Estimation examples

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Estimation examples

Part Data analysis
Chapter 4 of 9 in this part
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Questions 7
Definitions Margin of error · Welch interval

The definitions of an estimator, maximum likelihood, bias, and a confidence interval are in Estimation and Confidence intervals. A confidence interval is not a credible interval: after the sample is observed the endpoints are numbers, and the confidence is a property of the procedure. This page works the procedures on data, with the R that does the arithmetic, and adds the one interval the inference chapters leave out: Welch’s interval for two means with unequal variances.

From data to an estimate

Example 1 (A biased coin) A coin is tossed 1000 times and shows 570 heads.

Question. Estimate the probability of heads, and say whether p=0.5 is plausible.

Solution. Model the count as X\sim\operatorname{Bin}(1000,p) with p unknown. The probability p will never be known; the data give an estimate, \widehat p=570/1000=0.570. Its standard error is \sqrt{0.57\times0.43/1000}=0.0157, and the approximate 95% interval is

0.570\pm1.96\times0.0157=(0.539,\ 0.601).

R’s exact binomial interval agrees to three decimal places:

binom.test(570, 1000)$conf.int     # 0.5386 0.6009

The interval excludes 0.5, so a fair coin is not a plausible explanation of 570 heads.

Example 2 (Ten normal measurements) Ten measurements, modelled as a \operatorname{N}(\mu,\sigma^2) sample, are

77.01,\ 71.37,\ 77.15,\ 79.89,\ 76.46,\ 78.10,\ 77.18,\ 74.08,\ 75.88,\ 72.63.

Question. Estimate \mu and \sigma, and give a 95% interval for \mu.

Solution.

x <- c(77.01, 71.37, 77.15, 79.89, 76.46, 78.10, 77.18, 74.08, 75.88, 72.63)
mean(x); sd(x)              # 75.975  2.581
t.test(x)$conf.int          # 74.13  77.82

The estimates are \widehat\mu=\overline x=75.975 and s=2.58. The t interval uses t_{9;0.975}=2.262: 75.975\pm2.262\times2.58/\sqrt{10}=(74.13,\ 77.82). With only ten values the t quantile matters; the normal quantile 1.96 would give an interval about 13% too short.

Large samples

When the data are not known to be normal, the central limit theorem still makes \overline X approximately normal for a large sample, and s is close to \sigma.

Definition 1 (Margin of error) For a random sample with mean \mu and finite variance, the approximate confidence interval for \mu with confidence 1-\alpha is \overline X\pm z_{1-\alpha/2}\,s/\sqrt n. Its half-width, z_{1-\alpha/2}\,s/\sqrt n, is the margin of error.

Article: Margin of error →

The approximation needs care when the sample is small or the data are heavily skewed, which is what a histogram or a normal quantile plot checks. For a household fuel survey with n=51, \overline x=1136 and s=178 dollars, the approximate 95% interval is 1136\pm1.96\times178/\sqrt{51}=(1087,\ 1185). The exact normal-theory interval, with t_{50;0.975}=2.009, is (1086,\ 1186); at n=51 the two hardly differ.

The margin of error shrinks like 1/\sqrt n, so planning a sample size is a matter of solving for n: halving the margin takes four times the data. The calculation is in Confidence intervals.

Small normal samples

Example 3 (The volume of a drop) A burette reads 25.36 mL, then 22.84 mL after 50 drops, then 20.36 mL after another 50.

Question. Give a 95% interval for the volume of one drop.

Solution. The two batches of 50 drops have volumes x_1=2.52 and x_2=2.48 mL. Model them as a \operatorname{N}(\mu,\sigma^2) sample of size n=2, where \mu is the expected volume of 50 drops. Then \overline x=2.50, s=0.02\sqrt2, and the standard error is s/\sqrt2=0.02. With one degree of freedom t_{1;0.975}=12.71, so the interval for \mu is

2.50\pm12.71\times0.02=(2.25,\ 2.75).

Dividing by 50, a monotone change of scale, gives a 95% interval for one drop of (0.045,\ 0.055) mL. The huge quantile 12.71 is the price of estimating a standard deviation from two numbers.

