Answer to Question #268124 in Statistics and Probability for Ave

Question

It has been known that a fixed dose of a certain drug results to an average increase of pulse rate by at least 12 beats per minute with a standard deviation of 5. A group of 20 patients given the same dose showed the following increases: 15, 12, 10, 8, 14, 15, 16, 11, 7, 13, 9, 10, 12, 11, 9, 10, 17, 14, 15, 7. Is there proof to show that this group has a lower average increase of pulse rate than the ones in general? Use 0.05 level of significance.

Answer 1

$H_0:\mu=12$ , group has not a lower average increase of pulse rate than the ones in general

$H_a:\mu<12$ , group has a lower average increase of pulse rate than the ones in general

group mean:

$\overline{x}=\sum x_i/n=11.75$

group standard deviation:

$\sigma=\sqrt{\sum (x_i-\overline{x})^2/(n-1)}=3.02$

$t=\frac{\overline{x}-\mu}{\sigma/\sqrt n}=\frac{11.75-12}{5/\sqrt{20}}=-0.370$

$df=n-1=19$

from t-table for $\alpha=0.05$ and lower one-sided test we get critical value:

$t_{crit}=1.729$ for one-sided test

Since $|t|<t_{crit}$ we accept the null hypothesis, that the group has not a lower average increase of pulse rate than the ones in general.