Answer to Question #342238 in Statistics and Probability for sele

Question

The manufacturer of a new drug claims that it will lower

blood pressure by 13 points on the average. When the drug was

administered to 6 patients, the following drops in blood pressure

were registered: 12, 8, 15, 9, 10 and 16. Is the claim sustained at

the 0.05 level of significance.

Answer 1

Mean $\mu$ = $\dfrac{12+8+15+9+10+16}{6}=\dfrac{35}{3}$

Variance of population 

$$s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}=\dfrac{1}{6-1}((12-\dfrac{35}{3})^2$$

$$+(8-\dfrac{35}{3})^2+(15-\dfrac{35}{3})^2+(9-\dfrac{35}{3})^2$$

$$+(10-\dfrac{35}{3})^2+(16-\dfrac{35}{3})^2)=\dfrac{32}{3}$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{32}{3}}\approx3.266$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=13$

$H_a:\mu\not=13$

This corresponds to a two-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 $\alpha = 0.05,$ $df=n-1=5$ degrees of freedom, and the critical value for a two-tailed test is $t_c =2.570543.$The rejection region for this two-tailed test is $R = \{t:|t|> 2.570543\}.$

The t-statistic is computed as follows:

$$t=\dfrac{11.667-13}{3.266/\sqrt{6}}=-0.9997$$

Since it is observed that $|t| = 0.9997<2.570543=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed $df=5$ degrees of freedom, $t=-0.9997$ is $p=0.363349,$ and since $p=0.363349>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 13, at the $\alpha = 0.05$ significance level.