Answer to Question #268154 in Statistics and Probability for Axl

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

Let the random variable $X$ represent the average increase of pulse rate. The sample mean $(\bar{x})$ of the data given above is given as,

$\bar{x}=\sum(x)/n$ where, $n=20$.

$\bar{x}=235/20=11.75$ and $\sigma=5$

This question requires us to conduct an hypotheses test for the population mean.

The hypotheses to be tested are,

$H_0:\mu\geqslant 12$ against $H_1:\mu\lt 12$

Since the population variance is given, we apply the normal distribution to perform this test as follow s,

The test statistic is given as,

$Z^*=(\bar{x}-\mu)/(\sigma/\sqrt{n})=(11.75-12)/(5/\sqrt{20})=-0.25/ 1.118034=-0.2236068$

The test statistic $(Z^*)$ is compared with the table value at $\alpha=5\%$ level of significance.

The table value $Z_{0.05}$ is obtained by entering the following command in $R$,

> qnorm(0.05)

[1] -1.644854

The output of -1.644854 is the table value, that is, $Z_{0.05}=-1.644854$ and the null hypothesis is rejected if $Z^*\lt Z_{0.05}.$

Since $Z^*=-0.2236068\gt Z_{0.05}=-1.644854$, we fail to reject the null hypothesis and conclude that there is no enough proof to show that this group has a lower average increase of pulse rate than the ones in general at $5\%$ level of significance.