8 - Interval Estimation

A point estimator is a sample statistic used to estimate a population parameter. For examples, the sample mean \(\bar{x}\) is a point estimator for the population mean \(\mu\) and the sample proportion \(\bar{p}\) is a point estimator of the population proportion \(p\).

Because a point estimator cannot be expected to provide the exact value of a population parameter, an interval estimate is often computed by adding and subtracting the margin of error to and from the point estimate:

\[ \text{Point estimate } \pm \text{ Margin of error} \]

Using this, we can calculate interval estimates for the population mean \(\mu\) and the population proportion \(p\):

\[\begin{split} \bar{x} \ \pm \text{ Margin of error} \\ \bar{p} \ \pm \text{ Margin of error} \end{split}\]

8.1 Population Mean: \(\sigma\) Known

In order to develop an interval estimate of a population mean, either the population standard deviation \(\sigma\) or the sample standard deviation \(s\) must be known. However, in some applications, the population standard deviation \(\sigma\) can be estimated from large amounts of historical data, or it can be assumed to be operating correctly in some quality control processes. These cases are called \(\sigma\) known cases.

Margin of Error and the Interval Estimate

The standard error of the sampling mean can be calculated as follows given the population standard deviation \(\sigma\) is known:

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]

We can then use this to develop a 95% confidence interval about the mean by first calculating the margin of error using the knowledge that 95% of values fall within 1.96 standard deviations of a normal distribution:

\[ \text{Margin of error } = \ 1.96 \sigma_{\bar{x}} \]

Which gives us this 95% confidence interval:

\[ \bar{x} \ \pm \text{ Margin of error} \]

Note that in this process, we consider the 95% to be the confidence level and the value 0.95 to be the confidence coefficient.

The level of signifance \(\alpha\) is also used to describe the confidence:

\[ \alpha = 1- \text{Confidence coefficient} \]

All of this information can be generalized to find the interval estimate of a population mean for a case in which the population standard deviation \(\sigma\) is known:

\[ \bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \]

Note that these values are commonly used as confidence levels:

Confidence Level	\(\alpha\)	\(\alpha/2\)	\(z_{\alpha/2}\)
90%	.10	.05	1.645
95%	.05	.025	1.960
99%	.01	.005	2.576

8.2 Population Mean: \(\sigma\) Unknown

In most cases, we will not have a good estimate of the population standard deviation \(\sigma\). In these cases, we must use the sample to estimate both \(\mu\) and \(\sigma\). In this case, the interval estimate is based on the \(t\) distribution.

The \(t\) distibution depends on the degrees of freedom. Note that as the degrees of freedom approach infinity, the \(t\) distribution approaches the normal distribution. The \(t\) distribution used in interval estimation is \(t_{\alpha/2}\), where the upper area of the \(t\) distribution is equal to \(\alpha/2\).

Margin of Error and the Interval Estimate

The interval estimate of a population mean where the population standard deviation \(\sigma\) is unknown is as follows:

\[ \bar{x} \ \pm \ t_{\alpha/2} \frac{s}{\sqrt{n}} \]

where

• \(s\) is the sample standard deviation

• \(t_{\alpha/2}\) has \(n-1\) degrees of freedom

Recall that the sample standard deviation \(s\) is

\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]

8.3 Determining the Sample Size

Suppose we have a desired margin of error \(E\) that we wish to achieve. We could then determine the size of our sample to satisfy that margin of error:

\[\begin{split} \begin{align} E &= z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \\ n &= \frac{(z_{\alpha/2})^2 \sigma^2}{E^2} \end{align} \end{split}\]

Note that this requires us to know the population standard deviation \(\sigma\). However, we can still follow this method if we have a preliminary/planning value for \(\sigma\).

8.4 Population Proportion

Recall that the interval estimate for the population proportion \(p\) is given by:

\[ \bar{p} \ \pm \text{ Margin of error} \]

The mean of the sampling distribution of \(\bar{p}\) is the population proportion \(p\), and the standard error is:

\[ \sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}} \]

Because the sampling distribution of \(\bar{p}\) is normally distributed, we can find the margin of error:

\[ \text{Margin of error } = z_{\alpha/2} \sigma_{\bar{p}} = z_{\alpha/2} \sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \]

Brining this all together, we get the interval estimate:

\[ \bar{p} \ \pm \ z_{\alpha/2} \sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \]

Determining the Sample Size

We can solve for the required sample size for a desired margin of error like in the population mean case:

\[\begin{split} \begin{align} E &= z_{\alpha/2} \sqrt{\frac{\bar{p}(1-\bar{p})}{n}} \\ n &= \frac{(z_{\alpha/2})^2 \bar{p}(1-\bar{p})}{E^2} \end{align} \end{split}\]

However, since \(\bar{p}\) won’t be known until after the sample is collected, we must use an estimate \(p^*\):

\[\begin{split} \begin{align} E &= z_{\alpha/2} \sqrt{\frac{p^*(1-p^*)}{n}} \\ n &= \frac{(z_{\alpha/2})^2 p^*(1-p^*)}{E^2} \end{align} \end{split}\]