Prerequisite Self-Test

This self-test tells you where to start. It is twenty questions across calculus, probability and statistics, and light Python — the toolkit the camp assumes on Day 1. Take it closed-book, give yourself about ninety minutes, and write down a real answer for each before checking anything. The questions are fair but not soft; a few will make you think. That is the point. When you finish, grade yourself honestly against the Solutions section, then read the Routing table at the very end to decide what to do before Week 1.

Do not panic about a wrong answer here. The whole purpose of the test is to route you to the right starting point. The students you will follow through the book — Maya, Devon, Priya, Sam, and Leah — each missed something on a test like this. What mattered was what they did next.

Questions

Q1 (derivative). Let f(x) = 3x^4 - 5x^2 + 7x - 2. Find f'(x), and evaluate f'(1).

Q2 (chain rule). Let g(x) = e^{-x^2/2}. Find g'(x). At what value(s) of x is g'(x) = 0?

Q3 (product / quotient rule). A firm’s revenue is R(p) = p \cdot q(p) where p is price and quantity sold is q(p) = 100 - 4p. Find \dfrac{dR}{dp} and the price p that maximizes revenue.

Q4 (partial derivatives). Let h(x, y) = x^2 y + 3xy^3. Find \dfrac{\partial h}{\partial x} and \dfrac{\partial h}{\partial y}.

Q5 (optimization / first-order condition). Sam fits a line through one number: he wants the constant c that minimizes the sum of squared errors S(c) = \sum_{i=1}^{n}(y_i - c)^2 for a fixed dataset y_1, \dots, y_n. Set up the first-order condition and solve for the minimizing c. What familiar quantity is it?

Q6 (definite integral). Evaluate \displaystyle\int_{0}^{2} (3x^2 + 2x)\, dx.

Q7 (limit). Evaluate \displaystyle\lim_{x \to 0} \frac{\sin(3x)}{x}. Briefly say why.

Q8 (expectation and variance). A single roll of a fair six-sided die yields X \in \{1,2,3,4,5,6\}, each with probability 1/6. Compute \mathbb{E}[X] and \operatorname{Var}(X).

Q9 (linearity / scaling of variance). Priya has a random variable X with \mathbb{E}[X] = 4 and \operatorname{Var}(X) = 9. Let Y = 2X - 5. Find \mathbb{E}[Y] and \operatorname{Var}(Y).

Q10 (conditional probability / Bayes). A loan-fraud screen flags 5% of all applications. Among truly fraudulent applications it flags 90%; among legitimate ones it flags 4%. Suppose 2% of applications are truly fraudulent. Given that an application is flagged, what is the probability it is actually fraudulent? (Round to two decimals.)

Q11 (law of total expectation). Leah models a startup’s payoff. With probability 0.3 it is a “hit” with expected payoff $10M; with probability 0.7 it is a “miss” with expected payoff $1M. What is the unconditional expected payoff?

Q12 (binomial). Devon flags that a certain coin lands heads with probability 0.6. He flips it 10 times. What is the probability of exactly 7 heads? Give the formula and a numerical value (two decimals).

Q13 (normal distribution). Daily returns on a stock are approximately \mathcal{N}(\mu = 0.0005,\ \sigma = 0.02). Roughly what fraction of days have a return below -0.0395? (Use the 68–95–99.7 rule; no calculator needed.)

Q14 (t-test interpretation). A regression reports a coefficient \hat{\beta}_1 = 0.42 with standard error 0.15. State the approximate t-statistic, and explain in one or two sentences what it tells you about the null hypothesis \beta_1 = 0 at the usual 5% level.

Q15 (correlation vs. causation). Across cities, ice-cream sales and drowning deaths are strongly positively correlated. Does ice cream cause drowning? Name the most likely reason for the correlation and the term for the variable responsible.

Q16 (reading a confidence interval). A study reports that a job-training program raised earnings by $1,200 with a 95% confidence interval of [-200,\ 2{,}600]. Maya says “the program had no effect.” Is her statement supported? Explain what the interval does and does not let you conclude.

