Question

1. Acompanysellinglicensesofnewe-commercesoftwareadvertisedthatfirmsusingthissoftware could obtain, on average during the first year, a minimum yield (in cost savings) of 20 percent on average on their software investment. To disprove the claim, we checked with 10 different firms which used the software. These firms reported the following cost-saving yields (in percent) during the first year of their operations:

   {17.0, 19.2, 21.5, 18.6, 22.1, 14.9, 18.4, 20.1, 19.4, 18.9}.

   1. Do we have significant evidence to show that the software company’s claim of a minimum of 20 percent in cost savings was not valid? Test using α = 0.05.

   2. Compute a 95% confidence interval for the average cost-saving yield estimate.

Answer 1

Ho :   µ ≥ 20                  
Ha :   µ <   20       (Left tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   2.0712                  
Sample Size ,   n =    10                  
Sample Mean,    x̅ = ΣX/n =    19.0100                  
                          
degree of freedom=   DF=n-1=   9                  
                          
Standard Error , SE = s/√n =   2.0712   / √    10   =   0.654972      
t-test statistic= (x̅ - µ )/SE = (   19.010   -   20   ) /    0.65497   =   -1.51
                          
  
p-Value   =   0.0825 [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value>α, Do not reject null hypothesis                       
Conclusion: There is not enough evidence to reject  the claim

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Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   9          
't value='   tα/2=   2.2622   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   2.0712   / √   10   =   0.654972
margin of error , E=t*SE =   2.2622   *   0.65497   =   1.481651
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    19.01   -   1.481651   =   17.528349
Interval Upper Limit = x̅ + E =    19.01   -   1.481651   =   20.491651
95%   confidence interval is (   17.53   < µ <   20.49   )