Question

Total #of men's per Bag/ it of people 53 SS 56 54 57 58 S 8 13 in 3 2 1 59 Activity 14 [2 points): After checking many bags of M&M's, an M&M enthusiast believes that there should be about 58 plain M&M's per 1.69 oz. bag. Test this claim using the class data at the 0.05 level of significance. Hoi Activity 15 [2 points): Create a 95% confidence interval for the mean number of M&M's per bag, using the class data. Interpret your result. Chez Com

Answer 1

1)

Ho :   µ =   58                  
Ha :   µ ╪   58       (Two tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   1.5285                  
Sample Size ,   n =    45                  
Sample Mean,    x̅ = ΣX/n =    56.0667                  
                          
degree of freedom=   DF=n-1=   44                  
                          
Standard Error , SE = s/√n =   1.5285   / √    45   =   0.2279      
t-test statistic= (x̅ - µ )/SE = (   56.067   -   58   ) /    0.2279   =   -8.48
                          

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

2)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   44          
't value='   tα/2=   2.0154   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   1.5285   / √   45   =   0.227858
margin of error , E=t*SE =   2.0154   *   0.22786   =   0.459217
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    56.07   -   0.459217   =   55.607449
Interval Upper Limit = x̅ + E =    56.07   -   0.459217   =   56.525884
95%   confidence interval is (   55.61   < µ <   56.53   )

Thanks in advance!

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