# When to use t distribution versus normal distribution quantiles in constructing confidence interval for the mean

To construct confidence interval for the mean, we often use quantiles of standardized sample mean distribution. Here, I include a list of cases where I’d use quantiles of t-distribution versus quantiles of normal distribution for that purpose.

Note: the below text could be directly translated to answer when to use t-test versus z-test in testing hypothesis about the mean parameter.

# Standardized sample mean

Consider $X_1, \ldots, X_n$ – a sequence of i.i.d. random variables with mean $E(X_i) = \mu$ and variance $\text{var}(X_i) = \sigma^2$. To construct confidence intervals for $\mu$ parameter, we often use a standardized sample mean,

\begin{aligned} \frac{\overline{X}_n - \mu}{\sigma/\sqrt{n}}, \end{aligned}

or its version where $S_n$ – a consistent estimator of true standard deviation $\sigma$ – is used, $\frac{\overline{X}_n - \mu}{S_n/\sqrt{n}}$; the latter is common in practice as we typically do not know $\sigma$ and must estimate it from the data. Knowing distribution of a standardized sample mean allows us to construct confidence interval for a mean $\mu$ parameter.

## Example 1: constructing confidence interval for $\mu$ with $z$-quantiles

Assume $X_1, \ldots, X_n$ are i.i.d. $\sim N(\mu,\sigma^2)$ and $\sigma$ is known. Then we have an exact distributional result for a standardized sample mean,

\begin{aligned} \frac{\overline{X}_n - \mu}{\sigma/\sqrt{n}} \sim N(0,1). \end{aligned}

Let us denote $z_{ 1-\frac{\alpha}{2}}$ to be $(1-\frac{\alpha}{2})$-th quantile of standard normal distribution $N(0,1)$. Since $N(0,1)$ is symmetric around $0$, we have $z_{\frac{\alpha}{2}} = -z_{1-\frac{\alpha}{2}}$ and we can write

\begin{aligned} 1-\alpha = P\left(-z_{1-\frac{\alpha}{2}} \leq \frac{\overline{X}_n-\mu}{\sigma/\sqrt{n}} \leq z_{1-\frac{\alpha}{2}} \right) = P\left(\bar{X}_n-z_{1-\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{X}_n+z_{1-\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \right), \end{aligned}

which yields that $\left[ \bar{X}_n-z_{1-\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}, ; \bar{X}_n+z_{1-\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right]$ is a $(1-\alpha )$-confidence interval for a mean parameter $\mu$.

## Example 2: constructing confidence interval for $\mu$ with $t$-quantiles

Assume $X_1, \ldots, X_n$ are i.i.d. $\sim N(\mu,\sigma^2)$ and $\sigma$ is unknown. We use $S_n$ – a consistent sample estimator of true standard deviation – to approximate $\sigma$, and have an exact distributional result for a standardized sample mean,

\begin{aligned} \frac{\overline{X}_n - \mu}{S_n/\sqrt{n}} \sim t_{n-1}. \end{aligned}

Let us denote $t_{n-1,1-\frac{\alpha}{2}}$ to be a $(n-1,1-\frac{\alpha}{2})$-th quantile of $t$-distributuon with $n-1$ degrees of freedom. Since $t$ is symmetric around $0$, we have

\begin{aligned} 1-\alpha =P\left(-t_{n-1,1-\frac{\alpha}{2}} \leq \frac{\bar{X}_{n}-\mu}{S_{n} / \sqrt{n}} \leq t_{n-1,1-\frac{\alpha}{2}}\right) = P\left(\bar{X}_n-t_{n-1, 1-\frac{\alpha}{2}}\frac{S_n}{\sqrt{n}} \leq \mu \leq \bar{X}_n+t_{n-1, 1-\frac{\alpha}{2}} \frac{S_n}{\sqrt{n}} \right), \end{aligned}

which yields that $\left[ \bar{X}_n-t_{n-1, 1-\frac{\alpha}{2}} \frac{S_n}{\sqrt{n}}, ; \bar{X}_n+t_{n-1, 1-\frac{\alpha}{2}} \frac{S_n}{\sqrt{n}}\right]$ is a $(1-\alpha )$-confidence interval for a mean parameter $\mu$.

# Cases

In many cases, whether to use quantiles of $t$-student distribution versus standard normal distribution is based on:

• distribution of $X_1, \ldots, X_n$ variables,
• whether $\sigma$ is known or not (and we need to estimate it i.e. with $S_n$),
• what is sample size $n$.

Note: the below cases could be directly translated to answer when to use $t$-test versus $z$-test in testing hypothesis about the mean $\mu$ parameter, i.e. test for $H_0: \mu = \mu_0$ versus $H_1: \mu < \mu_0$, or $H_1: \mu \neq \mu_0$, or $H_1: \mu > \mu_0$.

## Case 1: observations from normal distribution, $\sigma$ known, any $n$

• Observations $X_1, \ldots, X_n$ are from normal $N(\mu, \sigma^2)$ distribution.
• $\sigma$ known.
• Any sample size $n$.

$\Rightarrow$ We have exact result that $\frac{\bar{X}_{n}-\mu}{\sigma / \sqrt{n}} \sim N(0,1)$ and hence we use quantiles of normal distribution in constructing the CI.

## Case 2: observations from normal distribution, $\sigma$ unknown, small $n$

• Observations $X_1, \ldots, X_n$ are from normal $N(\mu, \sigma^2)$ distribution.
• $\sigma$ unknown.
• Small sample size ($n \leq 50$).

