t-based Confidence Interval for the Mean

The 95% confidence interval provided by ( 7.4) is simple, but not very useful. Why? Because the population standard deviation $\sigma$ is usually unknown (if we knew it, we would likely also know the population average $\mu$, and have no need for an interval estimate.)

In practical applications, we replace the population standard deviation $\sigma$  in ( 7.2) by S, the standard deviation of the sample. However, this substitution changes the coverage probability $1 - \alpha$. Fortunately, there is a simple adjustment that allows us to maintain the desired coverage level $1 - \alpha$: replace the normal distribution critical value z by the slightly larger t-distribution critical value t. The resulting confidence interval is the primary result of this section.   

  

  where t is a critical value determined from the t_n-1 distribution in such a way that there is area $(1-\alpha)$ between t and -t.    

The value n-1 is called degrees of freedom , or df for short. It is a parameter of the t-curve in the sense that changing the value of n-1 changes the shape of the t-curve, though usually not by much. Here are appropriate t critical values for selected $1 - \alpha$ and n-1.

$$\begin{tabular}{l\vert ccccc} $1-\alpha$\space & .80 & .90 &... ... $t$ : \mbox{n-1=500} & 1.28 & 1.65 & 1.96 & 2.58 \end{tabular}$$

   

The t critical values are always larger than the z, and get progressively closer as n-1 gets larger (they are equal at $n=\infty$). For a 95% confidence interval, the t values are 2.06, 2.03, 2.01, 1.98, and 1.96 for respective sample sizes n= 26,36, 51, 101, and 501.

Recall that the term $S/\sqrt {n}$ in equation ( 7.5) is the (estimated) standard error of the mean. With .68 chance, $\overline{X}$ misses $\mu$ by less than this amount. To generalize, $\overline{X}$ misses $\mu$ by less than $t (S/\sqrt{n})$ with $1 - \alpha$ certainty. Thus, the term $t (S/\sqrt{n})$ is called the margin of error with confidence level $1 - \alpha$.

If $1-\alpha=.95$, then t is close to 2.0. For this reason, the 95% margin of error is often written as $2 (S/\sqrt{n})$. When working with a random sample, the exact critical value t is read from a table or calculator, and depends on the sample size. However, for sample size calculations (see next section), the approximate critical value 2.0 is typically used.

Example: Given the following GPA for 6 students: 2.80, 3.20, 3.75, 3.10, 2.95, 3.40
a. Calculate a 95% confidence interval for the population mean GPA.
b. If the confidence level is increased from 95% to 99% , will the length of the confidence interval increase, decrease, or remain the same?
c. If the confidence level is kept at 95% but the sample size is quadrupled to n=24
(i) do you expect the sample mean $\overline{X}$ to increase, decrease, or remain approximately the same?
(ii) do you expect the sample SD S to increase, decrease, or remain approximately the same?
(iii) do you expect the length of the confidence interval to increase, decrease, or remain approximately the same?

We end this section with a reminder that the confidence interval ( 7.5) works well ONLY when at least one of the following conditions hold: (i) the population histogram looks like the normal curve, or (ii) the sample size is relatively large ($n\geq 30$ is a frequently used thumb rule . The following table summarizes the performance of the t-intervals under four situations.

	Normal curve	Not normal curve
Small sample size [n<30]	Good	Poor
Larger sample sizes [$n\geq 30$]	Good	Fair