Question

Plain M&M: 15,14,15,15,14,16,15,14,15,16,16,14,15,14,15,15,14,17,17,15,15,17,14,16,14,16,17,13,16,16, 15, 17,6,14,16,15,13,15,17,15,17,14,15,14,16,14

Peanut Butter: 8,6,8,7,8,8,8,7,3,7,10,7,7,6,8,7,8,7,7,7,8,8,8,7,9,7,8,8,7,8,8,8,6,7,8,8,8,7,6,7,7,8,8,8,14,8,8,7,6

Skittles: 15,14,15,14,14,15,15,14,15,15,14,15,15,15,14,15,20,15,15,15,14,14,15,15,15,16,12,15,15,14, 14,15,14,15,14,14,14,13,16

**Assume that both Plain and Peanut M&M candies, and Skittles are normally distributed.

2. Suppose that M&M claims that each bag of Peanut M&Ms should be 18 grams and Plain M&Ms should be 13.5 grams. Suppose that Skittles claims that each bag contains 16 grams.
a. Test the claim that M&M is shorting its customers in servings of Peanut M&Ms.
b. Test the claim that M&M is overfilling Plain serving bags of M&Ms.
c. Test the claim that Skittles weights are different than stated.
d. Discuss your choice of ?.
i. Why did you choose the ? you did?
ii. If you had chosen a different ?, would it have affected your conclusion?

Answer 1

a)

Ho :   µ =   18                  
Ha :   µ <   18       (Left tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   1.3860                  
Sample Size ,   n =    49                  
Sample Mean,    x̅ = ΣX/n =    7.5306                  
                          
degree of freedom=   DF=n-1=   48                  
                          
Standard Error , SE = s/√n =   1.3860   / √    49   =   0.1980      
t-test statistic= (x̅ - µ )/SE = (   7.531   -   18   ) /    0.1980   =   -52.88

                          
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
Conclusion: There is enough evidence that M&M is shorting its customers in servings of Peanut M&Ms

b)

Ho :   µ =   13.5                  
Ha :   µ >   13.5       (Right tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   1.7506                  
Sample Size ,   n =    46                  
Sample Mean,    x̅ = ΣX/n =    14.9565                  
                          
degree of freedom=   DF=n-1=   45                  
                          
Standard Error , SE = s/√n =   1.7506   / √    46   =   0.2581      
t-test statistic= (x̅ - µ )/SE = (   14.957   -   13.5   ) /    0.2581   =   5.64

                          
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
  

Conclusion: There is enough evidence that M&M is overfilling Plain serving bags of M&Ms.

c)

Ho :   µ =   16                  
Ha :   µ ╪   16       (Two tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   1.1506                  
Sample Size ,   n =    39                  
Sample Mean,    x̅ = ΣX/n =    14.6923                  
                          
degree of freedom=   DF=n-1=   38                  
                          
Standard Error , SE = s/√n =   1.1506   / √    39   =   0.1842      
t-test statistic= (x̅ - µ )/SE = (   14.692   -   16   ) /    0.1842   =   -7.10
                          

                          
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
Conclusion: There is enough evidence that   Skittles weights are different than stated

d)

choice of ? = 0.01

e)

To avoid type 1 error

No it would not effect conclusion

THANKS

revert back for doubt

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