The Lumenary · Grades 9–12

Sampling and statistical reasoning

Name: ______________________

Date: ______________________

You will learn to

Distinguish a population from a sample and a parameter from a statistic, and identify sources of bias in collecting data
Use simulation to see how sample statistics vary, and explain why their spread shrinks in proportion to $1/\sqrt{n}$
Interpret a margin of error and a $95\%$ confidence interval correctly
Fit a least-squares line, interpret the correlation coefficient $r$ , and prove that $-1 \le r \le 1$
Explain why correlation alone does not establish causation, and how randomised experiments can provide evidence of it

1.
A news website asks its readers to click “yes” or “no”: “Should the city ban cars from the old town on Sundays?” $12\,000$ readers vote, and $71\%$ say no. What is the main problem with using this to estimate the opinion of all the city's residents?
- A.Voluntary response: readers choose whether to vote, so people with strong feelings — and readers of that website — are over-represented.
- B.The sample is too small: $12\,000$ is a tiny fraction of the city.
- C.Nothing: with $12\,000$ responses the margin of error is under $1$ percentage point.
- D.The question has only two answers.
2.
To estimate the proportion of households in a region that own a bicycle, which survey would you trust more?
- A.A simple random sample of $800$ households.
- B. $20\,000$ households who answered a questionnaire handed out at bicycle shops.
- C.They are equally good, since both are large.
3.
A random survey of $500$ people has a margin of error of about $4.4$ percentage points. By what factor must the sample size be multiplied to make the margin of error about half as large?
4.
From a random sample, a researcher computes a $95\%$ confidence interval for the proportion of adults in a country who use mobile banking: $0.52$ to $0.58$ . Which interpretation is correct (in the usual, frequentist sense)?
- A.The method used produces intervals that contain the true proportion for about $95\%$ of random samples; this interval was produced by that method.
- B.There is a $95\%$ probability that the true proportion is between $0.52$ and $0.58$ .
- C. $95\%$ of adults use mobile banking at a rate between $52\%$ and $58\%$ .
- D.If the survey were repeated, $95\%$ of the new estimates would fall between $0.52$ and $0.58$ .
5.
Fit the least-squares line to the five points $(1, 2), (2, 2), (3, 5), (4, 6), (5, 10)$ . What value of $y$ does the line predict at $x = 6$ ?
6.
For a data set, the correlation coefficient is $r = 0$ . What can you conclude?
- A.There is no linear association: the least-squares line is horizontal ( $\hat b = 0$ ). There may still be a strong non-linear relationship.
- B. $x$ and $y$ are unrelated.
- C.The points lie on a horizontal line.
- D.Half the points lie above the line $y = \bar y$ .
7.
Across $50$ countries, the number of mobile phones per person is strongly positively correlated with life expectancy. Which conclusion is best supported?
- A.The correlation may be explained by a confounder, such as national wealth, which tends to increase both; it does not by itself show that phones lengthen lives.
- B.Giving people mobile phones will increase life expectancy.
- C.Living longer causes people to buy more phones.
- D.Because the correlation is strong, it must be causal.
8.
In the proof that $\sum_i (x_i - c)^2 = \sum_i (x_i - \bar x)^2 + n(\bar x - c)^2$ , the expansion produces a cross term $2(\bar x - c)\sum_i (x_i - \bar x)$ . Why is it zero?
- A.Because $\sum_i (x_i - \bar x) = \sum_i x_i - n\bar x = 0$ , by the definition of the mean.
- B.Because $c = \bar x$ .
- C.Because each $x_i - \bar x$ is zero.
- D.Because squares are never negative.
9.
Data are x : Fin n → ℝ and mean x is their mean. Which Lean statement says that the mean minimises the sum of squared deviations?
- A.∀ c : ℝ, ∑ i, (x i - mean x) ^ 2 ≤ ∑ i, (x i - c) ^ 2
- B.∀ c : ℝ, ∑ i, (x i - c) ^ 2 ≤ ∑ i, (x i - mean x) ^ 2
- C.∃ c : ℝ, ∑ i, (x i - mean x) ^ 2 ≤ ∑ i, (x i - c) ^ 2
- D.∀ c : ℝ, ∑ i, |x i - mean x| ≤ ∑ i, |x i - c|
10.
Write fit(xs, ys) returning the pair (a, b) of the least-squares line $y = a + bx$ , and correlation(xs, ys) returning $r$ , using $S_{xx}$ , $S_{xy}$ and $S_{yy}$ as in the lesson.
Computational — Practice

