Answer to Question #268138 in Statistics and Probability for piper

Question

1.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

$n=20 \\ \bar{x} = \frac{15+12+...+15+7}{20}= 11.75 \\ \sigma = 5.0 \\ H_0: \mu ≥12 \\ H_1: \mu < 12$

Test-statistic

$Z = \frac{\bar{x} -\mu}{ \sigma / \sqrt{n}} \\ Z = \frac{11.75-12.0}{5.0/ \sqrt{20}} = -0.22$

P-value = P(Z< -0.22) = 0.4129

(from normal distribution z -table)

Since P-value is greater than the significance level, α = 0.05 , we failed to reject the null hypothesis

Decision :

fail to reject Ho

Conclusion :

There is not sufficient evidence to conclude that the group has a lower average increase of pulse rate than the ones in general.