8.2 A Confidence Interval When the Population Standard Deviation Is Unknown and Small Sample Case

In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close enough results. This is what we did in Example 8.4 above. The point estimate for the standard deviation, s, was substituted in the formula for the confidence interval for the population standard deviation. In this case the 80 observations are well above the suggested 30 observations to eliminate any bias from a small sample. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.

William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to "discover" what is called the Student's t-distribution. The name comes from the fact that Gosset wrote under the pen name "A Student."

Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and used the Student's t-distribution only for sample sizes of at most 30 observations.

If you draw a simple random sample of size n from a population with mean μ and unknown population standard deviation σ and calculate the t-score t = $\frac{\bar{x}–\mu }{\left(\frac{s}{\sqrt{n}}\right)}$, then the t-scores follow a Student's t-distribution with n – 1 degrees of freedom. The t-score has the same interpretation as the z-score. It measures how far in standard deviation units $\bar{x}$ is from its mean μ. For each sample size n, there is a different Student's t-distribution.

The degrees of freedom, n – 1, come from the calculation of the sample standard deviation s. Remember when we first calculated a sample standard deviation we divided the sum of the squared deviations by n − 1, but we used n deviations $(x–\overline{x}\text{values})$ to calculate s. Because the sum of the deviations is zero, we can find the last deviation once we know the other n – 1 deviations. The other n – 1 deviations can change or vary freely. We call the number n – 1 the degrees of freedom (df) in recognition that one is lost in the calculations. The effect of losing a degree of freedom is that the t-value increases and the confidence interval increases in width.

A probability table for the Student's t-distribution is used to calculate t-values at various commonly-used levels of confidence. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). When using a t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails. Notice that at the bottom the table will show the t-value for infinite degrees of freedom. Mathematically, as the degrees of freedom increase, the t-distribution approaches the standard normal distribution. You can find familiar Z-values by looking in the relevant alpha column and reading value in the last row.

A Student's t table (See Appendix A Statistical Tables) gives t-scores given the degrees of freedom and the right-tailed probability.

The Student's t-distribution has one of the most desirable properties of the normal distribution: it is symmetrical. What the Student's t-distribution does is spread out the horizontal axis so it takes a larger number of standard deviations to capture the same amount of probability. In reality there are an infinite number of Student's t-distributions, one for each adjustment to the sample size. As the sample size increases, the Student's t-distribution become more and more like the normal distribution. When the sample size reaches 30 the normal distribution is usually substituted for the Student's t because they are so much alike. This relationship between the Student's t-distribution and the normal distribution is shown in Figure 8.7.

Figure 8.7

This is another example of one distribution limiting another one, in this case the normal distribution is the limiting distribution of the Student's t when the degrees of freedom in the Student's t approaches infinity. This conclusion comes directly from the derivation of the Student's t-distribution by Mr. Gosset. He recognized the problem as having few observations and no estimate of the population standard deviation. He was substituting the sample standard deviation and getting volatile results. He therefore created the Student's t-distribution as a ratio of the normal distribution and Chi squared distribution. The Chi squared distribution is itself a ratio of two variances, in this case the sample variance and the unknown population variance. The Student's t-distribution thus is tied to the normal distribution, but has degrees of freedom that come from those of the Chi squared distribution. The algebraic solution demonstrates this result.

Restating the formula for a confidence interval for the mean for cases when the sample size is smaller than 30 and we do not know the population standard deviation, σ:

Here the point estimate of the population standard deviation, s has been substituted for the population standard deviation, σ, and t_ν,α has been substituted for Z_α. The Greek letter ν (pronounced nu) is placed in the general formula in recognition that there are many Student t_v distributions, one for each sample size. ν is the symbol for the degrees of freedom of the distribution and depends on the size of the sample. Often df is used to abbreviate degrees of freedom. For this type of problem, the degrees of freedom is ν = n-1, where n is the sample size. To look up a probability in the Student's t table we have to know the degrees of freedom in the problem.

Example 8.5

Problem

The average earnings per share (EPS) for 10 industrial stocks randomly selected from those listed on the Dow-Jones Industrial Average was found to be $\overline{X}$ = 1.85 with a standard deviation of s=0.395. Calculate a 99% confidence interval for the average EPS of all the industrials listed on the DJIA.

$$\overline{x}-t_{\mathrm{v,\alpha }}\left(\frac{s}{\sqrt{n}}\right)\le \mu \le \overline{x}+t_{\mathrm{v,\alpha }}\left(\frac{s}{\sqrt{n}}\right)$$

Solution

To help visualize the process of calculating a confident interval we draw the appropriate distribution for the problem. In this case this is the Student’s t because we do not know the population standard deviation and the sample is small, less than 30.

Figure 8.8

To find the appropriate t-value requires two pieces of information, the level of confidence desired and the degrees of freedom. The question asked for a 99% confidence level. On the graph this is shown where (1-α) , the level of confidence , is in the unshaded area. The tails, thus, have .005 probability each, α/2. The degrees of freedom for this type of problem is n-1= 9. From the Student’s t table, at the row marked 9 and column marked .005, is the number of standard deviations to capture 99% of the probability, 3.2498. These are then placed on the graph remembering that the Student’s t is symmetrical and so the t-value is both plus or minus on each side of the mean.

Inserting these values into the formula gives the result. These values can be placed on the graph to see the relationship between the distribution of the sample means, $\overline{X}$'s and the Student’s t-distribution.

$$\mu =\overline{X}\pm t_{\mathrm{\alpha /2,df=n-1}}\frac{s}{\sqrt{n}}=1.851\pm 3.2498\frac{0.395}{\sqrt{10}}=1.8551\pm 0.406$$

$$1.445\le \mu \le 2.257$$

We state the formal conclusion as :

With 99% confidence level, the average EPS of all the industries listed at DJIA is from $1.44 to $2.26.