Constructing a Confidence Interval for the Population Mean Section
To construct a confidence interval for a population mean, we're going to apply the same three steps as with the population proportion, but first, let's look at the two possible cases.
Case 1: $\sigma$ is known Section
In the previous lesson, we learned that if the population is normal with mean \(\mu\) and standard deviation, \(\sigma\), then the distribution of the sample mean will be Normal with mean \(\mu\) and standard error \(\frac{\sigma}{\sqrt{n}}\).
Following the similar idea to developing the confidence interval for \(p\), the \((1\alpha)\)100% confidence interval for the population mean \(\mu\) is...
\(P\left(\left\dfrac{\bar{x}\mu}{\dfrac{\sigma}{\sqrt{n}}}\right\le z_{\alpha/2}\right)=1\alpha\)
A little bit of algebra will lead you to...
\(P\left(\bar{x}z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\le \mu\le \bar{x}+z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\right)=1\alpha\)
In other words, the \((1\alpha)\)100% confidence interval for \(\mu\) is:
\(\bar{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\)
Notice for this case, the only condition we need is the population distribution to be normal.
Note!
The case where \(\sigma\) is known is unrealistic. We explain it here briefly because it reinforces what we have previously learned. We do not present examples in this case.Case 2: \(\sigma\) is unknown Section
When the population is normal or when the sample size is large then,
\(Z=\dfrac{\bar{x}\mu}{\dfrac{\sigma}{\sqrt{n}}}\)
where Z has a standard Normal distribution.
Usually, we don't know \(\sigma\), so what can we do?
Recall that if X comes from a normal distribution with mean, $\mu$, and variance, $\sigma^2$, or if $n\ge 30$, then the sampling distribution will be approximately normal with mean $\mu$ and standard error, \(SE(\bar{X})=\frac{\sigma}{\sqrt{n}}\)
One way to estimate \(\sigma\) is by \(s\), the standard deviation of the sample, and replace \(\sigma\) by \(s\) in the above Zequation. However, this new quotient no longer has a Zdistribution. Instead it has a tdistribution. We call the following a 'studentized' version of \(\bar{X}\):
\(t=\dfrac{\bar{X}\mu}{\dfrac{s}{\sqrt{n}}}\)
Constructing the Confidence Interval Section

CHECK THE CONDITIONS
One of the following conditions need to be satisfied:
 If the sample comes from a Normal distribution, then the sample mean will also be normal. In this case, \(\dfrac{\bar{x}\mu}{\frac{s}{\sqrt{n}}}\) will follow a \(t\)distribution with \(n1\) degrees of freedom.
 If the sample does not come from a normal distribution but the sample size is large (\(n\ge 30\)), we can apply the Central Limit Theorem and state that \(\bar{X}\) is approximately normal. Therefore, \(\dfrac{\bar{x}\mu}{\frac{s}{\sqrt{n}}}\) will follow a \(t\)distribution with \(n1\) degrees of freedom.

CONSTRUCT THE GENERAL FORM
 \((1\alpha)\)100% Confidence Interval for the Population Mean, \(\mu\)
 \(\bar{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}\)
where the tdistribution has \(df = n  1\). This interval is also known as the onesample tinterval for the population mean.

INTERPRET THE CONFIDENCE INTERVAL
We are \((1\alpha)100\%\) confident that the population mean, \(\mu\), is between \(\bar{x}t_{\alpha/2}\frac{s}{\sqrt{n}}\) and \(\bar{x}+t_{\alpha/2}\frac{s}{\sqrt{n}}\).
What if the conditions are not met? Section
What will you do if you cannot use the tinterval? What do we do when the above conditions are not satisfied?
 If you do not know if the distribution comes from a normally distributed population and the sample size is small (i.e \(n<30\)), you can use the Normal Probability Plot to check if the data come from a normal distribution.
 You may want to consider what is known as nonparametric statistical methods. A procedure such as the onesample Wilcoxon procedure. Lesson 11 introduces nonparametric statistical methods.