Question

5. A coffee shop claims that a five-ounce serving of its fresh-brewed coffee has mean caffeine content of 80 milligrams. A sample of 76 five-ounce servings finds mean caffeine content of 83 milligrams with a standard deviation of 12 milligrams. At the 5% level of significance, can we show that the mean caffeine content of all five-ounce servings is not 80 milligrams?

(a) State and test appropriate hypotheses. State conclusions.


(b) If an error was made in (a), what type was it?

(c) Find the p-vale of the test in (a).

Answer 1

a)

Ho :   µ =   80                  
Ha :   µ ╪   80       (Two tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s =    12.0000                  
Sample Size ,   n =    76                  
Sample Mean,    x̅ =   83.0000                  
                          
degree of freedom=   DF=n-1=   75                  
                          
Standard Error , SE = s/√n =   12.0000   / √    76   =   1.3765      
t-test statistic= (x̅ - µ )/SE = (   83.000   -   80   ) /    1.3765   =   2.18
                          

                          
p-Value   =   0.0324   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
Conclusion: There is enough evidence that   the mean caffeine content of all five-ounce servings is not 80 milligrams

b)

Type 1 error

c)

p-Value   =   0.0324   [Excel formula =t.dist(t-stat,df) ]      

Thanks in advance!

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