Q17 (sampling distribution). You draw a sample of n = 100 from a population with mean \mu and standard deviation \sigma = 20, and compute the sample mean \bar{X}. What is the standard deviation of \bar{X} (the standard error), and what happens to it if you increase n to 400?

Q18 (Python output — pandas/numpy). What does this print?

import numpy as np
a = np.array([1, 2, 3, 4])
print(a[a > 2].sum())

Q19 (vectorized vs. loop). Two snippets compute the same thing on a NumPy array x of length 1,000,000. Which is preferred in this book, and why?

# A
total = 0
for v in x:
    total += v ** 2

# B
total = (x ** 2).sum()

Q20 (groupby interpretation). Given a DataFrame df with columns sector and ret (a return), what does this produce, in words?

df.groupby("sector")["ret"].mean()

Solutions

Q1. Differentiate term by term using the power rule \frac{d}{dx}x^n = nx^{n-1}: f'(x) = 12x^3 - 10x + 7. At x = 1: f'(1) = 12 - 10 + 7 = \boxed{9}.

Q2. Write g(x) = e^{u} with u = -x^2/2, so u' = -x. The chain rule gives g'(x) = e^{u}\cdot u' = -x\,e^{-x^2/2}. This equals zero only when the factor x = 0 (the exponential is never zero), so g'(x) = 0 at \boxed{x = 0}. (This is the standard normal density up to a constant; its slope is zero at the peak.)

Q3. R(p) = p(100 - 4p) = 100p - 4p^2, so \frac{dR}{dp} = 100 - 8p. Setting this to zero: p = \boxed{12.5}. The second derivative is -8 < 0, confirming a maximum. (You could also use the product rule on p\cdot q(p) directly: R' = q(p) + p\,q'(p) = (100 - 4p) + p(-4) = 100 - 8p, same answer.)

Q4. Treat the other variable as a constant. \frac{\partial h}{\partial x} = 2xy + 3y^3, \qquad \frac{\partial h}{\partial y} = x^2 + 9xy^2.

Q5. Differentiate S(c) = \sum_i (y_i - c)^2 with respect to c and set to zero: S'(c) = \sum_i 2(y_i - c)(-1) = -2\sum_i (y_i - c) = 0 \;\Rightarrow\; \sum_i (y_i - c) = 0 \;\Rightarrow\; c = \frac{1}{n}\sum_i y_i. The second derivative S''(c) = 2n > 0 confirms a minimum. The minimizing c is the sample mean \bar{y} — the first and simplest least-squares estimate you will meet, and the reason the mean shows up everywhere in regression.

Q6. Antiderivative of 3x^2 + 2x is x^3 + x^2. Evaluate from 0 to 2: (2^3 + 2^2) - (0 + 0) = 8 + 4 = \boxed{12}.

Q7. \displaystyle\lim_{x\to 0}\frac{\sin(3x)}{x} = \boxed{3}. Reason: \frac{\sin(3x)}{x} = 3\cdot\frac{\sin(3x)}{3x}, and \frac{\sin(u)}{u}\to 1 as u\to 0. (Equivalently, by L’Hôpital, the ratio of derivatives is \frac{3\cos(3x)}{1}\to 3.)

Q8. \mathbb{E}[X] = \frac{1}{6}(1+2+3+4+5+6) = \frac{21}{6} = 3.5. For the variance, \mathbb{E}[X^2] = \frac{1}{6}(1+4+9+16+25+36) = \frac{91}{6} \approx 15.1667, so \operatorname{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 = \frac{91}{6} - 3.5^2 = 15.1667 - 12.25 = \boxed{\tfrac{35}{12} \approx 2.9167}.

Q9. Expectation is linear: \mathbb{E}[Y] = 2\mathbb{E}[X] - 5 = 2(4) - 5 = \boxed{3}. Variance ignores additive constants and scales by the square of the multiplier: \operatorname{Var}(Y) = 2^2\operatorname{Var}(X) = 4 \cdot 9 = \boxed{36}. (The -5 shifts the whole distribution but does not change its spread.)