$\Rightarrow$ We use $S_{n}$ to approximate $\sigma$. We have exact result that $\frac{\bar{X}_{n}-\mu}{S_{n} / \sqrt{n}} \sim t_{n-1}$ and hence we use quantiles of $t$ distribution with $n-1$ degrees of freedom in constructing the CI.

## Case 3: observations from normal distribution, $\sigma$ unknown, large $n$

• Observations $X_1, \ldots, X_n$ are from normal $N(\mu, \sigma^2)$ distribution.
• $\sigma$ unknown.
• Moderate to large sample size ($n > 50$).

$\Rightarrow$ We use $S_{n}$ to approximate $\sigma$. Because of $n$ large enough, Slutsky’s theorem asymptotic kicks in'' and allows to replace $\sigma$ with $S_n$ – a consistent estimator of true population standard deviation, and to write that $\frac{\bar{X}_{n}-\mu}{S_n / \sqrt{n}} \approx \sim N(0,1)$. Because of $n$ large enough, we assume $N(0,1)$ is approximated ($\approx$) well enough to use quantiles of normal distribution in constructing the CI.

$\Rightarrow$ Another way to think about this case is that, as in Case 2, we have an exact result that $\frac{\bar{X}_{n}-\mu}{S_{n} / \sqrt{n}} \sim t_{n-1}$, and with large $n$, quantiles of $t$-distribution with $n-1$ degrees of freedom are almost equvalent to quantiles of normal distribution.

## Case 4: observations from any distribution, $\sigma$ known, small $n$

• Observations $X_1, \ldots, X_n$ are from (other than normal) distribution of mean $E(X_i) = \mu$ and variance $\text{var}(X_i) = \sigma^2$ (for normally distributed $X_i$’s, see cases 1-3).
• $\sigma$ known.
• Small sample size ($n \leq 50$).

$\Rightarrow$ Use CLT to get that standardized sample mean is approximately normal, $\frac{\bar{X}_{n}-\mu}{\sigma / \sqrt{n}} \approx \sim N(0,1)$. Since there is CLT approximation and we have a small sample size, in practice, we typically use quantiles of $t$ distribution with $n-1$ degrees of freedom to get more conservative (wider) CI.

• Note: when $X_1, \ldots, X_n$ distribution of is very skewed (i.e. Poisson) it may be not plausible that CLT already kicks in'' and other techniques may be needed.

## Case 5: observations from any distribution, $\sigma$ known, large $n$

• Observations $X_1, \ldots, X_n$ are from (other than normal) distribution of mean $E(X_i) = \mu$ and variance $\text{var}(X_i) = \sigma^2$ (for normally distributed $X_i$’s, see cases 1-3).
• $\sigma$ known.
• Moderate to large sample size ($n > 50$).

$\Rightarrow$ Use CLT to get that standardized sample mean is approximately normal. We have $\frac{\bar{X}_{n}-\mu}{\sigma/ \sqrt{n}} \approx \sim N(0,1)$. Since we have a moderate to small sample size, we assume that CLT kicks in'' and the approximation ($\approx$) is good enough to use quantiles of normal distribution.

## Case 6: observations from any distribution, $\sigma$ unknown, small $n$

• Observations $X_1, \ldots, X_n$ are from (other than normal) distribution of mean $E(X_i) = \mu$ and variance $\text{var}(X_i) = \sigma^2$ (for normally distributed $X_i$’s, see cases 1-3).
• $\sigma$ unknown.
• Small sample size ($n \leq 50$).

$\Rightarrow$ Use CLT to get that standardized sample mean is approximately normal and we also use Slutsky’s theorem to replace $\sigma$ with $S_n$ – a consistent estimator of true population standard deviation. We have $\frac{\bar{X}_{n}-\mu}{S_n/ \sqrt{n}} \approx \sim N(0,1)$. Since there is CLT and Slutsky’s theorem approximation and we have a small sample size, in practice, we typically use quantiles of $t$ distribution with $n-1$ degrees of freedom to get more conservative (wider) CI.

• Note: two approximations are happening here!
• Note: when $X_1, \ldots, X_n$ distribution of is very skewed (i.e. Poisson) it may be not plausible that CLT already kicks in'' and other techniques may be needed.

## Case 7: observations from any distribution, $\sigma$ unknown, large $n$

• Observations $X_1, \ldots, X_n$ are from (other than normal) distribution of mean $E(X_i) = \mu$ and variance $\text{var}(X_i) = \sigma^2$ (for normally distributed $X_i$’s, see cases 1-3).
• $\sigma$ unknown.
• Moderate to large sample size ($n > 50$).

$\Rightarrow$ Use CLT to get that standardized sample mean is approximately normal and we also use Slutsky’s theorem to replace $\sigma$ with $S_n$ – a consistent estimator of true population standard deviation. We have $\frac{\bar{X}_{n}-\mu}{S_n/ \sqrt{n}} \approx \sim N(0,1)$. Since we have a moderate to small sample size, we assume that both CLT and Slutsky asymptotics kicks in'' and the approximation ($\approx$) is good enough to use quantiles of normal distribution.

# Disclaimer

The views, thoughts, and opinions expressed in the text belong solely to the author, and not necessarily to the author’s employer, organization, committee or other group or individual.

: Methods in Biostatistics with R. Ciprian Crainiceanu, Brian Caffo, John Muschelli (2019). Available online at https://leanpub.com/biostatmethods.