Question

In: Statistics and Probability

A systems analyst tests a new algorithm designed to work faster than the currently-used algorithm. Each...

A systems analyst tests a new algorithm designed to work faster than the currently-used algorithm. Each algorithm is applied to a group of 5454 sample problems. The new algorithm completes the sample problems with a mean time of 11.9011.90 hours. The current algorithm completes the sample problems with a mean time of 14.0714.07 hours. Assume the population standard deviation for the new algorithm is 5.1155.115 hours, while the current algorithm has a population standard deviation of 5.7365.736 hours. Conduct a hypothesis test at the 0.050.05 level of significance of the claim that the new algorithm has a lower mean completion time than the current algorithm. Let μ1μ1 be the true mean completion time for the new algorithm and μ2μ2 be the true mean completion time for the current algorithm.

Step 1 of 5: State the null and alternative hypotheses for the test.

Step 2 of 5: Compute the value of the test statistic. Round your answer to two decimal places.

Step 3 of 5: Find the p-value associated with the test statistic. Round your answer to four decimal places.

Step 4 of 5: Make the decision for the hypothesis test.

Step 5 of 5: State the conclusion of the hypothesis test.

Solutions

Expert Solution

1)

µ1 == New algorithm mean

µ2 ==== Current algorithm mean

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
  

2)


Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   successful                  
mean of sample 1,    x̅1=   11.90                  
standard deviation of sample 1,   s1 =    5.12                  
size of sample 1,    n1=   54                  
                          
Sample #2   ---->   unsuccessful                  
mean of sample 2,    x̅2=   14.07                  
standard deviation of sample 2,   s2 =    5.74                  
size of sample 2,    n2=   54                  
                          
difference in sample means =    x̅1-x̅2 =    11.9000   -   14.1   =   -2.17  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    N/A                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.0458                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -2.1700   -   0   ) /    1.05   =   -2.08

3)

  
Degree of freedom, DF=   n1+n2-2 =    104                  
t-critical value , t* =        -1.6596   (excel function: =t.inv(α,df)              
Decision:   | t-stat | > | critical value |, so, Reject Ho                      
p-value =        0.0202 [ excel function: =T.DIST(t stat,df) ]       

4)

   
Decision:     p-value <α , Reject null hypothesis  

5)
                          
There is enough evidence to  New algorithm mean is less than current algorithm mean

Please revert in case of any doubt.

Please upvote. Thanks in advance

  


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