One-Sample Inference Across Donors

Why the Donor Is the Sample

Single-cell measurements can provide detailed information about cells, but a one-sample inference across donors asks a different question: what is the average donor-level quantity in a population of donors? First, each donor contributes one summary, such as a donor-specific mean expression value or the proportion of that donor’s cells expressing a marker. The summaries—not the individual cells used to construct them—are the observations for this inference.

Let \(x_i\) be the summary for donor \(i\), for \(i=1,\ldots,n\). The sample size \(n\) is the number of donors with summaries. The target \(\mu_D\) is the population mean of the donor-level quantity as defined by the study’s estimand. For example, if each \(x_i\) is a donor’s proportion of marker-positive cells, \(\mu_D\) is the mean of those proportions across the target donor population. It is not automatically the proportion among all cells pooled together.

Definition: A one-sample t interval estimates a population mean \(\mu_D\) from independent donor-level observations \(x_1,\ldots,x_n\). A one-sample t test evaluates a null hypothesis about that mean, commonly \(H_0:\mu_D=\mu_0\), where \(\mu_0\) is a specified reference value. In both procedures, \(n\) counts donors.

Assumptions and the One-Sample t Procedures

The exact one-sample t procedures assume that the donor summaries are a random or otherwise representative sample from the target donor population, that different donors’ summaries are independent, and that the donor-level distribution is normal. These conditions concern the \(x_i\), not the distribution of individual cells. If donor summaries are not exactly normal, the t procedures may still be useful approximations when the donor count is adequate and the distribution is not strongly skewed or dominated by outliers. With only a few donors, that approximation is difficult to assess and can be fragile.

Calculate the sample mean \(\bar{x}\) and sample standard deviation \(s\) across donors. The standard error estimates the sampling variability of \(\bar{x}\) under the model. For a two-sided confidence interval with confidence level \(1-\alpha\), use the \(t\) critical value with \(n-1\) degrees of freedom:

\bar{x}\ \pm\ t_{1-\alpha/2,\ n-1}\frac{s}{\sqrt{n}}.

For a test of \(H_0:\mu_D=\mu_0\), the statistic is

t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}},\qquad df=n-1.

For a two-sided alternative \(H_a:\mu_D\ne\mu_0\), the p-value is the probability, under \(H_0\) and the model assumptions, of obtaining a t statistic at least as far from zero as the observed statistic. A small p-value is evidence against the null model; it is not the probability that the null is true.

Why the t Reference Distribution Applies

The justification makes explicit why the donor count enters both the standard error and the degrees of freedom. Under the exact model, the donor summaries are independent and \(X_i\sim N(\mu_D,\sigma^2)\). Therefore, the standardized sample mean \((\bar{x}-\mu_D)/(\sigma/\sqrt{n})\) has a standard normal distribution. Also, \((n-1)s^2/\sigma^2\) has a chi-square distribution with \(n-1\) degrees of freedom and is independent of the sample mean. Replacing the unknown \(\sigma\) with \(s\) forms a ratio of these quantities:

\frac{\bar{x}-\mu_D}{s/\sqrt{n}} = \frac{(\bar{x}-\mu_D)/(\sigma/\sqrt{n})} {\sqrt{\{(n-1)s^2/\sigma^2\}/(n-1)}} \sim t_{n-1}.

This pivot gives the test statistic under \(H_0\). For the interval, the central \(1-\alpha\) probability statement is \(P(-t_{1-\alpha/2,n-1}\le(\bar{x}-\mu_D)/(s/\sqrt{n})\le t_{1-\alpha/2,n-1})=1-\alpha\). Rearranging the inequalities to isolate \(\mu_D\) gives the stated interval. Its coverage is exact under the normal, independent donor model.

Worked Example: A 95% Interval for Mean Donor-Level Expression

Suppose six hypothetical donors each contribute one summary: the mean log2-normalized expression of a gene in a prespecified cell type. The values are 0.12, 0.18, 0.09, 0.21, 0.16, and 0.20 log2-expression units. Assume these donors are independent and representative, and that the donor-level summaries are reasonably modeled by a normal distribution.

The donor count is \(n=6\), not the number of cells measured. The mean is \(\bar{x}=(0.12+0.18+0.09+0.21+0.16+0.20)/6=0.1600\) log2-expression units. The squared deviations from the mean sum to 0.0110, so \(s=\sqrt{0.0110/(6-1)}=\sqrt{0.0022}=0.04690\) log2-expression units. Thus \(SE=s/\sqrt{n}=0.04690/\sqrt{6}=0.01915\) log2-expression units.

For a 95% interval, \(df=5\) and \(t_{0.975,5}\approx2.571\). Substitution gives

0.1600\pm 2.571(0.01915) =0.1600\pm0.04922 =(0.1108,\ 0.2092).

Under the assumptions, this procedure gives a 95% confidence interval for the mean donor-level expression in the target donor population, from 0.1108 to 0.2092 log2-expression units. The interval describes uncertainty about the population mean; it does not say that 95% of donors’ expression values lie within these endpoints.

Worked Example: Testing a Reference Proportion

For six hypothetical independent donors, suppose the donor-level proportions of cells expressing a marker are 0.20, 0.30, 0.40, 0.40, 0.50, and 0.60. Test \(H_0:\mu_D=0.30\) against \(H_a:\mu_D\ne0.30\) at \(\alpha=0.05\). Here, \(\mu_D\) is the mean donor-specific proportion, not a cell-weighted pooled proportion. Assume the donors represent the target population and use a normal model as a working approximation for the donor summaries.

