10.2 - T-Test: When Population Variance is Unknown
10.2 - T-Test: When Population Variance is UnknownNow 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?
Answer
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:
| 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!