Answer to Question #115879 in Statistics and Probability for Shane Jessica Ballera

Question

Hypothesis Testing

1. If the normal white blood cell is an average of 9,000 cells per cubic millimeter (cmm), it is significantly different from the white blood count (WBC) of 25 fruitarians in the outskirts of Visayas? Remember that their mean was 10,500 cells/cmm with a standard deviation of 3,000 cells/ cmm. Use 99% level of certainty.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=10500$

$H_1:\mu\not=10500$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a two-tailed test is $z_c=2.58.$

The rejection region for this two-tailed test is $R=\{z: |z|>2.58\}$

The z-statistic is computed as follows:

$$z={\bar{X}-\mu\over \sigma/\sqrt{n}}={9000-10500\over3000/\sqrt{25}}=-2.5$$

Since it is observed that $|z|=2.5<2.58=z_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 10500, at the 0.01 significance level.

Using the P-value approach: The p-value is $p=0.012419,$ and since $p=0.012419>0.01,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population $\mu$ is different than 10500, at the 0.01 significance level.