Answer key — Sampling and statistical reasoning

1.
The sample is a voluntary-response sample: readers decide for themselves whether to vote. People with strong opinions, and people who read that particular website (perhaps not representative of residents, and possibly not residents at all), are over-represented. This bias does not shrink with the number of votes, and the margin-of-error formula does not apply.
2.
The random sample. The bicycle-shop sample is biased — almost everyone there owns a bicycle — and no number of responses removes the bias. For the random sample, the margin of error is at most about $1.96\sqrt{0.25/800} \approx 0.035$ .
3.
The margin is proportional to $1/\sqrt n$ . Multiplying $n$ by $k$ divides the margin by $\sqrt k$ , so halving it needs $\sqrt k = 2$ , that is $k = 4$ : about $2000$ people.
4.
(a). The true proportion is fixed; the interval depends on the random sample. About $95\%$ of intervals constructed this way from random samples contain the true proportion — that is the long-run success rate of the method. Whether this one does, we cannot know.
5.
$\bar x = 3$ , $\bar y = 5$ , $S_{xx} = 4 + 1 + 0 + 1 + 4 = 10$ and $S_{xy} = (-2)(-3) + (-1)(-3) + 0 + (1)(1) + (2)(5) = 20$ . So $\hat b = 20/10 = 2$ and $\hat a = 5 - 2 \times 3 = -1$ : the line is $\hat y = -1 + 2x$ , which predicts $\hat y = 11$ at $x = 6$ . (Predicting beyond the range of the data — extrapolation — should be done with caution.)
6.
$r = 0$ means $S_{xy} = 0$ , so the least-squares slope $S_{xy}/S_{xx}$ is $0$ : there is no linear trend. A strong curved relationship is still possible, as $y = x^2$ on a symmetric range shows — so plot the data.
7.
(a). An observational correlation between countries is consistent with a confounder (wealth, education, infrastructure) driving both variables, and with other explanations. Establishing causation would need much more — ideally randomised evidence.
8.
$\sum_i (x_i - \bar x) = \sum_i x_i - n\bar x$ , and $n\bar x = \sum_i x_i$ by definition of the mean, so the sum is $0$ and the cross term vanishes for every $c$ . In Lean this is sum_sub_mean, used in the proofs of sum_sq_sub and sse_eq.
9.
(a): for every number $c$ , the sum of squared deviations from $c$ is at least the sum of squared deviations from the mean. (b) reverses the inequality, (c) is trivially true with the wrong quantifier, and (d) concerns absolute deviations, which the median minimises — for $0, 0, 3$ the mean $1$ gives $1 + 1 + 2 = 4$ , but $c = 0$ gives $3$ .

10.

import math


def _sums(xs, ys):
    n = len(xs)
    mx, my = sum(xs) / n, sum(ys) / n
    sxx = sum((x - mx) ** 2 for x in xs)
    syy = sum((y - my) ** 2 for y in ys)
    sxy = sum((x - mx) * (y - my) for x, y in zip(xs, ys))
    return mx, my, sxx, sxy, syy


def fit(xs, ys):
    mx, my, sxx, sxy, _ = _sums(xs, ys)
    b = sxy / sxx
    return my - b * mx, b


def correlation(xs, ys):
    _, _, sxx, sxy, syy = _sums(xs, ys)
    return sxy / math.sqrt(sxx * syy)

For the revision data the line is $\hat y = 2.2 + 0.6x$ and $r = 6/\sqrt{60} \approx 0.775$ ; points on a falling line give $r = -1$ , and the symmetric parabola gives $r = 0$ .