Question

A farmer takes a load of oranges to the local grocery store to be sold. The farmer claims the weight of the oranges will be at least 11 ounces each. You take a sample of 15 oranges and find a sample mean of 11.14 ounces. Test this claim for a 0.05 level of significance using excel. Standard deviation of 2.0.

Step 1: Let mu = population mean

Ho: mu =

Ha: mu>

Step 2: get z-score =norm.dist( )

The use and type this in excel =norm.s.inv(z-score from above)

Now make a decision: If the p-value is less than 05, you reject the null hypothesis. If no, we FAIL to reject.

Answer 1

1)

Ho :   µ =   11                  
Ha :   µ > 11       (Left tail test)

2)
                          
Level of Significance ,    α =    0.05                  
population std dev ,    σ =    2.0000                  
Sample Size ,   n =    15                  
Sample Mean,    x̅ =   11.1400                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   2.0000   / √    15   =   0.5164      
Z-test statistic= (x̅ - µ )/SE = (   11.140   -   11   ) /    0.5164   =   0.27

p-Value   =   0.3932
                                
critical z value, z* =       -1.645 (using =norm.s.inv)
           
Decision:   p-value>α, Fail to reject null hypothesis