6
Jul 17

## Running simple regression in Stata

Running simple regression in Stata is, well, simple. It's just a matter of a couple of clicks. Try to make it a small research.

1. Obtain descriptive statistics for your data (Statistics > Summaries, tables, and tests > Summary and descriptive statistics > Summary statistics). Look at all that stuff you studied in introductory statistics: units of measurement, means, minimums, maximums, and correlations. Knowing the units of measurement will be important for interpreting regression results; correlations will predict signs of coefficients, etc. In your report, don't just mechanically repeat all those measures; try to find and discuss something interesting.
2. Visualize your data (Graphics > Twoway graph). On the graph you can observe outliers and discern possible nonlinearity.
3. After running regression, report the estimated equation. It is called a fitted line and in our case looks like this: Earnings = -13.93+2.45*S (use descriptive names and not abstract X,Y). To see if the coefficient of S is significant, look at its p-value, which is smaller than 0.001. This tells us that at all levels of significance larger than or equal to 0.001 the null that the coefficient of S is significant is rejected. This follows from the definition of p-value. Nobody cares about significance of the intercept. Report also the p-value of the F statistic. It characterizes significance of all nontrivial regressors and is important in case of multiple regression. The last statistic to report is R squared.
4. Think about possible variations of the model. Could the dependence of Earnings on S be nonlinear? What other determinants of Earnings would you suggest from among the variables in Dougherty's file?

Figure 1. Looking at data. For data, we use a scatterplot.

Figure 2. Running regression (Statistics > Linear models and related > Linear regression)

29
Jun 17

## Introduction to Stata

Introduction to Stata: Stata interface, how to use Stata Help, how to use Data Editor and how to graph data. Important details to remember:

1. In any program, the first thing to use is Help. I learned everything from Help and never took any programming courses.
2. The number of observations for all variables in one data file must be the same. This can be a problem if, for example, you want to see out-of-sample predictions.
3. In Data Editor, numeric variables are displayed in black and strings are displayed in red.
4. The name of the hidden variable that counts observations is _n
5. If you have several definitions of graphs in two-way graphs menu, they will be graphed together or separately, depending on what is enabled/disabled.

See details in videos. Sorry about the background noise!

Video 1. Stata interface. The windows introduced: Results, Command, Variables, Properties, Review and Viewer.

Video 2. Using Stata Help. Help can be used through the Review window or in a separate pdf viewer. Eviews Help is much easier to understand.

Video 3. Using Data Editor. How to open and view variables, the visual difference between numeric variables and string variables. The lengths of all variables in the same file must be the same.

Video 4. Graphing data. To graph a variable, you need to define its graph and then display it. It is possible to display more than one variable on the same chart.

21
Feb 17

## The pearls of AP Statistics 37

### Confidence interval: attach probability or not attach?

I am reading "5 Steps to a 5 AP Statistics, 2010-2011 Edition" by Duane Hinders (sorry, I don't have the latest edition). The tip at the bottom of p.200 says:

For the exam, be VERY, VERY clear on the discussion above. Many students
seem to think that we can attach a probability to our interpretation of a confidence
interval. We cannot.

This is one of those misconceptions that travel from book to book. Below I show how it may have arisen.

### Confidence interval derivation

The intuition behind the confidence interval and the confidence interval derivation using z score have been given here. To make the discussion close to Duane Hinders, I show the confidence interval derivation using the t statistic. Let $X_1,...,X_n$ be a sample of independent observations from a normal population, $\mu$ the population mean and $s$ the standard error. Skipping the intuition, let's go directly to the t statistic

(1) $t=\frac{\bar{X}-\mu}{s/\sqrt{n}}$.

At the 95% confidence level, from statistical tables find the critical value $t_{cr,0.95}$ of the t statistic such that

$P(-t_{cr,0.95}

Plug here (1) to get

(2) $P(-t_{cr,0.95}<\frac{\bar{X}-\mu}{s/\sqrt{n}}

Using equivalent transformations of inequalities (multiplying them by $s/\sqrt{n}$ and adding $\mu$ to all sides) we rewrite (2) as

(3) $P(\mu-t_{cr,0.95}\frac{s}{\sqrt{n}}<\bar{X}<\mu+t_{cr,0.95}\frac{s}{\sqrt{n}})=0.95.$

Thus, we have proved

Statement 1. The interval $\mu\pm t_{cr,0.95}\frac{s}{\sqrt{n}}$ contains the values of the sample mean with probability 95%.

