Question

A gardener was interested to see how different types of fertilizer affect the yield of tomatoes in his garden. Two brands of fertilizer, A and B, were available for comparison. He had enough room to grow eleven plants in a single row, so he picked a random sample of five spots for A, and the remainder for B. The yield for each tomato plant in pounds was as follows:

Position 1 2 3 4 5 6 7 8 9 10 11

Fertilizer A A B B A B B B A A B

Yield 29.9 11.4 26.6 23.7 25.3 28.5 14.2 17.9 16.5 21.1 24.3

Part (e) The gardener is interested to know whether or not there is a meaningful difference between the yield of tomatoes grown using fertilizer A instead of B. State the null and alternative hypotheses, compute the appropriate test statistic, state its distribution and report the p-value. You should assume that variance of yields is the same for both fertilizer brands.

Part (f) Based on your computations, what conclusion would you draw about the two types of fertilizer?

Part (g) Horticulturally speaking, it would have been more convenient to use fertilizer A on the first five plants in the row, and fertilizer B for the remainder. The gardener was a statistician. Why did he eschew the more convenient option?

Answer 1

e)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   20.84                  
standard deviation of sample 1,   s1 =    7.25                  
size of sample 1,    n1=   5                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   22.53                  
standard deviation of sample 2,   s2 =    5.43                  
size of sample 2,    n2=   6                  
                          
difference in sample means =    x̅1-x̅2 =    20.8400   -   22.5   =   -1.69  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    6.3028                  
std error , SE =    Sp*√(1/n1+1/n2) =    3.8165                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -1.6933   -   0   ) /    3.82   =   -0.444
                          
Degree of freedom, DF=   n1+n2-2 =    9                  
  
p-value =        0.6677 (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis                      
                          

f)

There is not enough evidence that   there is a meaningful difference between the yield of tomatoes grown using fertilizer A instead of B.

g)

Farmer done this to avoid any bias for the study

THANKS

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