In: Statistics and Probability
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.
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.
..................
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).
..................
THANKS
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