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Estimating the Population Mean Using a Random Sample

Consider estimating the average GPA (call this $\mu$)  of the approximately 23,000 WMU undergraduates. In the absence of the complete database, we may wish to estimate $\mu$ by taking a random sample of, say, n=25 students and computing the sample average (call this $\overline{X}$ ). Suppose $\overline{X}$=3.05. Now, unless we got really lucky with the random sample, chances are $\mu$ is not equal to 3.05; i.e. the estimate missed. By how much?

On the average, a random variable misses the mean by one standard deviation. From the previous section, the standard deviation of $\overline{X}$ is $\sigma /\sqrt {n}$, where sigma is the population SD for individuals. Therefore, $\sigma /\sqrt {n}$ is the expected size of the miss when $\overline{X}$ is used to estimate $\mu$. Typically, $\sigma /\sqrt {n}$ is called the standard error of estimation of the mean  , or simply standard error of the mean. `Standard error' will often be denoted by the abbreviation SE throught this manuscript.

In the absence of the population database for 23000 WMU students, we do not know $\mu$ nor $\sigma$. Thus, we cannot calculate $\sigma /\sqrt {n}$. In this situation, $\sigma$ is typically replaced by the standard deviation S of the sample. The resulting quantity $S/\sqrt {n}$ is called the estimated standard error of the mean .

Returning to the GPA example, suppose that the sample of n=25 students yielded an average GPA of 3.05 and a standard deviation of .40. Then the WMU population average GPA is estimated as 3.05 with an estimated standard error of $S/\sqrt{n}= .40/\sqrt{25} = .08$.

\fbox{ \parbox{5.5in}{
\vspace*{1ex}
The population mean
$\mu$\space is estimat...
...error (SE) of the mean},
which is calculated as $S/\sqrt{n}$ .
\vspace*{1ex}
} }
     

Example 3. What is the average length of stay for undergraduate students at WMU?
a. Suppose 25 graduating students were randomly selected and asked about their length of stay. Suppose that the sample averaged 5.3 years, with an SD of 1.5 years. Then the WMU average stay is estimated as years give or take years or so.

b. A second sample of 100 students were interviewed. The mean and SD for the second sample were also 5.3 years and 1.5 years, respectively. Calculate an estimate for the WMU average stay and provide a standard error for your estimate.

Similar to the SE for proportions, the formula for the SE of the mean has the sample size (i) in the denominator, and (ii) inside the squareroot sign. Therefore, increasing the sample size by a factor of 4 makes the standard error decrease by a factor of $\sqrt{4}$.

\fbox{ \parbox{5.5in}{
\vspace*{1ex}
The standard error of the mean decreases like the squarerroot of the sample size.
\vspace*{1ex}
} }
 


next up previous contents index
Next: Exercises Up: Sampling Distribution of the Previous: Drawing from a Nonnormal

2003-09-08