Question

WE ARE TRAINING OUR EMPLOYEES THROUGH A SPECIFIC WORKSHOP HOPING TO LOWER THEIR MISTAKES DURING THEIR WORK. TO EVALUATE THE EFFECTIVENESS OF THE TRAINING WORKSHOP, WE TOOK A SAMPLE OF THE WORKERS AND MEASURED THEIR PERFORMANCES BEFORE AND AFTER THE WORKSHOP, BASED THE NUMBER OF MISTAKES THEY COMMIT IN A WEEK. THE FOLLOWING TABLE CONTAINS THE DATA:

EMPLOYEE BEFORE AFTER A 24  22 B 27 26 C 19 14 D 24 22 E 22 16 F 26 21 G 27 28    AT THE 0.05 LEVEL OF SIGNIFICANCE, IS THERE EVIDENCE OF A DIFFERENCE IN THE MEAN BETWEEN THE BEFORE AND AFTER WORKSHOP.   PLEASE SHOW HOW YOU CALCULATE THE STANDARD DEVIATION AND ALL WORK, GRAPHS AND EXPLANATION Expert Answer

Answer 1

= (2 + 1 + 5 + 2 + 6 + 5 + (-1))/7 = 2.857

s_d = sqrt(((2 - 2.857)^2 + (1 - 2.857)^2 + (5 - 2.857)^2 + (2 - 2.857)^2 + (6 - 2.857)^2 + (5 - 2.857)^2 + (-1 - 2.857)^2)/6) = 2.5448

The test statistic

  

DF = 7 - 1 = 6

At 0.05 significance level, the critical values are +/- t_{0.025, 6} = +/- 2.447

Since the test statistic value is greater than the positive critical value (2.97 > 2.447), so we should reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that there is a difference in the mean between the before and after workshop.