A Worked Correlation Interval Using Fisher’s Z

Why Work on a Transformed Scale?

A sample correlation $r$ always lies between $-1$ and $1$, while its sampling variability depends on the population correlation $\rho$. In particular, an interval built directly around $r$ using one fixed standard error can behave poorly near the boundaries. Fisher’s transformation converts correlations to an unbounded scale where a useful normal approximation has a nearly constant standard error. The interval is constructed there, then its two endpoints are transformed back to the correlation scale.

The result is a practical, approximate confidence interval—not an exact finite-sample guarantee. The calculations below use the usual Fisher approximation for independent observations from a bivariate normal population.

Definition and Interval Formula

Definition: For $-1<x<1$, Fisher’s $z$ transformation is $$ g(x)=\operatorname{atanh}(x)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right). $$ Its inverse is $g^{-1}(u)=\tanh(u)$. For a sample correlation $r$, the transformed estimate is $g(r)$; for population correlation $\rho$, the transformed parameter is $g(\rho)$.

Suppose $(X_i,Y_i)$, $i=1,\ldots,n$, are independent observations from a bivariate normal distribution with nonzero marginal variances and population correlation $\rho$, where $-1<\rho<1$. For $n>3$, the usual Fisher approximation treats $g(r)$ as approximately normal with mean $g(\rho)$ and standard error $1/\sqrt{n-3}$. This approximation is most useful when the sample is adequate for the normal approximation; $n>3$ makes the formula defined, but does not by itself ensure accurate coverage.

Approximate Fisher $z$ interval: For a two-sided confidence level $1-\alpha$, let $z_{1-\alpha/2}$ be the corresponding standard normal quantile. Calculate $$ g(r)\ \pm\ \frac{z_{1-\alpha/2}}{\sqrt{n-3}}, $$ then apply $\tanh$ separately to the lower and upper endpoints. The resulting interval estimates $\rho$.

The standard error belongs to the transformed scale. It is not a standard error for $r$, and it should not be used to add and subtract a margin directly from $r$. Keeping the two scales separate is essential to getting the interval right.

Worked Calculations

Worked Example: A 95% Interval When the Sample Correlation Is 0.80

Suppose $n=20$ independent bivariate-normal observations give $r=0.80$. Construct the usual approximate 95% interval for $\rho$.

Transform the estimate. $$ g(r)=\operatorname{atanh}(0.80) =\frac12\log\left(\frac{1.80}{0.20}\right) =\frac12\log(9) \approx 1.098612. $$

Find the standard error and margin. Here $n-3=17$, so the transformed-scale standard error is $$ \frac{1}{\sqrt{17}}\approx 0.242536. $$ Using $z_{0.975}=1.96$, the margin is $1.96/\sqrt{17}\approx 0.475370$. Thus the transformed endpoints are $$ 1.098612-0.475370=0.623242,\qquad 1.098612+0.475370=1.573982. $$

Transform back. Applying $\tanh$ to each endpoint gives $$ \tanh(0.623242)\approx 0.553,\qquad \tanh(1.573982)\approx 0.918. $$ The approximate 95% confidence interval is therefore $(0.553,\,0.918)$. The endpoints are not found by adding and subtracting $0.475370$ from $0.80$; that margin is on the transformed scale.

Worked Example: A Negative Sample Correlation

Suppose $n=30$ independent bivariate-normal observations give $r=-0.40$. Find the usual approximate 90% interval for $\rho$.

The transformed estimate is $$ g(-0.40)=\frac12\log\left(\frac{0.60}{1.40}\right) \approx -0.423649. $$ The standard error is $1/\sqrt{30-3}=1/\sqrt{27}\approx 0.192450$. Using $z_{0.95}\approx 1.645$, the margin is $$ 1.645(0.192450)\approx 0.316580. $$ The transformed interval is $$ -0.423649-0.316580=-0.740229,\qquad -0.423649+0.316580=-0.107069. $$

Back-transforming gives $$ \tanh(-0.740229)\approx -0.629,\qquad \tanh(-0.107069)\approx -0.107. $$ So the approximate 90% interval is $(-0.629,\,-0.107)$. Both limits remain negative, consistent with this transformed interval’s location relative to zero.

Worked Example: A More Moderate Correlation

Suppose $n=50$ independent bivariate-normal observations give $r=0.20$. Construct the usual approximate 95% interval.

First, $$ g(0.20)=\frac12\log\left(\frac{1.20}{0.80}\right) =\frac12\log(1.5) \approx 0.202733. $$ The standard error is $1/\sqrt{47}\approx 0.145865$, and the 95% margin is $$ 1.96(0.145865)\approx 0.285895. $$ Therefore the transformed limits are $0.202733-0.285895=-0.083162$ and $0.202733+0.285895=0.488628$.

