T-Test: When Population Variance is Unknown

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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.

waikikiExample

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: 

drawing 

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:

drawing

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: 

 minitab output

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!