The t and chi-squared quantiles come from R:

qt(0.975, df = c(1, 2, 5, 50, 100))   # 12.71  4.30  2.57  2.01  1.98
qchisq(c(0.025, 0.975), df = 24)      # 12.40  39.36

The t quantiles fall quickly towards the normal 1.96 as the degrees of freedom grow. The chi-squared quantiles give the interval for a variance, as in the aspirin example of Confidence intervals.

Two samples with unequal variances

The pooled interval for \mu_X-\mu_Y assumes the two populations have the same variance. Without that assumption, \overline X-\overline Y is still approximately normal with variance \sigma_X^2/n_1+\sigma_Y^2/n_2, and replacing each variance by its sample estimate gives a statistic close to a t distribution with fractional degrees of freedom.

Definition 2 (Welch interval) For independent normal samples of sizes n_1 and n_2 with sample variances s_X^2 and s_Y^2, the Welch interval for \mu_X-\mu_Y with approximate confidence 1-\alpha is

\overline x-\overline y\pm t_{\mathrm{df};\,1-\alpha/2}\sqrt{\frac{s_X^2}{n_1}+\frac{s_Y^2}{n_2}}, \qquad \mathrm{df}=\frac{\bigl(s_X^2/n_1+s_Y^2/n_2\bigr)^2}{\dfrac{(s_X^2/n_1)^2}{n_1-1}+\dfrac{(s_Y^2/n_2)^2}{n_2-1}}.

The degrees of freedom are the Welch–Satterthwaite approximation.

Article: Welch interval →

The effective degrees of freedom lie between the smaller of n_1-1 and n_2-1 and the pooled n_1+n_2-2. For non-normal data the interval is still approximately right when both samples are reasonably large, say 30 or more; for small, skewed samples a rank-based method such as the Wilcoxon rank-sum test is safer.

Example 4 (First- and third-year weights) 15 first-year students have mean weight 62.0 kg and standard deviation 15 kg; 10 third-year students have mean 71.5 kg and standard deviation 12 kg.

Question. Give a 95% interval for the difference in mean weights without assuming equal variances.

Solution. The standard error is \sqrt{15^2/15+12^2/10}=5.422, and the effective degrees of freedom are 22.10, so t_{22.10;0.975}=2.073. The interval is

-9.5\pm2.073\times5.422=(-20.74,\ 1.74).

It contains 0, so the data do not show a difference in mean weight. The pooled interval, which assumes equal variances, is (-21.24,\ 2.24) on 23 degrees of freedom, a little wider here because the larger sample has the larger variance. With the raw data, t.test(x, y) gives the Welch interval, R’s default, and t.test(x, y, var.equal = TRUE) the pooled one.

Proportions

For a count X\sim\operatorname{Bin}(n,p), the Wald interval \widehat p\pm z_{1-\alpha/2}\sqrt{\widehat p(1-\widehat p)/n} is reasonable when there are at least 5 successes and 5 failures. In a poll of 1000 voters, 227 support a party: \widehat p=0.227, comfortably above the threshold, and the 95% interval is (0.20,\ 0.25). For two groups, the interval for p_X-p_Y adds the two variances. For the nightmare survey, 55 of 160 men and 60 of 192 women, it is (-0.067,\ 0.130).

R’s prop.test does these calculations, but by default it applies a continuity correction, which widens the interval slightly:

prop.test(c(55, 60), c(160, 192), correct = FALSE)$conf.int   # -0.067  0.130

With correct = FALSE it reproduces the Wald interval for a difference.

Key idea

An estimate is a number; its interval says how far it could plausibly be from the parameter. Use the t quantile for a small normal sample, the normal quantile for a large one, Welch’s interval when two variances may differ, and check the success and failure counts before trusting an interval for a proportion.

Questions on this topic

Open a question to attempt it. Its worked answer is folded away at the bottom of the page.

Question Answer
Ten normal observations ✓
Household spending on fuel ✓
Narrowing a pilot interval ✓
Printer paper ✓
Hours of sleep ✓
A difference of proportions ✓
Variance of a fill volume ✓

See also