Applying the inverse transformation gives $$ \tanh(-0.083162)\approx -0.083,\qquad \tanh(0.488628)\approx 0.453. $$ The approximate 95% interval is $(-0.083,\,0.453)$. It includes zero, so these data do not rule out a zero population correlation at this approximate confidence level. That statement concerns the interval’s location; it does not make the approximation exact.

Why the Back-Transformation Works

The transformed interval is intended to cover $g(\rho)$. To turn it into limits for $\rho$, the endpoints must be mapped back in the correct order. The following fact justifies that step and clarifies what the resulting interval covers.

Theorem: Let $L$ and $U$ be finite transformed-scale limits with $L\le U$. For any $-1<\rho<1$, $$ L\le g(\rho)\le U \quad\Longleftrightarrow\quad \tanh(L)\le \rho\le\tanh(U). $$ Consequently, if a transformed-scale interval covers $g(\rho)$, its back-transformed interval covers $\rho$ on exactly the same outcomes.

Proof. Differentiating $g(x)$ for $-1<x<1$ gives $$ g'(x)=\frac{1}{1-x^2}>0. $$ Thus $g$ is strictly increasing, and its inverse $\tanh$ is also strictly increasing. Applying this increasing inverse to $L\le g(\rho)\le U$ preserves both inequalities and gives $\tanh(L)\le \rho\le\tanh(U)$. Conversely, applying the increasing function $g$ to the latter inequalities recovers $L\le g(\rho)\le U$, since $g(\tanh(u))=u$. The coverage events are therefore identical. $\square$

This equivalence explains why the transformed interval’s approximate coverage carries over to the back-transformed interval: the inverse transformation does not change which parameter values are covered on a given outcome. It does not make the normal approximation exact. The coverage is still approximate because the interval on the transformed scale is based on an approximate sampling distribution.

Useful Checks and Interpretation

Result: For a finite two-sided Fisher interval centered at $g(r)$, the back-transformed interval is ordered and contains $r$.

Derivation. Write $m=z_{1-\alpha/2}/\sqrt{n-3}$, which is positive for the usual two-sided confidence levels. The transformed limits are $g(r)-m$ and $g(r)+m$, so the lower limit is no larger than the upper limit. Since $\tanh$ is increasing, their back-transformed limits are also ordered. Further, $$ g(r)-m\le g(r)\le g(r)+m. $$ Applying $\tanh$ throughout gives $$ \tanh(g(r)-m)\le r\le\tanh(g(r)+m), $$ because $\tanh(g(r))=r$. Thus the interval contains the observed sample correlation. This is a useful arithmetic check, not a claim that the interval is narrow or that its coverage is exact.

The inverse transformation also explains why the interval need not be symmetric around $r$. Equal distances on the transformed scale generally do not correspond to equal distances on the correlation scale. In the first example, the limits $0.553$ and $0.918$ are not equally far from $0.80$, even though the transformed limits were centered at $g(0.80)$.

Common Mistakes and Limitations

Using the transformed margin on the original scale. The margin $z_{1-\alpha/2}/\sqrt{n-3}$ is added to and subtracted from $g(r)$, not from $r$. Apply $\tanh$ to both transformed endpoints afterward.
Back-transforming only the estimate. A confidence interval requires transforming each endpoint separately. Transforming the center alone does not produce interval limits.
Assuming every $n$ works. The stated standard error is real only for $n>3$. Even when $n>3$, the normal approximation may be inadequate for a small sample, so the nominal confidence level need not equal the actual coverage.
Treating the interval as exact. The construction relies on the approximate sampling distribution of Fisher’s transformed sample correlation under the bivariate-normal model.
Ignoring boundary cases. The transformation is finite only for $-1<r<1$. If the observed correlation is exactly $1$ or $-1$, $g(r)$ is infinite and this usual calculation cannot be applied as written.

In practice, report the interval as an approximate interval and state the sample correlation, sample size, confidence level, and modeling assumption. These details make the calculation interpretable and prevent a useful approximation from being mistaken for an exact result.

Check Your Understanding

Answer each question using the scale and assumptions relevant to the Fisher interval.

For $n=40$ and $r=0.50$, write the transformed estimate and approximate standard error before calculating any confidence limits.
Why must the normal margin be applied to $g(r)$ rather than directly to $r$?
Explain why back-transforming both transformed endpoints preserves the coverage event for $\rho$.
What does $n>3$ guarantee about the standard-error formula, and what does it not guarantee about coverage?
Why can a valid back-transformed Fisher interval be asymmetric around the observed sample correlation?