Two individual staff members were being observed performing identical activities in the Doctor's office. 25 random samples were taken. One of the Medical Assistants is a new employee. Medical Assistant has been with Dr. Deasley for several years. Medical Assistant is a new employee and has been with this medical practice for 9 months. We want to determine how Medical Assistant #2 performs when compared to Medical Assistant since she is a new employee. (SEE APPENDIX Ein this file for data set) Assume this is a one-sided t-test and the historical average of Medical Assistant is .0126 Please provide the following information based on your analysis of the two Medical Assistants HINT: Strongly suggested you calculate this by manual claculation. It takes only a couple minutes, and software has many options and types of t tests that might confuse you until you know the method! Medical Assistant Average Medical Assistant Standard Deviatio Null Hypothesis Alternative Hypothesis T-Test Statistic Determine Degrees of Freedom, Critical Value 0.01136 0.00912241 -3.24227 -1.711 Ho: Xbar = 0.0126 Xbar > 0.0126 Hint: the null hypothesis assumes the two means we are comparing are equal Hint: the alternative hypothesis assumes the two means we are comparing are not equal (two tail test), or that one mean is more than the other (one tail) n-1 Hint- because we are comparing the sample mean of only one sample (Assistant 2) to a single value (the historical average of Assistant 1), use a One Sample t Test. Hint: remember that the t-table will give a positive value, but if you are testing the left side of the distribution, the t-test statistic will be negative and you should puta minus in frontof the crtical value as well. Also, this isa one tail test so find the critical t in the column with 5% alpha in one tail. Use the t Table on page 85 of the Green Belt textbook. Statistical Conclusion for the null and alternative hypothesis. Since the Critical Value of T is less than the critical value, we reject the null hypothesis. Summary of my confidence level: 99% Hint: Do you reject, or fail to reject, the null hypothesis? What does this practicvally mean?