# 10.2 - T-Test: When Population Variance is Unknown

10.2 - 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 10-2 It is assumed that the mean systolic blood pressure is $$\mu$$ = 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?

The null hypothesis is $$H_0:\mu=120$$, and because there is no specific direction implied, the alternative hypothesis is $$H_A:\mu\ne 120$$. 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 $$\alpha=0.05$$ level if $$t\ge t_{0.025, 99}=1.9842$$ or if $$t\le t_{0.025, 99}=-1.9842$$. 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 $$\alpha=0.05$$ level if the $$p$$-value $$\le \alpha=0.05$$. In this case, the $$p$$-value is $$2 \times P(T_{99}>4.762)<2\times P(T_{99}>1.9842)=2(0.025)=0.05$$:

As expected, we reject the null hypothesis because $$p$$-value $$\le 0.01<\alpha=0.05$$.

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

Test of mu = 120 vs not = 120
N Mean StDev SE Mean 95% CI T P
100 130.100 21.210 2.121 (125.891, 134.309) 4.76 0.000

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 $$\mu$$ 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 $$130.1\pm 4.21$$. 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 $$\alpha=0.05$$ hypothesis test tells us to reject the null hypothesis that $$\mu=120$$:

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!

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