From a Genotype Slope to Uncertainty
A regression slope for genotype summarizes an estimated trait difference associated with a change in genotype dosage. It does not, by itself, show how precisely that difference has been estimated. Confidence intervals describe the range of effect sizes compatible with the data and model, while Wald tests assess whether the data provide evidence against a specified slope, often zero. These are related tools, but they answer different questions: one emphasizes plausible magnitudes, and the other measures evidence against a null value.
For an additive dosage analysis, let \(G_i\) be the number of effect alleles carried by participant \(i\), coded as 0, 1, or 2, and let \(Y_i\) be that participant’s quantitative trait. The simple linear model is
Here, \(\beta\) is the population mean change in the trait for each additional effect allele under the model. Its units are trait units per allele. The estimated slope, \(\hat{\beta}\), is the fitted association estimate, and \(SE(\hat{\beta})\) is its standard error. A positive estimate indicates a higher modeled mean trait value with each additional effect allele; a negative estimate indicates a lower modeled mean.
The standard error quantifies sampling uncertainty in the slope estimate; it is not the standard deviation of individual trait values. A small standard error means the estimate is relatively precise under the fitted model. The estimated slope and its standard error are the key inputs to the interval and test.
Model Conditions and the Inference Formulas
For the formulas below, assume the simple linear model is appropriate for the mean trait across genotype dosages, genotype dosage varies in the sample, and participants’ errors are independent with mean zero and constant variance. For the exact finite-sample \(t\) procedure, also assume normally distributed errors. In a genome-wide analysis, relatedness, population structure, or other systematic differences may require a different or adjusted model; an unadjusted slope need not represent a causal effect.
In the simple model with an intercept and one genotype predictor, the residual degrees of freedom are \(df=n-2\). The conventional two-sided \(100(1-\alpha)\%\) confidence interval is
The critical value \(t_{1-\alpha/2,\,df}\) is the quantile of the \(t\) distribution with \(df\) degrees of freedom that leaves probability \(\alpha/2\) in its upper tail. For example, a 95% interval uses \(\alpha=0.05\). In sufficiently large samples, the corresponding standard-normal critical value is often used as an approximation; the \(t\) formula is the exact conventional procedure under the stated normal-error assumptions.
To test a specified null slope \(\beta_0\), the Wald statistic is
For the usual test of no slope, \(H_0:\beta=0\), with a two-sided alternative \(H_a:\beta\ne 0\), this reduces to \(\hat{\beta}/SE(\hat{\beta})\). Under the classical linear model, this statistic follows a \(t\) distribution under \(H_0\), with \(df=n-2\). The two-sided p-value is the probability, assuming \(H_0\) and the model, of obtaining a statistic at least as far from zero as the observed one:
In large-sample settings the Wald statistic is commonly compared with a standard normal distribution, or its square with a chi-squared distribution with one degree of freedom. For a one-parameter test, squaring the statistic gives \(t_{\text{Wald}}^2\); the two-sided \(t\) test and its corresponding one-degree-of-freedom test express the same evidence in different forms.
Why the Interval and Test Agree
A two-sided test at level \(\alpha\) and a two-sided \(100(1-\alpha)\%\) confidence interval are dual procedures when they use the same estimate, standard error, model, and reference distribution. This follows by comparing the test statistic with its critical value. The null value is rejected exactly when
which is equivalent to
Rearranging, the null value lies outside the interval centered at \(\hat{\beta}\) with margin \(t_{1-\alpha/2,\,df}SE(\hat{\beta})\). Thus, for a two-sided test of \(H_0:\beta=0\) at \(\alpha=0.05\), zero is outside the 95% confidence interval exactly when the test rejects at the 0.05 level, apart from rounding.
Worked Example: A Positive Per-Allele Slope
Suppose a hypothetical study analyzes 120 unrelated participants. The quantitative trait is measured in millimeters of mercury, and dosage is the count of the effect allele. The fitted slope is \(\hat{\beta}=0.42\) mmHg per allele, with \(SE(\hat{\beta})=0.15\) mmHg per allele. Assume the model is linear across dosage, the errors are independent with roughly constant variance, and residual checks show no major departure from the normal-error assumption. Genotype dosage varies in the sample.