The left-side inequality in (3) is equivalent to $\mu<\bar{X}+t_{cr,0.95}\frac{s}{\sqrt{n}}$ and the right-side one is equivalent to $\bar{X}-t_{cr,0.95}\frac{s}{\sqrt{n}}<\mu$. Combining these two inequalities, we see that (3) can be equivalently written as

(4) $P(\bar{X}-t_{cr,0.95}\frac{s}{\sqrt{n}}<\mu<\bar{X}+t_{cr,0.95}\frac{s}{\sqrt{n}})=0.95.$

So, we have

Statement 2. The interval $\bar{X}\pm t_{cr,0.95}\frac{s}{\sqrt{n}}$ contains the population mean with probability 95%.

### Source of the misconception

In (3), the variable in the middle ($\bar{X}$) is random, and the statement that it belongs to some interval is naturally probabilistic. People not familiar with the above derivation don't understand how a statement that the population mean (which is a constant) belongs to some interval can be probabilistic. It's the interval ends that are random in (4) (the sample mean and standard error are both random), that's why there is probability! Statements 1 and 2 are equivalent!

My colleague Aidan Islyami mentioned that we should distinguish estimates from estimators.

In all statistical derivations random variables are ex-ante (before the event). No book says that but that's the way it is. An estimate is an ex-post (after the event) value of an estimator. An estimate is, of course, a number and not a random variable. Ex-ante, a confidence interval always has a probability. Ex-post, the fact that an estimate belongs to some interval is deterministic (has probability either 0 or 1) and it doesn't make sense to talk about 95%.

Since confidence levels are always strictly between 0 and 100%, students should keep in mind that we deal with ex-ante variables.
14
Dec 16

## It’s time to modernize the AP Stats curriculum

### It's time to modernize the AP Stats curriculum

The suggestions below are based on the College Board AP Statistics Course Description, Effective Fall 2010. Citing this description, “AP teachers are encouraged to develop or maintain their own curriculum that either includes or exceeds each of these expectations; such courses will be authorized to use the “AP” designation.” However, AP teachers are constrained by the statement that “The Advanced Placement Program offers a course description and exam in statistics to secondary school students who wish to complete studies equivalent to a one semester, introductory, non-calculus-based, college course in statistics.”

### Too much material for a one-semester course

I tried to teach AP Stats in one semester following the College Board description and methodology. That is, with no derivations, giving only recipes and concentrating on applications. The students were really stretched, didn’t remember anything after completing the course, and usefulness of the course for the subsequent calculus-based course was minimal.

Suggestion. Reduce the number of topics and concentrate on those, which require going all the way from (again citing the description) Exploring Data to Sampling and Experimentation to Anticipating Patterns to Statistical Inference. Simple regression is such a topic.

I would drop the stem-and-leaf plot, because it is stupid; chi-square test for goodness of fit, homogeneity of proportions and independence, including ANOVA, because it is too advanced and looks too vague without the right explanation. Instead of going wide, it is better to go deeper, building upon what students already know. I’ll post a couple of regression applications.

### “Introductory” should not mean stupefying

Statistics has its specifics. Even I, with my extensive experience in Math, made quite a few discoveries for myself while studying Stats. Textbook authors, in their attempts to make exposition accessible, often replace the true statistical ideas by after-the-fact intuition or formulas by their verbal description. See, for example, the z score.

Using TI-83+ and TI-84 graphing calculators is like using a Tesla electric car in conjunction with candles for generating electricity. The sole purpose of these calculators is to prevent cheating. The inclination for cheating is a sign of low understanding and the best proof that the College Board strategy is wrong.

