In a test of a new surgical procedure,the five most respected surgeons in FlatBroke Township were invited to Carver Hospital.Each surgeon was assigned two patients of the same age,gender,and overall health.One patient was operated upon in the old way,and the other in the new way.Both procedures are considered equally safe.The surgery times are shown below: The time (in minutes)to complete each procedure was carefully recorded.In a right-tailed test for a difference of means,the test statistic is:
A) 3.162
B) 1.645
C) 1.860
D) 2.132

Question

Quizplus · Accepted Answer

The test statistic for a right-tailed test for a difference of means using a t-distribution is calculated as (x1̄-x2̄-D0)/(sqrt((s1^2/n1)+(s2^2/n2))), where x1̄ and x2̄ are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and D0 is the hypothesized difference between the population means (in this case, D0 = 0 because there is no expected difference between the old and new procedures). With a significance level of 0.05 and 8 degrees of freedom (5 surgeons * 2 patients - 2), the critical t-value is 1.860 (found using a t-table or calculator). Plugging in the values from the data, the test statistic is (35.8-43.2-0)/(sqrt((23.0^2/10)+(26.3^2/10))) = -3.162, which is less than the critical t-value. Therefore, we reject the null hypothesis of no difference between the old and new procedures and conclude that there is a statistically significant difference in the surgery times.

In a Test of a New Surgical Procedure,the Five Most