Q10. Let F = fraudulent, + = flagged. We want P(F \mid +). Bayes: P(F\mid +) = \frac{P(+\mid F)P(F)}{P(+\mid F)P(F) + P(+\mid \text{not }F)P(\text{not }F)} = \frac{0.90 \cdot 0.02}{0.90\cdot 0.02 + 0.04\cdot 0.98} = \frac{0.018}{0.018 + 0.0392} = \frac{0.018}{0.0572} \approx \boxed{0.31}. Note the lesson: even with a screen that catches 90% of fraud, a flagged application is fraudulent only about 31% of the time, because fraud is rare to begin with. (The “flags 5% of all applications” figure in the prompt is a deliberate distractor — and a useful sanity check. The two conditional rates actually imply an overall flag rate of P(+) = 0.90\cdot0.02 + 0.04\cdot0.98 = 0.0572, i.e. about 5.7%, not exactly 5%. A careful reader notices the stated marginal is only approximate; either way it is not used in the Bayes calculation, which needs only the conditional rates and the base rate.)

Q11. Law of total expectation: \mathbb{E}[\text{payoff}] = P(\text{hit})\,\mathbb{E}[\text{payoff}\mid\text{hit}] + P(\text{miss})\,\mathbb{E}[\text{payoff}\mid\text{miss}] = 0.3(10) + 0.7(1) = 3 + 0.7 = \boxed{\$3.7\text{M}}.

Q12. Binomial with n = 10, p = 0.6, k = 7: P(X = 7) = \binom{10}{7}(0.6)^7(0.4)^3 = 120 \cdot 0.0279936 \cdot 0.064 \approx \boxed{0.21}.

Q13. The threshold -0.0395 is \mu - 2\sigma = 0.0005 - 2(0.02) = 0.0005 - 0.04 = -0.0395, i.e., two standard deviations below the mean. By the 68–95–99.7 rule, about 95% of mass lies within \pm 2\sigma, leaving 5% in the two tails, so about 2.5\% lies below -2\sigma. Answer: about 2.5% of days.

Q14. The t-statistic is t = \hat\beta_1 / \text{SE} = 0.42 / 0.15 = 2.8. Since |t| = 2.8 exceeds the usual critical value of about 1.96, we reject the null \beta_1 = 0 at the 5% level: the estimate is statistically distinguishable from zero. (It says nothing, by itself, about whether the effect is large or causal — only that it is unlikely to be zero given the data and the model.)

Q15. No — ice cream does not cause drowning. Both rise with hot weather / summer temperature, which drives more ice-cream buying and more swimming (hence more drownings). The temperature variable is a confounder (a common cause of both, also called a lurking or omitted variable). This is the central problem the camp’s middle weeks are built to solve.

Q16. Maya’s statement is not supported as stated. The point estimate is a $1,200 increase, and the 95% confidence interval [-200, 2600] means the data are consistent with effects ranging from a small negative to a large positive. Because the interval includes zero, we cannot reject the hypothesis of no effect at the 5% level — but “cannot reject zero” is not the same as “the effect is zero.” The interval is too wide to distinguish “no effect” from “a substantial $2,000 effect”; the honest conclusion is that the study is inconclusive / underpowered, not that the program did nothing.

Q17. The standard error of the sample mean is \sigma/\sqrt{n} = 20/\sqrt{100} = \boxed{2}. Increasing n to 400 gives 20/\sqrt{400} = 1, so the standard error halves. Note it falls with the square root of n: to cut it in half you need four times the data. This is the engine behind why bigger samples give more precise estimates.

Q18. The boolean mask a > 2 selects the elements 3 and 4; their sum is 3 + 4 =7.

Q19. B (the vectorized version) is preferred. Both compute \sum x_i^2, but B pushes the loop into NumPy’s compiled C implementation, so it is far faster on a million elements and is more readable. The Python-level for loop in A is slow and error-prone. Vectorize whenever you can; we treat explicit elementwise loops over large arrays as a code smell in this book.

Q20. It splits the rows of df into groups by their sector value, then computes the mean return (ret) within each sector, returning a Series indexed by sector. In words: the average return for each sector.

Routing table

Count your misses by cluster and follow the routing. “Missed” means you got it wrong or had to guess.

