Answer to Question #333215 in Statistics and Probability for Chanwin

Question

Medical literature tells us that our blood is mainly composed of red and white blood cells corpuscles and a normal human body must average 7250/mm³ of white blood cell counts.If a sample of 15 individuals chosen at random from a certain place has an average of 4850/mm³ with a standard deviation of 2500/mm³, would you say that the people in that place have low white blood cell counts?

A.Formulate the null and alternative hypothesis in sentence and in symbol.

B.Determine what test statistics to use.

C.Compute the test statistics.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge7250$

$H_1:\mu<7250$

This corresponds to a left-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 $\alpha=0.05,$ $df=n-1=15-1=14$ degrees of freedom, and the critical value for a left-tailed test is $t_c = - 1.76131.$

The rejection region for this left-tailed test is $R = \{t: t<- 1.76131\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{4850-7250}{2500/\sqrt{15}}\approx-3.718$$

Since it is observed that $t =-3.718<- 1.76131=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=14$ degrees of freedom, $t=-3.718$ is $p=0.001147,$ and since $p=0.001147<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 7250, at the $\alpha = 0.05$ significance level.