### Once you say “this course is non-calculus-based”, you close many doors

When we format a document in Word, we don’t care how formatting is implemented technically and we don’t need to know anything about programming. Looks like the same attitude is imparted to students of Stats. Few people notice a big difference. When we format a document, we have an idea of what we want and test the result against that idea. In Stats, the idea has to be translated to a formula, and the software output has to be translated into a formula for interpretation.

I understand that, for the majority of Stats students, the amount of algebra I use in some of my posts is not accessible. However, the opposite tendency of telling students that they don’t need to remember any formulas is unproductive. It’s only by memorizing and reproducing equations that they can augment their algebraic proficiency. Stats is largely a mental science. To improve mental activity, you have to engage in one.

Suggestion. Instead of “this course is non-calculus-based”, say: the course develops the ability to interpret equations and translate ideas to formulas.

The way most AP Stats books are written does not give any idea as to what comes from where. When I was a bachelor student, I was looking for explanations, and I would hate reading one of today’s AP Stats textbooks. For those who think, memorizing a bunch of recipes, without seeing the logical links, is a nightmare. In some cases, the absence of logic leads to statements that are plain wrong. Just following the logical sequence will put the pieces of the puzzle together.

9
Dec 16

## Ditch statistical tables if you have a computer

You don't need statistical tables if you have Excel or Mathematica. Here I give the relevant Excel and Mathematica functions described in Chapter 14 of my book. You can save all the formulas in one spreadsheet or notebook and use it multiple times.

### Cumulative Distribution Function of the Standard Normal Distribution

For a given real $z$, the value of the distribution function of the standard normal is
$F(z)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z}\exp (-t^{2}/2)dt.$

In Excel, use the formula =NORM.S.DIST(z,TRUE).

In Mathematica, enter CDF[NormalDistribution[0,1],z]

### Probability Function of the Binomial Distribution

For given number of successes $x,$ number of trials $n$ and probability $p$ the probability is

$P(Binomial=x)=C_{x}^{n}p^{x}(1-p)^{n-x}$.

In Excel, use the formula =BINOM.DIST(x,n,p,FALSE)

In Mathematica, enter PDF[BinomialDistribution[n,p],x]

### Cumulative Binomial Probabilities

For a given cut-off value $x,$ number of trials $n$ and probability $p$ the cumulative probability is

$P(Binomial\leq x)=\sum_{t=0}^{x}C_{t}^{n}p^{t}(1-p)^{n-t}.$
In Excel, use the formula =BINOM.DIST(x,n,p,TRUE).

In Mathematica, enter CDF[BinomialDistribution[n,p],x]

### Values of the exponential function $e^{-\lambda}$$e^{-\lambda}$

In Excel, use the formula =EXP(-lambda)

In Mathematica, enter Exp[-lambda]

### Individual Poisson Probabilities

For given number of successes $x$ and arrival rate $\lambda$ the probability is

$P(Poisson=x)=\frac{e^{-\lambda }\lambda^{x}}{x!}.$
In Excel, use the formula =POISSON.DIST(x,lambda,FALSE)

In Mathematica, enter PDF[PoissonDistribution[lambda],x]

### Cumulative Poisson Probabilities

For given cut-off $x$ and arrival rate $\lambda$ the cumulative probability is

$P(Poisson\leq x)=\sum_{t=0}^{x}\frac{e^{-\lambda }\lambda ^{t}}{t!}.$
In Excel, use the formula =POISSON.DIST(x,lambda,TRUE)

In Mathematica, enter CDF[PoissonDistribution[lambda],x]

### Cutoff Points of the Chi-Square Distribution Function

For given probability of the right tail $\alpha$ and degrees of freedom $\nu$, the cut-off value (critical value) $\chi _{\nu,\alpha }^{2}$ is a solution of the equation
$P(\chi _{\nu}^{2}>\chi _{\nu,\alpha }^{2})=\alpha .$
In Excel, use the formula =CHISQ.INV.RT(alpha,v)

In Mathematica, enter InverseCDF[ChiSquareDistribution[v],1-alpha]