If you missed…	What it signals	Do this before Week 1
Q1–Q7 (any 2 or more)	Calculus rust — derivatives, chain/product rule, partials, optimization, integrals, limits. Week 2’s inference and Week 3’s estimators lean on these constantly.	Work through Appendix A: Math Toolkit before Week 1. Maya routed here and it paid off.
Q8–Q17 (any 3 or more)	Probability/statistics gaps — expectation/variance, Bayes, total expectation, binomial/normal, t-tests, confidence intervals, sampling distributions, confounding. This is the conceptual spine of the whole camp.	Review the probability and inference sections of Appendix A, with special attention to confounding (Q15) and confidence-interval interpretation (Q16) — these recur in every causal-inference week.
Q18–Q20 (any 2 or more)	Python/`pandas`/NumPy fundamentals — boolean indexing, vectorization, `groupby`. You will write code from Day 1.	Set up your environment and work the fundamentals in Appendix B: Python & LaTeX Setup first. Devon skipped this and hit a wall — do not be Devon.
Q15 and Q16 specifically	Correlation-vs-causation and confidence-interval reading. Even strong students miss these; they are the conceptual hinges of empirical work.	You can start Week 1 on schedule, but flag these for your mentor in your first lab session — they are exactly what Weeks 3–6 are about.
Two or more clusters	Broad gaps across math, stats, and code.	Budget extra time on Appendices A and B before Week 1, and tell your mentor where you are starting from. There is no penalty for arriving honest about it; there is real cost to bluffing past it.
0–3 missed total, spread out	You are ready.	Skim Appendix A for the notation conventions and start Week 1. Keep the appendices handy as reference.

