Estimating the Population Mean Using Intervals

Problem revisited: Estimate the average GPA $\mu$  of the population of approximately 23000 WMU undergraduates.

Consider estimating the population average $\mu$  by the sample average of, say n=25 randomly selected students. Suppose this sample average, which we denote by $\overline{X}$, equals 3.05. Now chances are the true average $\mu$ is not equal to 3.05. For sure, the true WMU average GPA is between 1.00 and 4.00, and with high confidence between (2.50, 3.50); but what level of confidence do we have that it is between say, (2.75, 3.25) or (2.95, 3.15)? Even better, can we find an interval (a, b) which will contain $\mu$ with , say, 95% certainty?

Recall that if $\sigma$  is the standard deviation for individuals, then $\sigma /\sqrt {n}$ is the standard deviation for averages (also called the SE of the sample average). More precisely, n-member averages form an approximate normal histogram with mean $\mu$ and standard error $\sigma /\sqrt {n}$. This means that there is 68% likelihood that the sample average falls within $\sigma /\sqrt {n}$ of $\mu$, and 95% likelihood that the sample average falls within $2(\sigma/\sqrt{n})$ of $\mu$; see figure below.

Selected areas under histogram for $\overline{X}$

In statistical shorthand, we write

$$P( \overline{X}\;\; \mbox{is inside the interval } \mu \pm 2(\sigma/\sqrt{n})) = .95$$

where $P(\;\cdot\;)$ is read as ``the probability that $(\;\cdot\;)$".

Of course, the distance of $\overline{X}$ from $\mu$ is also the distance of $\mu$ from $\overline{X}$, so it is equally true that

$$P( \mu \;\; \mbox{is inside the interval } \overline{X} \pm 2(\sigma/\sqrt{n})) = .95.$$ (7.1)

Recall the question posed earlier: ``Can we find an interval which we know contains $\mu$  with 95% confidence?'' Answer: $\overline{X} \pm 2(\sigma/\sqrt{n})$.  

Example 1: Suppose that the SD of student GPA is .30. If a random sample of n=25 students has average GPA equal to 3.05, then we are 95% confident that $\mu$ is contained within $3.05 \pm 2(.30/\sqrt{25})$ or (2.93, 3.17).

Replacing $2(\sigma/\sqrt{n})$ by $1(\sigma/\sqrt{n})$ in Equation ( 7.1) results in a 68% confidence interval. In general, we can create an interval with any desired confidence level by replacing the multiplier with the appropriate standard normal percentile z (also called the z-critical value).   

$$P( \mu \;\;\mbox{is inside the interval } \overline{X} \pm z(\sigma/\sqrt{n})) = 1-\alpha$$ (7.2)

where z is determined from $1 - \alpha$ as follows:

The following table gives the appropriate critical value z for typical values of $1 - \alpha$.  

$$\begin{array}{c\vert cccccc} 1-\alpha & .68 & .80 & .90 & .... ...\hline z & 1.00 & 1.28 & 1.64 & 1.96 & 2.00 & 2.58 \end{array}$$ (7.3)

  

   

Example 1 (con't.):
a. Construct a 99% confidence interval for WMU GPA.
b. Construct a 75% confidence interval for WMU GPA.
c. Construct a 95.4% confidence interval for WMU GPA.
d. Construct a 95.0% confidence interval for WMU GPA.

Theoretically speaking, $\overline{X} \pm 2(\sigma/\sqrt{n})$ has a confidence level of 95.4%. Applied statisticians usually just round off and say 95%. We can, of course, choose the more precise critical value of 1.96 to get an exact 95% confidence level. However, normality of $\overline{X}$ is at best approximate anyway, so the precision seems spurious. However, the distinction is important enough to merit a reminder.