Inference for one mean

The preceding pages introduced confidence intervals, null distributions, and the Student t distribution. We now combine them for inference about one population mean. Suppose ten speakers each supply one vowel-duration difference, measured in milliseconds, between a target condition and a fixed reference condition.

D_1,\ldots,D_{10}.

The population mean difference \mu_D is the estimand, while the sample mean \overline{D} is the estimator. A one-sample t-test, together with its corresponding confidence interval, uses the sample to estimate the population mean and the uncertainty of that estimate. We first separate response spread from estimator spread, then account for estimating that spread, and only then calculate an interval and test.

Estimating the mean and response spread

For n independent speaker differences, the mean estimator is

\overline{D} =\frac{1}{n}\sum_{i=1}^nD_i.

The sample standard deviation is

s_D = \sqrt{ \frac{1}{n-1} \sum_{i=1}^n(D_i-\overline{D})^2 }.

The value s_D describes variation among speaker differences. Dividing it by \sqrt{n} gives the estimated standard error of their mean:

\widehat{\operatorname{SE}}(\overline{D}) =\frac{s_D}{\sqrt{n}}.

The distinction between these two spreads remains important: s_D concerns the responses, while s_D/\sqrt{n} concerns the estimator.

Accounting for estimating the spread

If the population standard deviation \sigma_D were known and the observations were normally distributed, standardizing with \sigma_D/\sqrt{n} would produce a standard normal variable. In practice, \sigma_D is unknown and s_D is estimated from the same data as \overline{D}.

Under independent normal sampling,

T = \frac{\overline{D}-\mu_D}{s_D/\sqrt{n}}

has a Student t distribution with

\nu=n-1

degrees of freedom. Estimating the mean imposes one constraint on the deviations because they must sum to zero:

\sum_{i=1}^n(D_i-\overline{D})=0.

Only n-1 deviations can vary freely. This is the source of the degrees of freedom in the one-sample t-test.

The t distribution introduced earlier has heavier tails than the standard normal. Those heavier tails represent the additional uncertainty from estimating the population spread. As the sample grows, s_D becomes more stable and the t distribution approaches the standard normal.

Working a confidence interval by hand

Suppose the observed summary is

n=10, \qquad \overline{d}=18\text{ ms}, \qquad s_D=20\text{ ms}.

The estimated standard error is

\widehat{\operatorname{SE}}(\overline{D}) =\frac{20}{\sqrt{10}} \approx6.3249\text{ ms}.

For a 95% interval with 9 degrees of freedom, the .975 t quantile is approximately 2.262. The interval is

\overline{d} \mathbin{\pm} t_{.975,9}\widehat{\operatorname{SE}}(\overline{D}),

which becomes

18\mathbin{\pm}2.262(6.3249) \approx[3.69,32.31]\text{ ms}.

This interval targets the population mean difference. It does not contain 95% of speaker differences.

Testing the zero reference value

For H_0:\mu_D=0, the observed statistic is

t_{\mathrm{obs}} =\frac{18-0}{6.3249} \approx2.846.

The two-sided tail probability is approximately .0192.

Code

n <- 10
mean_difference <- 18
sd_difference <- 20
se_difference <- sd_difference / sqrt(n)
df <- n - 1
t_observed <- mean_difference / se_difference
critical_value <- qt(.975, df = df)
interval <- mean_difference + c(-1, 1) * critical_value * se_difference
p_value <- 2 * pt(-abs(t_observed), df = df)

stopifnot(abs(se_difference - 6.324555) < .000001)
stopifnot(abs(p_value - 0.019213) < .000001)
c(t = t_observed, lower = interval[1], upper = interval[2], p = p_value)

With a vector named duration_difference, R carries out the same procedure with

t.test(duration_difference, mu = 0)

The hand calculation and software call should agree. Writing the target, estimator, and standard error first makes it possible to diagnose a disagreement.

Preserving the independent unit

The ten D_i values must be the independent units relevant to the population claim. Ten tokens from one speaker do not substitute for ten sampled speakers, since token rows are not independent speaker observations.

If each speaker contributes several tokens, averaging within speaker can produce one difference per speaker. That aggregation may discard information, but it preserves the analysis unit needed by this procedure. A later multilevel model can retain token-level observations while representing speaker dependence.

Problem-set reading

PS1, Task 3: paired-mean inference applies one-sample t inference to within-speaker differences. It is included because paired inference becomes ordinary one-sample inference once the response has been constructed at the correct unit.

Check your understanding

Why is the standard error s_D/\sqrt{n} rather than s_D?
Why does the t distribution have n-1 degrees of freedom here?
Which population quantity does [3.69,32.31] target?
What goes wrong if the ten differences come from ten tokens produced by one speaker?

The following pages show how design determines the response supplied to an inferential procedure. They begin with paired observations.

