Confidence intervals

A single number is a poor answer to the question 'What is the true value?' If the transit survey finds a mean commute of 25.1 minutes, the true city-wide mean is almost certainly not exactly 25.1. What we can say is that the true mean is probably within some range around 25.1, and we can say how confident we are in that range. A confidence interval does exactly this. It is the most useful single tool of statistical inference, and it is what lies behind every margin of error you read in the news.

The idea

Lesson 9 showed that a sample statistic, such as a sample mean or proportion, varies around the true parameter, and that about 95 percent of samples give a statistic within about two standard deviations of the parameter. Turn that statement around. If the statistic is usually within two standard deviations of the parameter, then the parameter is usually within two standard deviations of the statistic. So the interval 'statistic plus or minus about two standard deviations' will capture the parameter in about 95 percent of samples.

Every confidence interval in this course has the same form: estimate plus or minus margin of error. The estimate is our best single guess, usually a sample mean or proportion. The margin of error is a critical value, which depends on the confidence level, times the standard error, which measures the variability of the estimate. The confidence level, usually 95 percent, describes how often the method captures the true parameter over many samples.

What 95 percent confidence means

The confidence level is a property of the method, not of a single interval. Once we have computed a particular interval, say 23.8 to 26.5 minutes, the true mean is either in it or not; we just do not know which. What we know is that the method we used produces intervals that capture the true parameter 95 percent of the time. Saying 'we are 95 percent confident that the true mean commute is between 23.8 and 26.5 minutes' is shorthand for that.

# What "95% confidence" means: repeat the sampling many times and count how often intervals capture p.
import random, math
random.seed(141)
p_true, n, reps = 0.40, 300, 1000
captured = 0
examples = []
for i in range(reps):
    phat = sum(random.random() < p_true for _ in range(n)) / n
    me = 1.96 * math.sqrt(phat * (1 - phat) / n)
    hit = phat - me <= p_true <= phat + me
    captured += hit
    if i < 5:
        examples.append(f"({phat - me:.3f}, {phat + me:.3f}) {'captures' if hit else 'MISSES'} 0.40")
print("\n".join(examples))
print(f"intervals capturing the true p: {captured} of {reps} = {captured / reps:.1%}")

(0.364, 0.476) captures 0.40
(0.374, 0.486) captures 0.40
(0.348, 0.459) captures 0.40
(0.358, 0.469) captures 0.40
(0.315, 0.425) captures 0.40
intervals capturing the true p: 938 of 1000 = 93.8%

In this run about 94 percent of the intervals captured the true proportion, close to the advertised 95 percent. The simple interval tends to fall slightly short of its nominal level in moderate samples, one reason statisticians sometimes prefer adjusted versions, but the idea is the same: the confidence level is a long-run success rate.

Common misreadings are worth naming. A 95 percent confidence interval does not mean that 95 percent of the individuals fall in the interval; it is about a parameter, such as a mean, not about individual values. It does not mean there is a 95 percent probability that this particular interval contains the parameter in the sense of a coin not yet tossed; the parameter is fixed and the interval either contains it or not. And it says nothing about bias: if the sample was badly chosen, the interval can be confidently wrong.

A confidence interval for a proportion

For a population proportion p, the estimate is the sample proportion \(\hat{p}\), and its standard error is \(SE = \sqrt{\hat{p}(1 - \hat{p})/n}\). The 95 percent confidence interval is \(\hat{p} \pm 1.96\,SE\). The critical value 1.96 comes from the standard normal distribution: 95 percent of the area lies between -1.96 and 1.96. For other confidence levels use 1.645 for 90 percent and 2.576 for 99 percent.

The interval is reliable when the sample is a random sample from a population at least ten times larger and when there are at least 10 successes and 10 failures in the sample. With smaller counts, an adjusted method, such as adding two successes and two failures before computing (sometimes called the plus-four method), works much better.

# 95% confidence interval for a proportion: 248 of 400 residents support a new bus route.
import math
from statistics import NormalDist
x, n = 248, 400
phat = x / n
se = math.sqrt(phat * (1 - phat) / n)
for level in (0.90, 0.95, 0.99):
    z = NormalDist().inv_cdf(1 - (1 - level) / 2)
    print(f"{level:.0%} CI: {phat:.3f} +/- {z:.3f} x {se:.4f} = ({phat - z * se:.3f}, {phat + z * se:.3f})")

90% CI: 0.620 +/- 1.645 x 0.0243 = (0.580, 0.660)
95% CI: 0.620 +/- 1.960 x 0.0243 = (0.572, 0.668)
99% CI: 0.620 +/- 2.576 x 0.0243 = (0.557, 0.683)

In context: we are 95 percent confident that between about 57 and 67 percent of all residents support the new route. Note the trade-off between confidence and precision: a 99 percent interval is wider than a 95 percent one, which is wider than a 90 percent one. You cannot have high confidence and a narrow interval without collecting more data.

Polls and the margin of error

For a poll, the margin of error is the plus-or-minus part of the interval. Pollsters usually report the margin for a sample proportion of 0.5, because that gives the largest possible margin for a given sample size: \(1.96\sqrt{0.5 \times 0.5/n} \approx 1/\sqrt{n}\). That is why a poll of about 1,000 people reports a margin of error of about plus or minus 3 percentage points, 19 times out of 20, and a poll of about 2,500 reports about 2 points.

