A new drug has been developed for a group of people and the researchers are worried that it might have an effect on pulse rate. This particular group of people has a known mean pulse rate of 82 bpm (beats per minute). A sample of 36 people taking this experimental drug is taken from the population. The mean pulse rate of the sample is 84 bpm with a standard deviation of 5 bpm.
Assume all samples are simple random samples and α (alpha) is taken to be 0.05.
To determine if the new drug has a significant effect on pulse rate, we can conduct a hypothesis test.
Null hypothesis (H0): The new drug does not have an effect on pulse rate.
Alternate hypothesis (Ha): The new drug does have an effect on pulse rate.
We will use a one-sample t-test to test the hypothesis. The formula for the test statistic is:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Given:
Population mean (μ) = 82 bpm
Sample mean (x̄) = 84 bpm
Sample standard deviation (s) = 5 bpm
Sample size (n) = 36
Level of significance (α) = 0.05
Calculating the test statistic:
t = (84 - 82) / (5 / sqrt(36))
= 2 / (5 / 6)
= 2 * (6 / 5)
= 2.4
The degrees of freedom for this test is (sample size - 1) = 36 - 1 = 35.
To determine if the test statistic is significant at the 0.05 level of significance, we need to compare it with the critical value from the t-distribution table.
Looking up the critical value for a one-tailed test at the 0.05 level of significance with 35 degrees of freedom, we find the critical value to be approximately 1.6909.
Since the test statistic (2.4) is greater than the critical value (1.6909), we reject the null hypothesis. This means there is sufficient evidence to conclude that the new drug has a significant effect on pulse rate for this group of people.