# T-Test: When Population Variance is Unknown

Printer-friendly version

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 H0μ = 120, and because there is no specific direction implied, the alternative hypothesis is HAμ ≠ 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 ≥ t0.025,99 = 1.9842 or if t  t0.025,99 = −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(T99 >4.762) < 2 × P(T99 >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!