The mean is \(\bar{x}=2.40/6=0.4000\). The squared deviations from 0.40 sum to 0.1000, so \(s=\sqrt{0.1000/5}=0.1414\), and \(SE=0.1414/\sqrt{6}=0.05774\). The observed statistic is

t=\frac{0.4000-0.3000}{0.05774}=1.732,\qquad df=5.

The two-sided p-value is approximately 0.1438, which is greater than 0.05. The test does not reject \(H_0\) at the 5% level: these six donor summaries do not provide sufficient evidence of a difference between the population mean donor proportion and 0.30. This result does not establish that the mean is exactly 0.30. With only six donors, the normal approximation also deserves caution; bounded proportions and influential donors can make the t result sensitive to the data’s shape.

Worked Example: Evidence Against a Reference Mean

Consider five hypothetical independent donors with donor-level gene summaries of 1.2, 1.3, 1.4, 1.5, and 1.6 log2-expression units. Test \(H_0:\mu_D=1.0\) against \(H_a:\mu_D\ne1.0\). Assume representative sampling and an approximately normal donor-level distribution.

The mean is \(\bar{x}=7.0/5=1.400\). The squared deviations from 1.4 sum to 0.1000, so \(s=\sqrt{0.1000/4}=0.1581\), and \(SE=0.1581/\sqrt{5}=0.07071\). Consequently,

t=\frac{1.400-1.000}{0.07071}=5.657,\qquad df=4.

The two-sided p-value is approximately 0.004812. Because this is below 0.05, reject \(H_0\) at the 5% level. Under the assumptions, the data provide evidence that the population mean donor-level summary differs from 1.0 log2-expression units. The test alone does not show that a change in expression causes a biological outcome, and its small donor sample makes the normality assumption important.

Diagnostics, Alternatives, and Limitations

A t calculation does not repair a mismatch between the study design and the inference. Check that each donor contributes only one observation to this analysis and that the donors can reasonably be treated as independent. Donors recruited from a narrow or nonrepresentative group may not support inference to a broader population, regardless of the p-value. Inspect a dot plot of the donor summaries for skewness, gaps, and outliers. With few donors, such plots reveal conspicuous problems but cannot certify normality.

The t procedure gives equal weight to each donor summary. That is appropriate when the estimand is the average donor-level quantity and the donor summaries are the observations of interest. It can be a poor approximation if some summaries are much noisier because of very different numbers of cells, or if measurement error varies substantially across donors. More cells may improve a donor’s summary, but they do not create more independent donors. A cell-level test that counts cells as independent observations instead uses the wrong sample size when cells within donors are correlated, and can understate uncertainty.

The one-sample t test is often preferable to a z test because the population standard deviation \(\sigma\) is usually unknown and must be estimated by \(s\). A z procedure presumes a known \(\sigma\), a condition rarely justified for donor summaries. When there are severe outliers or strong skewness, a t procedure may be unreliable; a robust or nonparametric procedure may be considered, but it answers inference under different assumptions and is not automatically a solution to few donors. If the scientific question compares two groups, a one-sample procedure against a fixed reference does not replace an analysis designed for that comparison.

For a single prespecified outcome, the reported p-value pertains to that one test. If many genes or outcomes are tested, interpreting each unadjusted p-value at the same threshold does not control the overall rate of false positives. The mean difference \(\bar{x}-\mu_0\), in the outcome’s units, should also be reported so that statistical evidence is not mistaken for biological importance.

Common Mistakes

Using the cell count as \(n\). The wrong calculation treats every cell as an independent observation. Use the number of independent donors whose summaries enter the analysis.
Interpreting a nonsignificant result as proof of no difference. Failure to reject means the data did not provide sufficient evidence against the specified null at the chosen level; it does not establish equality.
Reading a confidence interval as a range containing most donors. The t interval estimates the population mean, not the spread of individual donor values.
Assuming the formula guarantees validity. Independence, a relevant donor population, and a suitable donor-level distribution remain substantive requirements.

Key takeaway: A one-sample t interval or test summarizes uncertainty across independent donor-level observations. The number of donors determines \(n\) and \(df\); the cell count does not.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

Eight donors each contribute one summary, based on different numbers of cells. What is \(n\) for a one-sample t analysis?
For \(n=4\), \(\bar{x}=2.3\), \(s=0.4\), and \(H_0:\mu_D=2.0\), calculate the t statistic and degrees of freedom.
A 95% interval for a donor-level mean proportion is 0.4278 to 0.6122. What does this interval estimate, and what does it not claim?
What concern arises if the donors were recruited from one unusually selected clinic but the conclusion is intended to apply to all donors?

Solutions. (1) \(n=8\), because the eight donors supply the independent summaries; unequal cell counts do not change this count. (2) \(SE=0.4/\sqrt{4}=0.2\), so \(t=(2.3-2.0)/0.2=1.5\), with \(df=4-1=3\). (3) It estimates the population mean donor-level proportion under the model; it is not an interval containing 95% of donors’ proportions. (4) The sample may not represent the intended donor population, limiting generalization even if the t calculation is correct.