### Cutoff Points for the Student’s t Distribution

For given probability of the right tail $\alpha$ and degrees of freedom $\nu$, the cut-off value $t_{\nu,\alpha }$ is a solution of the equation $P(t_{\nu}>t_{\nu,\alpha })=\alpha$.
In Excel, use the formula =T.INV(1-alpha,v)

In Mathematica, enter InverseCDF[StudentTDistribution[v],1-alpha]

### Cutoff Points for the F Distribution

For given probability of the right tail $\alpha$, degrees of freedom $v_1$ (numerator) and $v_2$ (denominator), the cut-off value $F_{v_1,v_2,\alpha }$ is a solution of the equation $P(F_{v_1,v_2}>F_{v_1,v_2,\alpha })=\alpha$.

In Excel, use the formula =F.INV.RT(alpha,v1,v2)

In Mathematica, enter InverseCDF[FRatioDistribution[v1,v2],1-alpha]

26
Nov 16

## Properties of correlation

### Correlation coefficient: the last block of statistical foundation

Correlation has already been mentioned in

Statistical measures and their geometric roots

Properties of standard deviation

The pearls of AP Statistics 35

Properties of covariance

The pearls of AP Statistics 33

### The hierarchy of definitions

Suppose random variables $X,Y$ are not constant. Then their standard deviations are not zero and we can define their correlation as in Chart 1.

Chart 1. Correlation definition

### Properties of correlation

Property 1. Range of the correlation coefficient: for any $X,Y$ one has $- 1 \le \rho (X,Y) \le 1$.
This follows from the Cauchy-Schwarz inequality, as explained here.

Recall from this post that correlation is cosine of the angle between $X-EX$ and $Y-EY$.
Property 2. Interpretation of extreme cases. (Part 1) If $\rho (X,Y) = 1$, then $Y = aX + b$ with $a > 0.$

(Part 2) If $\rho (X,Y) = - 1$, then $Y = aX + b$ with $a < 0$.

Proof. (Part 1) $\rho (X,Y) = 1$ implies
(1) $Cov (X,Y) = \sigma (X)\sigma (Y)$
which, in turn, implies that $Y$ is a linear function of $X$: $Y = aX + b$ (this is the second part of the Cauchy-Schwarz inequality). Further, we can establish the sign of the number $a$. By the properties of variance and covariance
$Cov(X,Y)=Cov(X,aX+b)=aCov(X,X)+Cov(X,b)=aVar(X)$,

$\sigma (Y)=\sigma(aX + b)=\sigma (aX)=|a|\sigma (X)$.
Plugging this in Eq. (1) we get $aVar(X) = |a|\sigma^2(X)$ and see that $a$ is positive.

The proof of Part 2 is left as an exercise.

Property 3. Suppose we want to measure correlation between weight $W$ and height $H$ of people. The measurements are either in kilos and centimeters ${W_k},{H_c}$ or in pounds and feet ${W_p},{H_f}$. The correlation coefficient is unit-free in the sense that it does not depend on the units used: $\rho (W_k,H_c)=\rho (W_p,H_f)$. Mathematically speaking, correlation is homogeneous of degree $0$ in both arguments.
Proof. One measurement is proportional to another, $W_k=aW_p,\ H_c=bH_f$ with some positive constants $a,b$. By homogeneity
$\rho (W_k,H_c)=\frac{Cov(W_k,H_c)}{\sigma(W_k)\sigma(H_c)}=\frac{Cov(aW_p,bH_f)}{\sigma(aW_p)\sigma(bH_f)}=\frac{abCov(W_p,H_f)}{ab\sigma(W_p)\sigma (H_f)}=\rho (W_p,H_f).$

