Question

In: Statistics and Probability

A garden seed wholesaler wishes to test the claim that tomato seeds germinate faster when each...

A garden seed wholesaler wishes to test the claim that tomato seeds germinate faster when each individual seed is "pelletized" within a coating of corn starch. The table below shows the germination times, in days, of six pelletized seeds. The table also shows the germination times in days of six un-coated seeds (the controls).

Pelletized: 8 6 7 8 10 7

Control: 11 8 9 10 7 11

Can you conclude that the mean germination time for pelletized seeds is less than the mean for the un-pelletized seeds? Use the α = 0.05 level of significance.

Solutions

Expert Solution

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 <   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   1          
mean of sample 1,    x̅1=   7.67          
standard deviation of sample 1,   s1 =    1.366260102          
size of sample 1,    n1=   6          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   9.333          
standard deviation of sample 2,   s2 =    1.63          
size of sample 2,    n2=   6          
                  
difference in sample means = x̅1-x̅2 =    7.667   -   9.3333   =   -1.6667
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.8692          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -1.6667   /   0.8692   ) =   -1.9174
                  
Degree of freedom, = 9          
                  
                  
t-critical value , t* =        -1.8331   (excel function: =t.inv(α,df)      
Decision:   | t-stat | > | critical value |, so, Reject Ho              
p-value =        0.04371   [ excel function: =T.DIST(t stat,df) ]       
Conclusion:     p-value<α , Reject null hypothesis              

yes,  you conclude that the mean germination time for pelletized seeds is less than the mean for the un-pelletized seeds


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