Question

An article presents a study comparing the effectiveness of a video system that allows a crane operator to see the lifting point while operating the crane with the old system in which the operator relies on hand signals from a tagman. Three different lifts, A, B, and C were studied. Lift A was of little difficulty, lift B was of moderate difficulty, and lift C was of high difficulty. Each lift was performed several times, both with the new video system and with the old tagman system. The time (in seconds) required to perform each lift was recorded. The following table present the mean, standard deviation, and sample size.

Low Difficulty Mean Standard Tagman 47.79 2.19 14 Video 47.55 2.65 40 Deviation Sample Size Moderate Difficulty Mean Standard Tagman 67.33 6.26 12 Video 59 5.59 24 Deviation Sample Size Standard Sample Size High Difficulty Mean Deviation Tagman 107.71 17.02 17 Video 84.52 13.51 29

Can you conclude that the mean time to perform a lift of moderate difficulty is less when using the video system than when using the tagman system? Find the P-value and state a conclusion.

Since  (Choose between) (0.0005 < P <0.001 , 0.001 < P < 0.005 , 0.025 < P < 0.05 , 0.05 < P < 0.10 , P < 0.0005 , P > 0.40 , 0.01 < P < 0.025 , 0.10 < P < 0.25 , 0.005 < P < 0.01 , 0.25 < P < 0.40) we (choose between) ( cannot , can ) conclude that the mean time to perform a lift of moderate difficulty is less with the video system than with the tagman system.

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   67.330                  
standard deviation of sample 1,   s1 =    6.260                  
size of sample 1,    n1=   12                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   59.000                  
standard deviation of sample 2,   s2 =    5.590                  
size of sample 2,    n2=   24                  
                          
difference in sample means =    x̅1-x̅2 =    67.3300   -   59.0   =   8.33  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.8152                  
std error , SE =    Sp*√(1/n1+1/n2) =    2.0560                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   8.3300   -   0   ) /    2.06   =   4.0516
                          
Degree of freedom, DF=   n1+n2-2 =    34                        
p-value =        0.00014 [excel function: =T.DIST.RT(t stat,df) ]              

P < 0.0005

we can conclude that the mean time to perform a lift of moderate difficulty is less with the video system than with the tagman system.

Please let me know in case of any doubt.

Thanks in advance!

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