Two cautions apply. The margin of error for a subgroup, such as young adults or one province or state, is much larger than for the whole sample, because the subgroup sample is smaller. And the margin of error covers only random sampling variability. It does not cover non-response bias, coverage problems, question wording or last-minute changes of opinion, which is why polls sometimes miss by more than their stated margin. Many online panels are not random samples at all; reputable pollsters that use them describe their results with a credibility interval or similar term and explain their weighting.

A confidence interval for a mean

For a population mean \(\mu\), the estimate is the sample mean \(\bar{x}\), and its standard error is \(SE = s/\sqrt{n}\), using the sample standard deviation s because we do not know the population standard deviation \(\sigma\). Replacing \(\sigma\) with s adds extra uncertainty, especially in small samples. To account for it, we use a critical value from the t distribution instead of the normal distribution.

The t distributions, discovered by William Gosset while working on quality control at the Guinness brewery in Dublin in the early 1900s, look like the standard normal curve but with heavier tails. There is a different t distribution for each sample size, identified by its degrees of freedom, df = n - 1. With few degrees of freedom the tails are much heavier and the critical values larger; as n grows, the t distribution approaches the standard normal. For df = 9 the 95 percent critical value is 2.262; for df = 29 it is 2.045; for df = 99 it is 1.984, already close to 1.96.

The one-sample t interval is \(\bar{x} \pm t^*\,\frac{s}{\sqrt{n}}\), where \(t^*\) is the critical value with n - 1 degrees of freedom. It requires a random sample and a population distribution that is roughly normal, or a large enough sample for the central limit theorem to apply. In practice: for n under 15, use it only if the data look close to normal with no outliers; for n of 15 to 40, it is fine unless there are outliers or strong skewness; for n over 40, it works well even for skewed data.

# One-sample t interval for the mean commute time (invented random sample of 400 residents).
import random, statistics as st, math
from scipy import stats            # scipy.stats provides t-distribution critical values
random.seed(141)
times = [random.gammavariate(3.2, 7.7) for _ in range(400)]
n, xbar, s = len(times), st.mean(times), st.stdev(times)
se = s / math.sqrt(n)
tstar = stats.t.ppf(0.975, df=n - 1)
print(f"n = {n}, sample mean = {xbar:.2f} min, s = {s:.2f} min, SE = {se:.3f}")
print(f"t* (df = {n - 1}) = {tstar:.3f}")
print(f"95% CI for the mean commute: ({xbar - tstar * se:.2f}, {xbar + tstar * se:.2f}) minutes")
# A small sample of 12 gives a much wider interval
small = times[:12]
t12 = stats.t.ppf(0.975, df=11)
m12, se12 = st.mean(small), st.stdev(small) / math.sqrt(12)
print(f"with only 12 residents: t* = {t12:.3f}, 95% CI = ({m12 - t12 * se12:.1f}, {m12 + t12 * se12:.1f}) minutes")

n = 400, sample mean = 25.13 min, s = 13.93 min, SE = 0.696
t* (df = 399) = 1.966
95% CI for the mean commute: (23.76, 26.50) minutes
with only 12 residents: t* = 2.201, 95% CI = (17.3, 31.0) minutes

The survey of 400 pins down the mean commute to within about one and a half minutes either way. The same calculation with only 12 residents gives an interval several times as wide, both because the standard error is larger and because the critical value is larger. Note that commute times are skewed to the right, but with 400 observations the central limit theorem makes the interval trustworthy; with 12, the skewness would be a real concern.

Choosing a sample size

Before collecting data, you can decide how large a sample you need to achieve a desired margin of error. For a proportion, solving \(m = z^*\sqrt{p(1-p)/n}\) for n gives \(n = \left(\frac{z^*}{m}\right)^2 p(1 - p)\). If you have no idea what p is, use 0.5, which gives the largest, safest answer. For a mean, \(n = \left(\frac{z^* \sigma}{m}\right)^2\), using an estimate of \(\sigma\) from a pilot study or earlier research. Always round up.

# Sample size needed for a desired margin of error (95% confidence).
import math
z = 1.96
for m in (0.05, 0.03, 0.02, 0.01):
    n = math.ceil((z / m) ** 2 * 0.25)
    print(f"proportion, margin +/- {m:.0%}: n = {n:,}")
sigma = 15          # minutes, rough SD of commute times from a pilot
for m in (3, 2, 1):
    print(f"mean commute, margin +/- {m} min: n = {math.ceil((z * sigma / m) ** 2):,}")

proportion, margin +/- 5%: n = 385
proportion, margin +/- 3%: n = 1,068
proportion, margin +/- 2%: n = 2,401
proportion, margin +/- 1%: n = 9,604
mean commute, margin +/- 3 min: n = 97
mean commute, margin +/- 2 min: n = 217
mean commute, margin +/- 1 min: n = 865

The square-root law from lesson 9 appears again: each halving of the margin of error roughly quadruples the sample size and, with it, the cost. A city planning a transit survey might reasonably settle for a margin of plus or minus 5 percentage points with about 400 respondents rather than plus or minus 1 point with nearly 10,000.