line to the transformed data, and then transform back.

The typical case is exponential growth (bacteria, populations, money with interest): take the natural logarithm of $y$.

1. Compute $\ln y$ for every point (on the calculator: L3 = ln(L2)).
2. Check that $x$ versus $\ln y$ looks linear and its residual plot shows no pattern.
3. Fit the line: $\widehat{\ln y}=a+bx$ (LinReg(a+bx) L1, L3).
4. Predict, then back-transform: the line gives $\ln \hat{y}$, so $\hat{y}=e^{a+bx}$.

Example. The model is $\widehat{\ln y}=0.5+0.4x$. For $x=5$: $\ln \hat{y}=0.5+0.4\cdot 5=2.5$, so $\hat{y}=e^{2.5}\approx 12.18$. Forgetting the last step (answering $2.5$) is the classic mistake.

On the exam you will rarely transform raw data yourself; more often you get the output for transformed data and are asked whether the transformation helped (compare the scatterplots, residual plots and $r^2$ values).

Unit 2 Practice 1

Fifteen multiple-choice questions and four free-response questions on two-variable data, adapted from Barron's AP Q&A Statistics, 5 Steps to a 5 and released AP exams.

Multiple-Choice Questions

Free-Response Questions

FRQ 1 (interpreting regression output). Modelled on AP Statistics 2022, FRQ 1, and 2018, FRQ 1. A biologist measured the length (cm) and mass (g) of 12 lizards of one species, whose lengths ranged from 10 to 24 cm. The scatterplot showed a strong, positive, linear relationship, and regression software gave:

Predictor      Coef      SE Coef      T        P
Constant     -18.40       ...        ...      ...
Length         1.92       ...        ...      ...

S = 3.10     R-Sq = 84.0%

(a) Write the equation of the least-squares regression line, defining the variables, and interpret the slope in context.

(b) Interpret $r^2$ and find the correlation $r$.

(c) One lizard is 18 cm long and has a mass of 12 g. Find its residual. Does the line overestimate or underestimate its mass?

(d) Would it be reasonable to use the line to predict the mass of a 45 cm lizard of a related species? Explain.

FRQ 2 (two categorical variables). Modelled on AP Statistics 2024, FRQ 2. Students at two schools were asked which kind of book they read most. School A surveyed 200 students: 45% said fiction, 35% nonfiction and 20% comics. School B surveyed 600 students: 30% fiction, 30% nonfiction and 40% comics.

(a) Describe how segmented bar graphs comparing the two schools would look.

(b) The librarian of school A says: "More of our students read fiction than at school B." Is this correct for the students surveyed? Explain.

(c) Is there an association between school and preferred kind of book for these students? Justify.

FRQ 3 (the line from summary statistics). Modelled on the free-response problems of Barron's AP Q&A Statistics, Chapter 5. For a random sample of students, the hours of sleep per night have mean 7.2 and standard deviation 0.9, and their grade point averages (GPA) have mean 3.1 and standard deviation 0.45. The correlation is $r=0.6$.

(a) Find the least-squares regression line for predicting GPA from hours of sleep.

(b) Predict the GPA of a student who sleeps 8 hours.

(c) A student sleeps one standard deviation more than the mean. How many standard deviations above the mean GPA is the predicted GPA? What is this effect called?

(d) A counsellor wants to predict hours of sleep from GPA and uses the line from (a), solved for sleep. Find the correct line and explain why the two differ.

FRQ 4 (choosing a model). Modelled on AP Statistics 2014, FRQ 6, and the transformation problems in 5 Steps to a 5. The number of users (thousands) of a new app was recorded for 12 months. Two models were fitted, with month $x=1,2,\dots ,12$:

• Model I: $\hat{y}=-4.1+2.3x$, with $r^2=0.91$; the residual plot is clearly U-shaped.
• Model II: $\widehat{\ln y}=0.85+0.21x$, with $r^2=0.99$; the residual plot shows random scatter around 0.

(a) Which model is more appropriate? Give two reasons.

(b) Use the better model to predict the number of users in month 10.

(c) By what factor does Model II predict the number of users grows each month?

Unit 2 Practice 2

A second set of fifteen multiple-choice questions and three free-response questions on Unit 2.

Multiple-Choice Questions

Free-Response Questions

FRQ 5 (predicting height). Modelled on AP Statistics 2015, FRQ 5. For a random sample of 30 adults, forearm length (cm) and height (cm) were measured. The scatterplot is linear, and the output is:

Predictor      Coef
Constant      52.30
Forearm        4.60

S = 4.80     R-Sq = 72.0%

(a) Predict the height of an adult with a forearm 26 cm long.

(b) One adult in the sample has a 26 cm forearm and is 180 cm tall. Find and interpret the residual.

(c) Interpret $s=4.80$ in context.

(d) If the heights had been measured in inches instead (1 inch = 2.54 cm), how would the slope, $r^2$ and $r$ change?

FRQ 6 (an influential point). Modelled on AP Statistics 2012, FRQ 1. A magazine rated 14 pairs of headphones on a 0–100 quality scale and recorded their prices (euros). With all 14 pairs, the regression line is $\widehat{\text{rating}}=22+0.12(\text{price})$ with $r=0.62$. One pair is far more expensive than the others (400 euros) and has a rating of only 50. Without it, the line is $\widehat{\text{rating}}=15+0.30(\text{price})$ with $r=0.85$.

(a) Describe the effect of the expensive pair on the slope and on the correlation.

(b) What kind of point is the expensive pair? Explain.

(c) Interpret the slope of the line without the expensive pair.

FRQ 7 (association in a two-way table). Modelled on AP Statistics 2014, FRQ 1. A school asked 240 students how they usually get to school.

(a) What proportion of grade 12 students come by car?

(b) Give the conditional distributions of travel mode for each grade, and describe how segmented bar graphs of them would look.

(c) Is there an association between grade and travel mode for these students? Explain.

Unit 2 Practice 3

A third set of fifteen multiple-choice questions and three free-response questions on Unit 2.

Multiple-Choice Questions

Free-Response Questions

FRQ 8 (regression output). Modelled on AP Statistics 2017, FRQ 1. Foresters measured the trunk diameter (cm) and the age (years) of 25 trees of one species. The output for predicting age from diameter is:

Predictor      Coef
Constant       3.20
Diameter       2.15

S = 6.40     R-Sq = 79.0%

(a) Interpret the slope in context.

(b) A tree with a diameter of 30 cm is 70 years old. Find its residual. Is the tree older or younger than predicted?

(c) Find the correlation between diameter and age.

(d) A forester says: "The intercept tells us that a tree with zero diameter is 3.2 years old." Comment.

FRQ 9 (the line by hand). Modelled on the two-variable free-response problems of Barron's AP Q&A Statistics, Chapter 5. Five students recorded hours of practice $x$ and the number of correct answers $y$ on a short quiz: $(1,2)$, $(2,4)$, $(3,5)$, $(4,4)$, $(5,5)$.

(a) Find the least-squares regression line, using $b={\displaystyle \frac{\sum (x-\overline{x})(y-\overline{y})}{\sum (x-\overline{x}{)}^2}}$ and $a=\overline{y}-b\overline{x}$.

(b) Find all five residuals and check that they add up to 0.

(c) Find $r^2$ and $s$.