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.
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.
