Question

Among 157 African-American men, the mean systolic blood pressure was 146 mm

Hg with a standard deviation of 27. We wish to know if on the basis of these data,

we may conclude that the mean systolic blood pressure for a population of African-

American is greater than 140.

• Setup the null and alternate hypothesis (1 mark)

• Determine the type of the test (1 mark)

• Use α=0.01, conduct the test and accept or reject the hypothesis on the basis of

the test. (Given Z_0.99=2.33) (3 marks)

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le140$

$H_1:\mu>140$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=156$ and the critical value for a right-tailed test is $t_c = 2.350489.$

The rejection region for this right-tailed test is $R = \{t:t>2.350489\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{146-140}{27/\sqrt{157}}=2.7844$$

Since it is observed that $t=2.7844>2.350489=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=156$ degrees of freedom, $t=2.7844$ is $p= 0.003013,$ and since $p= 0.003013<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 140, at the $\alpha = 0.01$ significance level.