8.5 - Sample Means and Variances

Let's now spend some time clarifying the distinction between a population mean and a sample mean, and between a population variance and a sample variance.

Situation

Suppose we are interested in determining \(\mu\), the mean number of hours slept nightly by American college students. Because the population of American college students is so large, we can't possibly record the number of hours slept by each American college student.

How can we determine the value of the population mean if the population is too large to measure it?
We could take a random sample of American college students, calculate the average for the students in the sample, and use that sample mean as an estimate of the population mean. Similarly, we could calculate the sample variance and use it to estimate the population variance \(\sigma^2\)

Let's take a look!

Now, all we need to do is define the sample mean and sample variance!

Sample Mean

The sample mean, denoted \(\bar{x}\) and read “x-bar,” is simply the average of the \(n\) data points \(x_1, x_2, \ldots, x_n\):

\(\bar{x}=\dfrac{x_1+x_2+\cdots+x_n}{n}=\dfrac{1}{n} \sum\limits_{i=1}^n x_i\)

The sample mean summarizes the "location" or "center" of the data.

Example 8-18 Section

A random sample of 10 American college students reported sleeping 7, 6, 8, 4, 2, 7, 6, 7, 6, 5 hours, respectively. What is the sample mean?

Solution

The sample mean is:

\(\bar{x}=\dfrac{7+6+8+4+2+7+6+7+6+5}{10}=5.8\)

Sample Variance

The sample variance, denoted \(s^2\) and read "s-squared," summarizes the "spread" or "variation" of the data:

\(s^2=\dfrac{(x_1-\bar{x})^2+(x_2-\bar{x})^2+\cdots+(x_n-\bar{x})^2}{n-1}=\dfrac{1}{n-1}\sum\limits_{i=1}^n (x_i-\bar{x})^2\)

Sample Standard Deviation

The sample standard deviation, denoted \(s\) is simply the positive square root of the sample variance. That is:

\(s=\sqrt{s^2}\)

Example 8-19 Section

A random sample of 10 American college students reported sleeping 7, 6, 8, 4, 2, 7, 6, 7, 6, 5 hours, respectively. What is the sample standard deviation?

Solution

The sample variance is:

\(s^2=\dfrac{1}{9}\left[(7-5.8)^2+(6-5.8)^2+\cdots+(5-5.8)^2\right]=\dfrac{1}{9}(27.6)=3.067\)

Therefore, the sample standard deviation is:

\(s=\sqrt{3.067}=1.75\)

Theorem

An easier way to calculate the sample variance is:

\(s^2=\dfrac{1}{n-1}\left[\sum\limits_{i=1}^n x^2_i-n{\bar{x}}^2\right]\)

Proof

Example 8-20 Section

A random sample of 10 American college students reported sleeping 7, 6, 8, 4, 2, 7, 6, 7, 6, 5 hours, respectively. What is the sample standard deviation?

Solution

The sample variance is:

\(s^2=\dfrac{1}{9}\left[(7^2+6^2+\cdots+6^2+5^2)-10(5.8)^2\right]=3.067\)

Therefore, the sample standard deviation is:

\(s=\sqrt{3.067}=1.75\)

We will get a better feel for what the sample standard deviation tells us later on in our studies. For now, you can roughly think of it as the average distance of the data values \(x_1, x_2, \ldots, x_n\) from their sample mean.