Question

Write a confidence interval and hypothesis test using the word problem below.

I am currently building a home gym in my garage and am not buying calibrated weight plates to reduce the overall budget. I want to know if the average weight of a 45lb steel plate is the same as a 45-lb concrete filled plate. I weighed seven steel plates and find weights of 44.8, 45.3, 45.0, 44.9, 44.7, 45.1, and 45.0lbs. I weighed seven concrete filled plates and find weights of 44.5, 44.6, 44.8, 45.4, 45.3, 45.1 and 44.7. Assume a random sample is drawn.

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   44.97                  
standard deviation of sample 1,   s1 =    0.20                  
size of sample 1,    n1=   7                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   44.91                  
standard deviation of sample 2,   s2 =    0.35                  
size of sample 2,    n2=   7                  
                          
difference in sample means =    x̅1-x̅2 =    44.9714   -   44.9   =   0.06  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.2862                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.1530                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   0.0571   -   0   ) /    0.15   =   0.374
                          
Degree of freedom, DF=   n1+n2-2 =    12                  
  
p-value =        0.715260   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis  

                  
SO,  the average weight of a 45lb steel plate is the same as a 45-lb concrete filled plate.

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Degree of freedom, DF=   n1+n2-2 =    12              
t-critical value =    t α/2 =    2.1788   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.2862              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.1530              
margin of error, E = t*SE =    2.1788   *   0.1530   =   0.3333  
                      
difference of means =    x̅1-x̅2 =    44.9714   -   44.914   =   0.0571
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    0.0571   -   0.3333   =   -0.276
Interval Upper Limit=   (x̅1-x̅2) + E =    0.0571   +   0.3333   =   0.390
CI (-0.276 , 0.390).

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