Final answer:
To determine if the new medical procedure is better than the old one, we use a t-test to compare the sample mean of the new procedure to the mean of the established procedure. By calculating the test statistic and comparing it to the critical t-value, we can determine if there is evidence to suggest that the new procedure is better. In this case, the calculated t-value is less than the critical t-value, so we reject the null hypothesis and conclude that the new procedure is better than the old procedure.
Step-by-step explanation:
In this question, we are comparing the recovery time of an established medical procedure to a new procedure. We are given the recovery time of the established procedure, which is 25 weeks. We are also given the sample mean and standard deviation of the new procedure, which are 21.3 weeks and 6.35 weeks, respectively.
To determine if the new procedure is better than the old procedure at α=0.01, we need to compare the sample mean of the new procedure to the mean of the established procedure. Since we do not have the population standard deviation, we need to use the t-distribution.
The null hypothesis (H0) is that the new procedure is not better than the old procedure, or in other words, the mean recovery time of the new procedure is greater than or equal to the mean recovery time of the established procedure. The alternative hypothesis (Ha) is that the new procedure is better than the old procedure, or in other words, the mean recovery time of the new procedure is less than the mean recovery time of the established procedure.
To test the hypotheses, we need to calculate the test statistic using the formula:
t = (sample mean - population mean) / (sample standard deviation / √(sample size))
Substituting the given values into the formula, we get:
t = (21.3 - 25) / (6.35 / √25) = -9.1
Since the absolute value of the critical t-value is 2.49 and our calculated t-value is -9.1, which is less than -2.49, we reject the null hypothesis and conclude that there is evidence to suggest that the new procedure is better than the old procedure at α=0.01. In other words, the mean recovery time of the new procedure is significantly less than the mean recovery time of the established procedure.