# Prerequisite Self-Test This self-test tells you where to start. It is twenty questions across calculus, probability and statistics, and light Python — the toolkit the camp assumes on Day 1. Take it closed-book, give yourself about ninety minutes, and write down a real answer for each before checking anything. The questions are fair but not soft; a few will make you think. That is the point. When you finish, grade yourself honestly against the **Solutions** section, then read the **Routing table** at the very end to decide what to do before Week 1. Do not panic about a wrong answer here. The whole purpose of the test is to route you to the right starting point. The students you will follow through the book — Maya, Devon, Priya, Sam, and Leah — each missed something on a test like this. What mattered was what they did next. --- ## Questions **Q1 (derivative).** Let $f(x) = 3x^4 - 5x^2 + 7x - 2$. Find $f'(x)$, and evaluate $f'(1)$. **Q2 (chain rule).** Let $g(x) = e^{-x^2/2}$. Find $g'(x)$. At what value(s) of $x$ is $g'(x) = 0$? **Q3 (product / quotient rule).** A firm's revenue is $R(p) = p \cdot q(p)$ where $p$ is price and quantity sold is $q(p) = 100 - 4p$. Find $\dfrac{dR}{dp}$ and the price $p$ that maximizes revenue. **Q4 (partial derivatives).** Let $h(x, y) = x^2 y + 3xy^3$. Find $\dfrac{\partial h}{\partial x}$ and $\dfrac{\partial h}{\partial y}$. **Q5 (optimization / first-order condition).** Sam fits a line through one number: he wants the constant $c$ that minimizes the sum of squared errors $S(c) = \sum_{i=1}^{n}(y_i - c)^2$ for a fixed dataset $y_1, \dots, y_n$. Set up the first-order condition and solve for the minimizing $c$. What familiar quantity is it? **Q6 (definite integral).** Evaluate $\displaystyle\int_{0}^{2} (3x^2 + 2x)\, dx$. **Q7 (limit).** Evaluate $\displaystyle\lim_{x \to 0} \frac{\sin(3x)}{x}$. Briefly say why. **Q8 (expectation and variance).** A single roll of a fair six-sided die yields $X \in \{1,2,3,4,5,6\}$, each with probability $1/6$. Compute $\mathbb{E}[X]$ and $\operatorname{Var}(X)$. **Q9 (linearity / scaling of variance).** Priya has a random variable $X$ with $\mathbb{E}[X] = 4$ and $\operatorname{Var}(X) = 9$. Let $Y = 2X - 5$. Find $\mathbb{E}[Y]$ and $\operatorname{Var}(Y)$. **Q10 (conditional probability / Bayes).** A loan-fraud screen flags 5% of all applications. Among truly fraudulent applications it flags 90%; among legitimate ones it flags 4%. Suppose 2% of applications are truly fraudulent. Given that an application is flagged, what is the probability it is actually fraudulent? (Round to two decimals.) **Q11 (law of total expectation).** Leah models a startup's payoff. With probability 0.3 it is a "hit" with expected payoff \$10M; with probability 0.7 it is a "miss" with expected payoff \$1M. What is the unconditional expected payoff? **Q12 (binomial).** Devon flags that a certain coin lands heads with probability 0.6. He flips it 10 times. What is the probability of exactly 7 heads? Give the formula and a numerical value (two decimals). **Q13 (normal distribution).** Daily returns on a stock are approximately $\mathcal{N}(\mu = 0.0005,\ \sigma = 0.02)$. Roughly what fraction of days have a return below $-0.0395$? (Use the 68–95–99.7 rule; no calculator needed.) **Q14 (t-test interpretation).** A regression reports a coefficient $\hat{\beta}_1 = 0.42$ with standard error $0.15$. State the approximate $t$-statistic, and explain in one or two sentences what it tells you about the null hypothesis $\beta_1 = 0$ at the usual 5% level. **Q15 (correlation vs. causation).** Across cities, ice-cream sales and drowning deaths are strongly positively correlated. Does ice cream cause drowning? Name the most likely reason for the correlation and the term for the variable responsible. **Q16 (reading a confidence interval).** A study reports that a job-training program raised earnings by \$1,200 with a 95% confidence interval of $[-200,\ 2{,}600]$. Maya says "the program had no effect." Is her statement supported? Explain what the interval does and does not let you conclude. **Q17 (sampling distribution).** You draw a sample of $n = 100$ from a population with mean $\mu$ and standard deviation $\sigma = 20$, and compute the sample mean $\bar{X}$. What is the standard deviation of $\bar{X}$ (the standard error), and what happens to it if you increase $n$ to $400$? **Q18 (Python output — pandas/numpy).