--- title: "Inference for one mean" --- The preceding pages introduced [confidence intervals](confidence-intervals.qmd), [null distributions](null-hypotheses-and-test-statistics.qmd), and the [Student t distribution](../random-variables-and-distributions/student-t-distribution.qmd). We now combine them for inference about one population mean. Suppose ten speakers each supply one vowel-duration difference, measured in milliseconds, between a target condition and a fixed reference condition. $$ D_1,\ldots,D_{10}. $$ The population mean difference $\mu_D$ is the estimand, while the sample mean $\overline{D}$ is the estimator. A [**one-sample t-test**](https://appliedstatisticsforlinguists.org/bwinter_stats_proofs.pdf#page=308), together with its corresponding confidence interval, uses the sample to estimate the population mean and the uncertainty of that estimate. We first separate response spread from estimator spread, then account for estimating that spread, and only then calculate an interval and test. ## Estimating the mean and response spread For $n$ independent speaker differences, the mean estimator is $$ \overline{D} =\frac{1}{n}\sum_{i=1}^nD_i. $$ The sample standard deviation is $$ s_D = \sqrt{ \frac{1}{n-1} \sum_{i=1}^n(D_i-\overline{D})^2 }. $$ The value $s_D$ describes variation among speaker differences. Dividing it by $\sqrt{n}$ gives the estimated standard error of their mean: $$ \widehat{\operatorname{SE}}(\overline{D}) =\frac{s_D}{\sqrt{n}}. $$ The distinction between these two spreads remains important: $s_D$ concerns the responses, while $s_D/\sqrt{n}$ concerns the estimator. ## Accounting for estimating the spread If the population standard deviation $\sigma_D$ were known and the observations were normally distributed, standardizing with $\sigma_D/\sqrt{n}$ would produce a standard normal variable. In practice, $\sigma_D$ is unknown and $s_D$ is estimated from the same data as $\overline{D}$. Under independent normal sampling, $$ T = \frac{\overline{D}-\mu_D}{s_D/\sqrt{n}} $$ has a Student t distribution with $$ \nu=n-1 $$ degrees of freedom. Estimating the mean imposes one constraint on the deviations because they must sum to zero: $$ \sum_{i=1}^n(D_i-\overline{D})=0. $$ Only $n-1$ deviations can vary freely. This is the source of the degrees of freedom in the one-sample t-test. The [t distribution introduced earlier](../random-variables-and-distributions/student-t-distribution.qmd) has heavier tails than the standard normal. Those heavier tails represent the additional uncertainty from estimating the population spread. As the sample grows, $s_D$ becomes more stable and the t distribution approaches the standard normal. ## Working a confidence interval by hand Suppose the observed summary is $$ n=10, \qquad \overline{d}=18\text{ ms}, \qquad s_D=20\text{ ms}. $$ The estimated standard error is $$ \widehat{\operatorname{SE}}(\overline{D}) =\frac{20}{\sqrt{10}} \approx6.3249\text{ ms}. $$ For a 95% interval with $9$ degrees of freedom, the $.975$ t quantile is approximately $2.262$. The interval is $$ \overline{d} \mathbin{\pm} t_{.975,9}\widehat{\operatorname{SE}}(\overline{D}), $$ which becomes $$ 18\mathbin{\pm}2.262(6.3249) \approx[3.69,32.31]\text{ ms}. $$ This interval targets the population mean difference. It does not contain 95% of speaker differences. ## Testing the zero reference value For $H_0:\mu_D=0$, the observed statistic is $$ t_{\mathrm{obs}} =\frac{18-0}{6.3249} \approx2.846. $$ The two-sided tail probability is approximately $.0192$. ```{r} #| label: verify-one-sample-t #| echo: true n <- 10 mean_difference <- 18 sd_difference <- 20 se_difference <- sd_difference / sqrt(n) df <- n - 1 t_observed <- mean_difference / se_difference critical_value <- qt(.975, df = df) interval <- mean_difference + c(-1, 1) * critical_value * se_difference p_value <- 2 * pt(-abs(t_observed), df = df) stopifnot(abs(se_difference - 6.324555) < .000001) stopifnot(abs(p_value - 0.019213) < .000001) c(t = t_observed, lower = interval[1], upper = interval[2], p = p_value) ``` With a vector named `duration_difference`, R carries out the same procedure with ```r t.test(duration_difference, mu = 0) ``` The hand calculation and software call should agree. Writing the target, estimator, and standard error first makes it possible to diagnose a disagreement. ## Preserving the independent unit The ten $D_i$ values must be the independent units relevant to the population claim. Ten tokens from one speaker do not substitute for ten sampled speakers, since token rows are not independent speaker observations. If each speaker contributes several tokens, averaging within speaker can produce one difference per speaker. That aggregation may discard information, but it preserves the analysis unit needed by this procedure. A later multilevel model can retain token-level observations while representing speaker dependence. ::: {.callout-note title="Problem-set reading"} [PS1, *Task 3: paired-mean inference*](../problem-sets/ps1/ps1.qmd#task-3-paired-mean-inference-30-points) applies one-sample t inference to within-speaker differences. It is included because paired inference becomes ordinary one-sample inference once the response has been constructed at the correct unit. ::: ## Check your understanding 1. Why is the standard error $s_D/\sqrt{n}$ rather than $s_D$? 2. Why does the t distribution have $n-1$ degrees of freedom here? 3. Which population quantity does $[3.69,32.31]$ target? 4. What goes wrong if the ten differences come from ten tokens produced by one speaker? The following pages show how design determines the response supplied to an inferential procedure. They begin with [paired observations](paired-observations.qmd).