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.

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.

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.

	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.

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