** What does this print? ```python import numpy as np a = np.array([1, 2, 3, 4]) print(a[a > 2].sum()) ``` **Q19 (vectorized vs. loop).** Two snippets compute the same thing on a NumPy array `x` of length 1,000,000. Which is preferred in this book, and why? ```python # A total = 0 for v in x: total += v ** 2 # B total = (x ** 2).sum() ``` **Q20 (groupby interpretation).** Given a DataFrame `df` with columns `sector` and `ret` (a return), what does this produce, in words? ```python df.groupby("sector")["ret"].mean() ``` --- ## Solutions **Q1.** Differentiate term by term using the power rule $\frac{d}{dx}x^n = nx^{n-1}$: $$f'(x) = 12x^3 - 10x + 7.$$ At $x = 1$: $f'(1) = 12 - 10 + 7 = \boxed{9}$. **Q2.** Write $g(x) = e^{u}$ with $u = -x^2/2$, so $u' = -x$. The chain rule gives $g'(x) = e^{u}\cdot u' = -x\,e^{-x^2/2}$. This equals zero only when the factor $x = 0$ (the exponential is never zero), so $g'(x) = 0$ at $\boxed{x = 0}$. (This is the standard normal density up to a constant; its slope is zero at the peak.) **Q3.** $R(p) = p(100 - 4p) = 100p - 4p^2$, so $\frac{dR}{dp} = 100 - 8p$. Setting this to zero: $p = \boxed{12.5}$. The second derivative is $-8 < 0$, confirming a maximum. (You could also use the product rule on $p\cdot q(p)$ directly: $R' = q(p) + p\,q'(p) = (100 - 4p) + p(-4) = 100 - 8p$, same answer.) **Q4.** Treat the other variable as a constant. $$\frac{\partial h}{\partial x} = 2xy + 3y^3, \qquad \frac{\partial h}{\partial y} = x^2 + 9xy^2.$$ **Q5.** Differentiate $S(c) = \sum_i (y_i - c)^2$ with respect to $c$ and set to zero: $$S'(c) = \sum_i 2(y_i - c)(-1) = -2\sum_i (y_i - c) = 0 \;\Rightarrow\; \sum_i (y_i - c) = 0 \;\Rightarrow\; c = \frac{1}{n}\sum_i y_i.$$ The second derivative $S''(c) = 2n > 0$ confirms a minimum. The minimizing $c$ is the **sample mean** $\bar{y}$ — the first and simplest least-squares estimate you will meet, and the reason the mean shows up everywhere in regression. **Q6.** Antiderivative of $3x^2 + 2x$ is $x^3 + x^2$. Evaluate from $0$ to $2$: $$(2^3 + 2^2) - (0 + 0) = 8 + 4 = \boxed{12}.$$ **Q7.** $\displaystyle\lim_{x\to 0}\frac{\sin(3x)}{x} = \boxed{3}$. Reason: $\frac{\sin(3x)}{x} = 3\cdot\frac{\sin(3x)}{3x}$, and $\frac{\sin(u)}{u}\to 1$ as $u\to 0$. (Equivalently, by L'Hôpital, the ratio of derivatives is $\frac{3\cos(3x)}{1}\to 3$.) **Q8.** $\mathbb{E}[X] = \frac{1}{6}(1+2+3+4+5+6) = \frac{21}{6} = 3.5$. For the variance, $\mathbb{E}[X^2] = \frac{1}{6}(1+4+9+16+25+36) = \frac{91}{6} \approx 15.1667$, so $$\operatorname{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 = \frac{91}{6} - 3.5^2 = 15.1667 - 12.25 = \boxed{\tfrac{35}{12} \approx 2.9167}.$$ **Q9.** Expectation is linear: $\mathbb{E}[Y] = 2\mathbb{E}[X] - 5 = 2(4) - 5 = \boxed{3}$. Variance ignores additive constants and scales by the square of the multiplier: $\operatorname{Var}(Y) = 2^2\operatorname{Var}(X) = 4 \cdot 9 = \boxed{36}$. (The $-5$ shifts the whole distribution but does not change its spread.) **Q10.** Let $F$ = fraudulent, $+$ = flagged. We want $P(F \mid +)$. Bayes: $$P(F\mid +) = \frac{P(+\mid F)P(F)}{P(+\mid F)P(F) + P(+\mid \text{not }F)P(\text{not }F)} = \frac{0.90 \cdot 0.02}{0.90\cdot 0.02 + 0.04\cdot 0.98} = \frac{0.018}{0.018 + 0.0392} = \frac{0.018}{0.0572} \approx \boxed{0.31}.$$ Note the lesson: even with a screen that catches 90% of fraud, a flagged application is fraudulent only about 31% of the time, because fraud is rare to begin with. (The "flags 5% of all applications" figure in the prompt is a deliberate distractor — and a useful sanity check. The two conditional rates actually imply an overall flag rate of $P(+) = 0.90\cdot0.02 + 0.04\cdot0.98 = 0.0572$, i.e. about 5.7%, not exactly 5%. A careful reader notices the stated marginal is only approximate; either way it is not used in the Bayes calculation, which needs only the conditional rates and the base rate.) **Q11.** Law of total expectation: $\mathbb{E}[\text{payoff}] = P(\text{hit})\,\mathbb{E}[\text{payoff}\mid\text{hit}] + P(\text{miss})\,\mathbb{E}[\text{payoff}\mid\text{miss}] = 0.3(10) + 0.7(1) = 3 + 0.7 = \boxed{\$3.7\text{M}}.$ **Q12.** Binomial with $n = 10$, $p = 0.6$, $k = 7$: $$P(X = 7) = \binom{10}{7}(0.6)^7(0.4)^3 = 120 \cdot 0.0279936 \cdot 0.064 \approx \boxed{0.21}.$$ **Q13.** The threshold $-0.0395$ is $\mu - 2\sigma = 0.0005 - 2(0.02) = 0.0005 - 0.04 = -0.0395$, i.e., two standard deviations below the mean. By the 68–95–99.7 rule, about 95% of mass lies within $\pm 2\sigma$, leaving 5% in the two tails, so about $2.5\%$ lies below $-2\sigma$. Answer: **about 2.5% of days.