Question

the new drug was administered to 15 patients, the following

drops in systolic blood pressure were registered: 12, 8, 15, 9, 10, 11,

14, 13, 6, 9, 14, 12, 18, 14 and 16 points. Assuming that the drops in

systolic blood pressure are normally distributed.

(a) Construct a 99% confidence interval for the mean drops in blood

pressure.

(b) The manufacturer of a new drug claims that it will lower blood

pressure by 13 points on the average. Is the claim sustained at

the 0.01 level of significance?

Answer 1

$$12, 8, 15, 9, 10, 11, 14, 13, 6, 9, 14, 12, 18, 14,16$$

$$\sum x_i=181, n=15$$

$$\bar{x}=\dfrac{181}{15}\approx12.067$$

$$\sum x_i^2=2333$$

$$s^2=\dfrac{1}{n-1}\big(\sum x_i^2-\dfrac{1}{n}(\sum x_i)^2\big)$$

$$=\dfrac{1}{15-1}\big(2333-\dfrac{1}{15}(181)^2\big)\approx10.6381$$

$$s=\sqrt{s^2}=\sqrt{10.6381}\approx3.2616$$

(a) The provided sample mean is $\bar{x}=12.067$ and the sample standard deviation is $s=3.2616.$ The size of the sample is $n=15$ and the required confidence level is $99\%.$

The number of degrees of freedom are $df=n-1=15-1=14,$ and the significance level is $\alpha=0.01.$ Based on the provided information, the critical t-value for $\alpha=0.01$ and $df=14$ degrees of freedom is $t_c=2.9768.$

The 99% confidence for the population mean $\mu$ is computed using the following expression

$$CI=\big(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}}\big)$$

$$=\big(12.067-2.9768\times\dfrac{3.2616}{\sqrt{15}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}}\big)$$

$$=(9.560, 14.574)$$

(b) The provided sample mean is $\bar{x}=12.067$ and the sample standard deviation is $s=3.2616,$ and the size of the sample is $n=15.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq13$

$H_1:\mu<13$

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

The number of degrees of freedom are $df=n-1=15-1=14,$ and the significance level is $\alpha=0.01.$ Based on the provided information, the critical t-value for $\alpha=0.01$ and $df=14$ degrees of freedom is $t_c=-2.624.$

The rejection region for this left-tailed test is $R=\{t:t<-2.624\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{12.067-13}{3.2616/\sqrt{15}}=-1.1082$$

Since it is observed that $t=-1.1082>-2.624=t_c,$ 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 less than 13, at the 0.01 significance level.

Use the P-value approach.

The P-value $t=-1.1082, df=14, \alpha=0.01,$ left-tailed, is $p=0.1433.$ Since $p=0.1433>0.01=\alpha,$ 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 less than 13, at the 0.01 significance level.

The claim is sustained at the 0.01 level of significance.