# T-Test: When Population Variance is Unknown

Now that, for purely pedagogical reasons, we have the unrealistic situation (of a known population variance) behind us, let's turn our attention to the realistic situation in which both the population mean and population variance are unknown.

### Example

It is assumed that the mean systolic blood pressure is *μ* = 120 mm Hg. In the Honolulu Heart Study, a sample of *n* = 100 people had an average systolic blood pressure of 130.1 mm Hg with a standard deviation of 21.21 mm Hg. Is the group significantly different (with respect to systolic blood pressure!) from the regular population?

**Solution.** The null hypothesis is *H*_{0}: *μ* = 120, and because there is no specific direction implied, the alternative hypothesis is *H*_{A}: *μ* ≠ 120. In general, we know that if the data are normally distributed, then:

\(T=\dfrac{\bar{X}-\mu}{S/\sqrt{n}}\)

follows a *t*-distribution with *n*−1 degrees of freedom. Therefore, it seems reasonable to use the test statistic:

\(T=\dfrac{\bar{X}-\mu_0}{S/\sqrt{n}}\)

for testing the null hypothesis \(H_0:\mu=\mu_0\) against any of the possible alternative hypotheses \(H_A:\mu \neq \mu_0\), \(H_A:\mu<\mu_0\), and \(H_A:\mu>\mu_0\). For the example in hand, the value of the test statistic is:

\[t=\dfrac{130.1-120}{21.21/\sqrt{100}}=4.762\]

The critical region approach tells us to reject the null hypothesis at the *α* = 0.05 level if *t ≥ t*_{0.025,99} = 1.9842 or if *t ≤ t*_{0.025,99} = −1.9842. Therefore, we reject the null hypothesis because *t* = 4.762 > 1.9842, and therefore falls in the rejection region:

Again, as always, we draw the same conclusion by using the *P*-value approach. The *P*-value approach tells us to reject the null hypothesis at the *α* = 0.05 level if the *P*-value ≤ *α* = 0.05. In this case, the *P*-value is 2 × P(T_{99} >4.762) < 2 × P(T_{99} >1.9842) = 2(0.025) = 0.05:

As expected, we reject the null hypothesis because *P*-value ≤ 0.01 < *α* = 0.05.

Again, we'll learn how to ask Minitab to conduct the *t*-test for a mean *μ* in a bit, but this is what the Minitab output for this example looks like:

By the way, the decision to reject the null hypothesis is consistent with the one you would make using a 95% confidence interval. Using the data, a 95% confidence interval for the mean *μ* is:

\[\bar{x}\pm t_{0.025,99}\left(\dfrac{s}{\sqrt{n}}\right)=130.1 \pm 1.9842\left(\dfrac{21.21}{\sqrt{100}}\right)\]

which simplifies to 130.1 ± 4.21. That is, we can be 95% confident that the mean systolic blood pressure of the Honolulu population is between 125.89 and 134.31 mm Hg. How can a population living in a climate with consistently sunny 80 degree days have elevated blood pressure?!

Anyway, the critical region approach for the *α* = 0.05 hypothesis test tells us to reject the null hypothesis that *μ* = 120:

if \(t=\dfrac{\bar{x}-\mu_0}{s/\sqrt{n}}\geq 1.9842\) or if \(t=\dfrac{\bar{x}-\mu_0}{s/\sqrt{n}}\leq -1.9842\)

which is equivalent to rejecting:

if \[\bar{x}-\mu_0 \geq 1.9842\left(\dfrac{s}{\sqrt{n}}\right)\] or if \[\bar{x}-\mu_0 \leq -1.9842\left(\dfrac{s}{\sqrt{n}}\right)\]

which is equivalent to rejecting:

if \[\mu_0 \leq \bar{x}-1.9842\left(\dfrac{s}{\sqrt{n}}\right)\] or if \[\mu_0 \geq \bar{x}+1.9842\left(\dfrac{s}{\sqrt{n}}\right)\]

which, upon inserting the data for this particular example, is equivalent to rejecting:

if \[\mu_0 \leq 125.89\] or if \[\mu_0 \geq 134.31\]

which just happen to be (!) the endpoints of the 95% confidence interval for the mean. Indeed, the results are consistent!