For a 95% interval, \(df=120-2=118\), and the \(t\) critical value is approximately 1.980. The margin of error is
The interval is therefore
Under the fitted model, the interval estimates a positive change in mean trait value of about 0.12 to 0.72 mmHg for each additional effect allele. It is a confidence interval produced by a procedure with 95% long-run coverage under its assumptions; it does not mean there is a 95% probability that this fixed slope lies in this particular calculated interval.
For the two-sided test of \(H_0:\beta=0\), the Wald statistic is
With 118 degrees of freedom, the two-sided p-value is approximately 0.006. This is evidence against a zero slope under the model. The estimate of 0.42 mmHg per allele describes the effect’s size; the p-value describes how unusual a statistic at least this extreme would be if the null and model were correct. Neither result alone establishes that the association is causal or clinically important.
Worked Example: An Imprecise Estimate Is Not Proof of No Effect
In a second hypothetical study, 100 unrelated participants have a quantitative trait measured in mmol/L. The estimated additive slope is \(0.08\) mmol/L per allele, and its standard error is \(0.10\) mmol/L per allele. Assume dosage varies, a linear mean is reasonable, and the independent, constant-variance normal-error model is adequate for these data. Here \(df=100-2=98\).
A 95% confidence interval uses a critical value of approximately 1.984. Its margin is
Thus,
For the test of \(H_0:\beta=0\), the statistic and two-sided p-value are approximately
The interval includes zero, and the test does not reject at the 0.05 level. The data are compatible with a modest negative slope, no slope, or a positive slope over the displayed range. They do not establish that the effect is exactly zero. Whether the range includes effects important for the trait depends on subject-matter context, not on the p-value alone.
Statistical Evidence and Effect Magnitude
A p-value is affected by both the estimated slope and its standard error. As sample size increases, the standard error can decrease, so a small slope may yield a small p-value. Conversely, a potentially important slope can remain uncertain in a small or noisy sample. For this reason, report the slope and its units, its confidence interval, and the test result rather than treating statistical significance as a measure of effect size.
Worked Example: Strong Evidence for a Small Slope
A hypothetical study of 10,000 unrelated participants estimates a slope of \(0.020\) trait units per allele, with standard error \(0.006\) trait units per allele. Assume dosage varies and that the linear model’s independence, constant-variance, and large-sample inference conditions are appropriate. The two-sided test of no slope gives
Using a 95% normal-based interval, with critical value 1.96, gives
This interval excludes zero and the test provides evidence against a zero slope. Yet the estimated change is only 0.020 trait units per additional allele. Whether that magnitude matters depends on the trait’s scale and the scientific or clinical question. A small p-value should not be translated into “a large effect.”
Common Mistakes and Practical Reporting
A frequent mistake is to interpret a nonsignificant result as proof that there is no genetic effect. In the second example, the p-value is not small, but the interval spans both negative and positive slopes. The correction is to describe the uncertainty and the range of effects still compatible with the model, not to accept \(H_0\) as established.
Another mistake is to describe a 95% confidence interval as a 95% probability statement about the fixed parameter, or to treat a p-value as the probability that \(H_0\) is true. The confidence level describes the long-run performance of the interval procedure. The p-value is calculated under the null hypothesis and model; it is not the probability that the null is correct.
A clear report gives the dosage coding, the trait units, the estimated slope and its standard error, the interval, and the test result. For example: “The estimated mean trait change was 0.42 mmHg per additional effect allele (95% CI 0.12 to 0.72; \(p=0.006\)).” This reports the direction, magnitude, uncertainty, and evidence without claiming causation. In large association analyses, the p-value should also be interpreted in the context of the testing plan; a nominal p-value alone does not describe the strength of evidence after accounting for many tested variants.
Check Your Understanding
Use the model and distinctions developed above to answer these questions.
- A slope is estimated as \(-0.30\) units per allele with standard error \(0.10\). What is the Wald statistic for testing a zero slope, and what does its sign indicate?
- A 95% confidence interval for a slope includes zero. What does that imply about the corresponding two-sided test at the 0.05 level when both use the same model and standard error?
- Why does a large p-value not establish that the genotype has no effect?
- State the units of an additive genotype slope when the trait is measured in centimeters and dosage counts effect alleles.
- Can a very small p-value establish that an estimated per-allele effect is practically important? Explain.