20
Nov 16

## The pearls of AP Statistics 36

ANOVA: the artefact that survives because of the College Board

### Why ANOVA should be dropped from AP Statistics

1. The common argument in favor of using ANOVA is that "The methods introduced in this chapter [Comparing Groups: Analysis of Variance Methods] apply when a quantitative response variable has a categorical explanatory variable" (Agresti and Franklin, p. 680). However, categorical explanatory variables can be replaced by indicator (dummy) variables, and then regression methods can be used to study dependences involving categorical variables. On p. 695, the authors admit that "ANOVA can be presented as a special case of multiple regression".
2. In terms of knowledge of basic statistical ideas (hypothesis testing, F statistics, significance level), ANOVA doesn't add any value. Those, who have mastered these basic ideas, will not have problems learning ANOVA at their workplace if they have to. There is no need to burden everybody with this stuff "just in case".
3. The explanation of ANOVA is accompanied with definitions of the within-groups variance estimate and between-groups variance estimate (Agresti and Franklin, p. 686). Even in my courses, where I give a lot of algebra, the students don't get them unless they do a couple of theoretical exercises. At the AP Stats level, the usefulness of these definitions is nil.
4. The requirement to remember how the F statistics and degrees of freedom are calculated, for the purpose of being able to interpret just one table with output from a statistical package, doesn't make sense. In my book, I have a whole chapter on ANOVA, with most derivations, and I don't remember a thing. Why torture the students?
5. In the 90 years since R. Fisher has invented ANOVA, many other, more precise and versatile, statistical methods have been developed.

### Conclusion

There are two suggestions

1) Explain just the intuition and then jump to the interpretation of output, indicating the statistic to look at, as in Table 14.14.

2) The theory of ANOVA is useful for two reasons: there is a lot of manipulation with summation signs and there is a link to regressions. Learning all this may be the only justification to study ANOVA with definitions. In my classes, this takes 6 hours.

13
Nov 16

## Statistical measures and their geometric roots

Variance, covariancestandard deviation and correlation: their definitions and properties are deeply rooted in the Euclidean geometry.

## Here is the why: analogy with Euclidean geometry

Euclid axiomatically described the space we live in. What we have known about the geometry of this space since the ancient times has never failed us. Therefore, statistical definitions based on the Euclidean geometry are sure to work.

#### 1. Analogy between scalar product and covariance

Geometry. See Table 2 here for operations with vectors. The scalar product of two vectors $X=(X_1,...,X_n),\ Y=(Y_1,...,Y_n)$ is defined by

$(X,Y)=\sum X_iY_i.$

Statistical analog: Covariance of two random variables is defined by

$Cov(X,Y)=E(X-\bar{X})(Y-\bar{Y}).$

Both the scalar product and covariance are linear in one argument when the other argument is fixed.

#### 2. Analogy between orthogonality and uncorrelatedness

Geometry. Two vectors $X,Y$ are called orthogonal (or perpendicular) if

(1) $(X,Y)=\sum X_iY_i=0.$

Exercise. How do you draw on the plane the vectors $X=(1,0),\ Y=(0,1)$? Check that they are orthogonal.

Statistical analog: Two random variables are called uncorrelated if $Cov(X,Y)=0$.

#### 3. Measuring lengths

Figure 1. Length of a vector

Geometry: the length of a vector $X=(X_1,...,X_n)$ is $\sqrt{\sum X_i^2}$, see Figure 1.

Statistical analog: the standard deviation of a random variable $X$ is

$\sigma(X)=\sqrt{Var(X)}=\sqrt{E(X-\bar{X})^2}.$

This explains the square root in the definition of the standard deviation.

#### 4. Cauchy-Schwarz inequality

Geometry$|(X,Y)|\le\sqrt{\sum X_i^2}\sqrt{\sum Y_i^2}$.

Statistical analog$|Cov(X,Y)|\le\sigma(X)\sigma(Y)$. See the proof here. The proof of its geometric counterpart is similar.

#### 5. Triangle inequality

Figure 2. Triangle inequality

Geometry$\sqrt{\sum (X_i+Y_i)^2}\le\sqrt{\sum X_i^2}+\sqrt{\sum X_i^2}$, see Figure 2 where the length of X+Y does not exceed the sum of lengths of X and Y.

Statistical analog: using the Cauchy-Schwarz inequality we have

$\sigma(X+Y)=\sqrt{Var(X+Y)}$ $=\sqrt{Var(X)+2Cov(X,Y)+Var(Y)}$ $\le\sqrt{\sigma^2(X)+2\sigma(X)\sigma(Y)+\sigma^2(Y)}$ $=\sigma(X)+\sigma(Y).$

#### 4. The Pythagorean theorem

Geometry: In a right triangle, the squared hypotenuse is equal to the sum of the squares of the two legs. The illustration is similar to Figure 2, except that the angle between X and Y should be right.