** **Q14.** The $t$-statistic is $t = \hat\beta_1 / \text{SE} = 0.42 / 0.15 = 2.8$. Since $|t| = 2.8$ exceeds the usual critical value of about $1.96$, we **reject the null** $\beta_1 = 0$ at the 5% level: the estimate is statistically distinguishable from zero. (It says nothing, by itself, about whether the effect is *large* or *causal* — only that it is unlikely to be zero given the data and the model.) **Q15.** No — ice cream does not cause drowning. Both rise with **hot weather / summer temperature**, which drives more ice-cream buying *and* more swimming (hence more drownings). The temperature variable is a **confounder** (a common cause of both, also called a lurking or omitted variable). This is the central problem the camp's middle weeks are built to solve. **Q16.** Maya's statement is **not supported as stated.** The point estimate is a \$1,200 *increase*, and the 95% confidence interval $[-200, 2600]$ means the data are consistent with effects ranging from a small negative to a large positive. Because the interval includes zero, we *cannot reject* the hypothesis of no effect at the 5% level — but "cannot reject zero" is not the same as "the effect is zero." The interval is too wide to distinguish "no effect" from "a substantial \$2,000 effect"; the honest conclusion is that the study is **inconclusive / underpowered**, not that the program did nothing. **Q17.** The standard error of the sample mean is $\sigma/\sqrt{n} = 20/\sqrt{100} = \boxed{2}$. Increasing $n$ to $400$ gives $20/\sqrt{400} = 1$, so the standard error **halves**. Note it falls with the square root of $n$: to cut it in half you need *four times* the data. This is the engine behind why bigger samples give more precise estimates. **Q18.** The boolean mask `a > 2` selects the elements `3` and `4`; their sum is `3 + 4 = `**`7`**. **Q19.** **B (the vectorized version) is preferred.** Both compute $\sum x_i^2$, but B pushes the loop into NumPy's compiled C implementation, so it is far faster on a million elements and is more readable. The Python-level `for` loop in A is slow and error-prone. Vectorize whenever you can; we treat explicit elementwise loops over large arrays as a code smell in this book. **Q20.** It splits the rows of `df` into groups by their `sector` value, then computes the **mean return (`ret`) within each sector**, returning a Series indexed by sector. In words: the average return for each sector. --- ## Routing table Count your misses by cluster and follow the routing. "Missed" means you got it wrong or had to guess. | If you missed… | What it signals | Do this before Week 1 | |----------------|-----------------|-----------------------| | **Q1–Q7 (any 2 or more)** | Calculus rust — derivatives, chain/product rule, partials, optimization, integrals, limits. Week 2's inference and Week 3's estimators lean on these constantly. | Work through **Appendix A: Math Toolkit** before Week 1. Maya routed here and it paid off. | | **Q8–Q17 (any 3 or more)** | Probability/statistics gaps — expectation/variance, Bayes, total expectation, binomial/normal, $t$-tests, confidence intervals, sampling distributions, confounding. This is the conceptual spine of the whole camp. | Review the probability and inference sections of **Appendix A**, with special attention to confounding (Q15) and confidence-interval interpretation (Q16) — these recur in every causal-inference week. | | **Q18–Q20 (any 2 or more)** | Python/`pandas`/NumPy fundamentals — boolean indexing, vectorization, `groupby`. You will write code from Day 1. | Set up your environment and work the fundamentals in **Appendix B: Python & LaTeX Setup** first. Devon skipped this and hit a wall — do not be Devon. | | **Q15 and Q16 specifically** | Correlation-vs-causation and confidence-interval reading. Even strong students miss these; they are *the* conceptual hinges of empirical work. | You can start Week 1 on schedule, but flag these for your mentor in your first lab session — they are exactly what Weeks 3–6 are about. | | **Two or more clusters** | Broad gaps across math, stats, and code. | Budget extra time on **Appendices A and B** before Week 1, and tell your mentor where you are starting from. There is no penalty for arriving honest about it; there is real cost to bluffing past it. | | **0–3 missed total, spread out** | You are ready. | Skim **Appendix A** for the notation conventions and start Week 1. Keep the appendices handy as reference. |