10.2: The One Mean T Procedure

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In One Mean Z Procedure, one major assumption was made that is rarely satisfied - it is very unlikely for anyone to know the population standard deviation without knowing the population mean. So, what do we do in a real-life scenario?

Naturally when a population parameter is not known we can use a sample statistic as an unbiased estimate. It is very natural to use the sample standard deviation, $s$, instead of the unknown population standard deviation, $\sigma$.

As a result, the sample mean of a sample of size $n$ is not normally distributed anymore but rather has another well-studied distribution called Student t-distribution with $n-1$ degrees of freedom.

From the practical point of view, the effects of this change are very simple – we can still use the same procedure but need to replace $\sigma$ and $z_{\alpha/2}$ with $s$ and $t_{\alpha/2}$ respectively.

Try It Yourself!

For the procedure to work the following assumptions must be made:

The sample is simple random.
The CLT must be applicable, that is at least one of the following must be true:
- The sample size, $n$, is greater than 30.
- The population is normally distributed.
The population standard deviation, $\sigma$, is unknown.

Try It Yourself!

Consider the following example.

Example $\PageIndex{1}$

$An image of a person contracting a virus and developing symptoms.$

An incubation period is a time between when you contract a virus and when your symptoms start. Assume that the population of incubation periods for a novel coronavirus is normally distributed. By surveying randomly selected local hospitals, a researcher was able to obtain the following sample of incubation periods of 10 patients:

6, 2, 4, 3, 3, 7, 5, 8, 5, 7

Use the sample to construct and interpret a 90% confidence interval for the average incubation period of the novel coronavirus.

Solution

First, let’s check if all necessary assumptions are satisfied:

The sample is simple random.
The population is normally distributed thus the CLT for sample means is applicable.
The population standard deviation, $\sigma$, is unknown.

Also, note that the average and the standard deviation of the sample are respectively

$\bar{x}_i=\dfrac{6+2+4+3+3+7+5+8+5+7}{10}=5$ and $s=2$

We will use the following template to construct and interpret the confidence interval:

Table $\PageIndex{1.1}$: Template of the procedure to construct a confidence interval for the population mean from a sample when the population standard deviation is unknown.
Unknown Parameter	$\mu$, population mean
Point Estimate	$\bar{x}_i$, the mean of a random sample of size $n$
Confidence Level (of the point estimate)	0%
Confidence Level (for a new interval estimate)	$(1-\alpha)100\%$ $\alpha$ $\alpha/2$
Critical Value(s)	$t_{\alpha/2}$ with $df=n-1$
Margin of Error	$ME=t_{\alpha/2}\dfrac{s}{\sqrt{n}}$
Lower Bound	$LB=\bar{x}_i-ME$
Upper Bound	$UB=\bar{x}_i+ME$
Interpretation	We are $(1-\alpha)100\%$ confident that the unknown population mean $\mu$ is between LB and UB.

In our example, we perform the procedure (fill out the table) in the following way:

Table $\PageIndex{1.2}$: The summary of the procedure to construct a 90% confidence interval for the mean incubation period of a novel coronavirus from a sample of size 10 when the population standard deviation is unknown.
	Example	Template
Unknown Parameter	$\mu$, the mean incubation period	$\mu$, population mean
Point Estimate	$\bar{x}_i=5$, the mean incubation period in the sample of size $n=10$	$\bar{x}_i$, the mean of a random sample of size $n$
Confidence Level (of the point estimate)	0%	0%
Confidence Level (for a new interval estimate)	$CL=90\%$ $\alpha=0.10$ $\alpha/2=0.05$	$(1-\alpha)100\%$ $\alpha$ $\alpha/2$
Critical Value(s)	When $df=9$: $t_{\alpha/2}=t_{0.05}=1.83$	$t_{\alpha/2}$ with $df=n-1$
Margin of Error	$ME=1.83\cdot\dfrac{2}{\sqrt{10}}=1.16$	$ME=t_{\alpha/2}\dfrac{s}{\sqrt{n}}$
Lower Bound	$LB=5-1.16=3.84$	$LB=\bar{x}_i-ME$
Upper Bound	$UB=5+1.16=6.16$	$UB=\bar{x}_i+ME$
Interpretation	We are $90\%$ confident that the average incubation period of the novel coronavirus is between 3.84 and 6.16 days.	We are $(1-\alpha)100\%$ confident that the unknown population mean $\mu$ is between LB and UB.

We are $90\%$ confident that the average incubation period of the novel coronavirus is between 3.84 and 6.16 days.

Try It Yourself! $\PageIndex{1}$

Try It Yourself! $\PageIndex{2}$

Try It Yourself! $\PageIndex{3}$

Try It Yourself! $\PageIndex{4}$

Try It Yourself! $\PageIndex{5}$

This page titled 10.2: The One Mean T Procedure was last modified on Sat, 16 Aug 2025 20:22:41 GMT and is shared under a not declared license and was authored, remixed, and/or curated by Edward Pineda-Castro.

	Example	Template
Unknown Parameter	\(\mu\), the mean incubation period	\(\mu\), population mean
Point Estimate	\(\bar{x}_i=5\), the mean incubation period in the sample of size \(n=10\)	\(\bar{x}_i\), the mean of a random sample of size \(n\)
Confidence Level (of the point estimate)	0%	0%
Confidence Level (for a new interval estimate)	\(CL=90\%\) \(\alpha=0.10\) \(\alpha/2=0.05\)	\((1-\alpha)100\%\) \(\alpha\) \(\alpha/2\)
Critical Value(s)	When \(df=9\): \(t_{\alpha/2}=t_{0.05}=1.83\)	\(t_{\alpha/2}\) with \(df=n-1\)
Margin of Error	\(ME=1.83\cdot\dfrac{2}{\sqrt{10}}=1.16\)	\(ME=t_{\alpha/2}\dfrac{s}{\sqrt{n}}\)
Lower Bound	\(LB=5-1.16=3.84\)	\(LB=\bar{x}_i-ME\)
Upper Bound	\(UB=5+1.16=6.16\)	\(UB=\bar{x}_i+ME\)
Interpretation	We are \(90\%\) confident that the average incubation period of the novel coronavirus is between 3.84 and 6.16 days.	We are \((1-\alpha)100\%\) confident that the unknown population mean \(\mu\) is between LB and UB.

Contents

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Try It Yourself!

Try It Yourself!

Try It Yourself!

Example \(\PageIndex{1}\)

Solution

Try It Yourself! \(\PageIndex{1}\)

Try It Yourself! \(\PageIndex{2}\)

Try It Yourself! \(\PageIndex{3}\)

Try It Yourself! \(\PageIndex{4}\)

Try It Yourself! \(\PageIndex{5}\)