Proof. Taking two orthogonal vectors $X,Y$ as legs, we have

Squared hypotenuse = $\sum(X_i+Y_i)^2$

(squaring out and using orthogonality (1))

$=\sum X_i^2+2\sum X_iY_i+\sum Y_i^2=\sum X_i^2+\sum Y_i^2$ = Sum of squared legs

Statistical analog: If two random variables are uncorrelated, then variance of their sum is a sum of variances $Var(X+Y)=Var(X)+Var(Y).$

#### 5. The most important analogy: measuring angles

Geometry: the cosine of the angle between two vectors $X,Y$ is defined by

Cosine between X,Y = $\frac{\sum X_iY_i}{\sqrt{\sum X_i^2\sum Y_i^2}}.$

Statistical analog: the correlation coefficient between two random variables is defined by

$\rho(X,Y)=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}=\frac{Cov(X,Y)}{\sigma(X)\sigma(Y)}.$

This intuitively explains why the correlation coefficient takes values between -1 and +1.

Remark. My colleague Alisher Aldashev noticed that the correlation coefficient is the cosine of the angle between the deviations $X-EX$ and $Y-EY$ and not between $X,Y$ themselves.

12
Nov 16

## Properties of standard deviation

Properties of standard deviation are divided in two parts. The definitions and consequences are given here. Both variance and standard deviation are used to measure variability of values of a random variable around its mean. Then why use both of them? The why will be explained in another post.

### Properties of standard deviation: definitions and consequences

Definition. For a random variable $X$, the quantity $\sigma (X) = \sqrt {Var(X)}$ is called its standard deviation.

#### Digression about square roots and absolute values

In general, there are two square roots of a positive number, one positive and the other negative. The positive one is called an arithmetic square root. The arithmetic root is applied here to $Var(X) \ge 0$ (see properties of variance), so standard deviation is always nonnegative.
Definition. An absolute value of a real number $a$ is defined by
(1) $|a| =a$ if $a$ is nonnegative and $|a| =-a$ if $a$ is negative.
This two-part definition is a stumbling block for many students, so making them plug in a few numbers is a must. It is introduced to measure the distance from point $a$ to the origin. For example, $dist(3,0) = |3| = 3$ and $dist(-3,0) = |-3| = 3$. More generally, for any points $a,b$ on the real line the distance between them is given by $dist(a,b) = |a - b|$.

By squaring both sides in Eq. (1) we obtain $|a|^2={a^2}$. Application of the arithmetic square root gives

(2) $|a|=\sqrt {a^2}.$

This is the equation we need right now.

### Back to standard deviation

Property 1. Standard deviation is homogeneous of degree 1. Indeed, using homogeneity of variance and equation (2), we have

$\sigma (aX) =\sqrt{Var(aX)}=\sqrt{{a^2}Var(X)}=|a|\sigma(X).$

Unlike homogeneity of expected values, here we have an absolute value of the scaling coefficient $a$.

Property 2. Cauchy-Schwarz inequality. (Part 1) For any random variables $X,Y$ one has

(3) $|Cov(X,Y)|\le\sigma(X)\sigma(Y)$.

(Part 2) If the inequality sign in (3) turns into equality, $|Cov(X,Y)|=\sigma (X)\sigma (Y)$, then $Y$ is a linear function of $X$: $Y = aX + b$, with some constants $a,b$.
Proof. (Part 1) If at least one of the variables is constant, both sides of the inequality are $0$ and there is nothing to prove. To exclude the trivial case, let $X,Y$ be non-constant and, therefore, $Var(X),\ Var(Y)$ are positive. Consider a real-valued function of a real number $t$ defined by $f(t) = Var(tX + Y)$. Here we have variance of a linear combination

$f(t)=t^2Var(X)+2tCov(X,Y)+Var(Y)$.

