Wald interval

Definition taught in Confidence intervals, part of Statistical inference

Wald interval

Type	Definition
Taught in	Confidence intervals
Part	Statistical inference
Worked examples	4
Questions using it	1

For X\sim\operatorname{Bin}(n,p) with \widehat p=X/n, the Wald interval with approximate confidence 1-\alpha is

\widehat p\pm z_{1-\alpha/2}\sqrt{\frac{\widehat p(1-\widehat p)}{n}}.

Examples

Nightmares in two groups

Worked example from Confidence intervals

Of 160 men, 55 report frequent nightmares; of 192 women, 60 do.

Question. Give an approximate 95% confidence interval for the difference p_1-p_2 in the population proportions, and interpret it.

Solution. \widehat p_1=55/160=0.344 and \widehat p_2=60/192=0.313, a difference of 0.031. The standard error is

\sqrt{\frac{0.344\times0.656}{160}+\frac{0.313\times0.688}{192}}=0.0503,

so the interval is 0.031\pm1.96\times0.0503=(-0.067,\ 0.130). It contains 0. The data are consistent with no difference between men and women, and also with a difference of several percentage points either way.

A biased coin

Worked example from Estimation examples

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.

A proportion in a poll

Question page: A proportion in a poll

In a sample of 1000 registered voters, 227 say they will vote for a given party. Using the Wald approximation and z_{0.975}=1.960, give an approximate 95% confidence interval for the population proportion.

Worked answer (click to show)

\widehat p=227/1000=0.227. The estimated standard error is

\sqrt{\frac{0.227\times 0.773}{1000}}\approx 0.01325.

The interval is 0.227\pm 1.960\times 0.01325, that is (0.201,\,0.253).

Sizing a poll

Question page: Sizing a poll

How many voters should a poll sample so that the 95% Wald interval for a proportion has half-width at most 0.02, whatever the true proportion? How does the answer change if earlier polls show the proportion is near 0.1?

Worked answer (click to show)

The half-width is 1.96\sqrt{p(1-p)/n}. Since p(1-p)\le1/4 for every p, requiring 1.96\sqrt{1/(4n)}\le0.02 guarantees the bound:

n\ge\frac{1.96^2}{4\times0.02^2}=2401.

If p is near 0.1, then p(1-p)\approx0.09, and n\ge1.96^2\times0.09/0.02^2=865 is enough. Knowing the proportion is far from 1/2 cuts the required sample by almost two thirds, but if the guess is wrong the interval will be wider than planned.

Used in

• Taught in Confidence intervals, where the derivation and examples are.
• Also used in Estimation examples
• Questions that use it (1) — Sizing a poll
• All 6 questions on Confidence intervals

See also

Confidence interval · Pivot · One-sample t pivot · Pooled variance · Two-sample t pivot · Variance-ratio F pivot

Glossary and index