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A computer company conducted a \"new and improved\" course designed to train its

ID: 3441669 • Letter: A

Question

A computer company conducted a "new and improved" course designed to train its service representatives in learning to repair personal computers. One hundred trainees were split into two groups on a random basis: 50 took the customary course and 50 took the new course. At the end of 6 weeks, all 100 trainees were given the same final examination. Using chi-square, test the null hypothesis that the new course was no better than the customary course in terms of teaching service representatives to repair personal computers. What do your results indicate?

Explanation / Answer

Doing an Expected Value Chart,          
          
17   17  
24   24  
9   9  
          
Using chi^2 = Sum[(O - E)^2/E],          
          
chi^2 =    0.776143791      
          
With df = (a - 1)(b - 1), where a and b are the number of categories of each variable,          
          
a =    2      
b =    3      
          
df =    2      
          
Thus, the critical value is          
          
significance level =    0.05      
          
chi^2(critical) =    5.991464547      
          
Also, the p value is          
          
P =    0.67836357      
          
Thus, comparing chi^2 and chi^2(crit) [or, p and significance level], we   FAIL TO REJECT THE NULL HYPOTHESIS.      
          
Thus, there is no significant evidence that the "new and improved course" was any different than the old course.

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