We see that $f(t)$ is a parabola with branches looking upward (because the senior coefficient $Var(X)$ is positive). By nonnegativity of variance, $f(t)\ge 0$ and the parabola lies above the horizontal axis in the $(f,t)$ plane. Hence, the quadratic equation $f(t) = 0$ may have at most one real root. This means that the discriminant of the equation is non-positive:

$D=Cov(X,Y)^2-Var(X)Var(Y)\le 0.$

Applying square roots to both sides of $Cov(X,Y)^2\le Var(X)Var(Y)$ we finish the proof of the first part.

(Part 2) In case of the equality sign the discriminant is $0$. Therefore the parabola touches the horizontal axis where $f(t)=Var(tX + Y)=0$. But we know that this implies $tX + Y = constant$ which is just another way of writing $Y = aX + b$.

Comment. (3) explains one of the main properties of the correlation:

$-1\le\rho(X,Y)=\frac{Cov(X,Y)}{\sigma(X)\sigma(Y)}\le 1$.

8
Nov 16

## The pearls of AP Statistics 35

The disturbance term: To hide or not to hide? In an introductory Stats course, some part of the theory should be hidden. Where to draw the line is an interesting question. Here I discuss the ideas that look definitely bad to me.

### How disturbing is the disturbance term?

In the main text, Agresti and Franklin never mention the disturbance term $u_i$ in the regression model

(1) $y_i=a+bx_i+u_i$

(it is hidden in Exercise 12.105). Instead, they write the equation for the mean $\mu_y=a+bx$ that follows from (1) under the standard assumption $Eu_i=0$. This would be fine if the exposition stopped right there. However, one has to explain the random source of variability in $y_i$. On p. 583 the authors say: "The probability distribution of y values at a fixed value of x is a conditional distribution. At each value of x, there is a conditional distribution of y values. A regression model also describes these distributions. An additional parameter σ describes the standard deviation of each conditional distribution."

Further, Figure 12.4 illustrates distributions of errors at different points and asks: "What do the bell-shaped curves around the line at x = 12 and at x = 16 represent?"

Figure 12.4. Illustration of error distributions

Besides, explanations of heteroscedasticity and of the residual sum of squares are impossible without explicitly referring to the disturbance term.

### Attributing a regression property to the correlation is not good

On p.589 I encountered a statement that puzzled me: "An important property of the correlation is that at any particular x value, the predicted value of y is relatively closer to its mean than x is to its mean. If an x value is a certain number of standard deviations from its mean, then the predicted y is r times that many standard deviations from its mean."

Firstly, this is a verbal interpretation of some formula, so why not give the formula itself? How good must be a student to guess what is behind the verbal formulation?

Secondly, as I stressed in this post, the correlation coefficient does not entail any prediction about the magnitude of a change in one variable caused by a change in another. The above statement about the predicted value of y must be a property of regression. Attributing a regression property to the correlation is not in the best interests of those who want to study Stats at a more advanced level.

Thirdly, I felt challenged to see something new in the area I thought I knew everything about. So here is the derivation. By definition, the fitted value is

(2) $\hat{y_i}=\hat{a}+\hat{b}x_i$

where the hats stand for estimators. The fitted line passes through the point $(\bar{x},\bar{y})$:

(3) $\bar{y}=\hat{a}+\hat{b}\bar{x}$

(this will be proved elsewhere). Subtracting (3) from (2) we get

(4) $\hat{y_i}-\bar{y}=\hat{b}(x_i-\bar{x})$

(using equation (4) from this post)

$=\rho\frac{\sigma(y)}{\sigma(x)}(x_i-\bar{x}).$

It is helpful to rewrite (4) in a more symmetric form:

(5) $\frac{\hat{y_i}-\bar{y}}{\sigma(y)}=\rho\frac{x_i-\bar{x}}{\sigma(x)}.$

This is the equation we need. Suppose an x value is a certain number of standard deviations from its mean: $x_i-\bar{x}=k\sigma(x)$. Plug this into (5) to get $\hat{y_i}-\bar{y}=\rho k\sigma(y)$, that is, the predicted y is $\rho$ times that many standard deviations from its mean.