Question

Because of the rising costs of industrial accidents, many chemical, mining, and manufacturing firms have instituted safety courses. Employees are encouraged to take these courses designed to heighten safety awareness. A company is trying to decide which one of two courses to institute. To help make a decision, eight employees take Course 1 and another eight take Course 2. Each employee takes a test that is graded out of a possible 25 points. The safety test results are shown below. Assume that the scores are normally distributed. The company wants to know if the marks from Course 1 are less than the marks from Course 2. A) State the Null and Alternative Hypotheses. B) Is this a left-tailed, right-tailed, or two-tailed test? C) Run an appropraite t-test (use 5% level of significance); use cell F26 for your output. D) Change correct p-value to a % and highlight yellow. Make a decision based on your t-test results.

Course 1	Course 2
14	20
21	18
17	22
14	15
17	23
19	21
20	19
16	15

Answer 1

a)

Ho :	µ1 - µ2 =	0
Ha :	µ1-µ2 <	0

b)

since Ha :µ1-µ2 < 0

so left tail test

c)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   successful                  
mean of sample 1,    x̅1=   17.25                  
standard deviation of sample 1,   s1 =    2.60                  
size of sample 1,    n1=   8                  
                          
Sample #2   ---->   unsuccessful                  
mean of sample 2,    x̅2=   19.13                  
standard deviation of sample 2,   s2 =    3.00                  
size of sample 2,    n2=   8                  
                          
difference in sample means =    x̅1-x̅2 =    17.2500   -   19.1   =   -1.88  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.8078                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.4039                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -1.8750   -   0   ) /    1.40   =   -1.336

d)

  p-value =        0.1015 [ excel function: =T.DIST(t stat,df) ]           

=10.15%

    Degree of freedom, DF=   n1+n2-2 =    14              
t-critical value , t* =        -1.7613   (excel function: =t.inv(α,df)     

   
Decision:   | t-stat | < | critical value |, so, Do not Reject Ho