Question

8. A sheep farmer suspects that when their animals are given a feed with a higher nitrogen content they tend to produce a greater weight of wool. To test this claim the farmer gave a random selection of 17 of their sheep a low nitrogen content feed and gave another random selection of 14 of their sheep a high nitrogen content feed. In the low nitrogen group it was found that the sheep produced an average of 10.9 pounds of wool with a standard deviation of 2.7 pounds. In the high nitrogen group it was found that the sheep produced an average of 15.4 pounds of wool with a standard deviation 3.5 pounds. [2 marks] Use these results to calculate an 80% confidence interval for the true mean difference in wool weight produced between the two groups. (You may order your two groups however you wish, but make your choice clear.) If needed, you may assume that the weight of wool produced under both types of feed is normally distributed. (b) (1 mark] Using your confidence interval from part (a), is it reasonable to say that the weight of wool produced between the low nitrogen group and the high nitrogen group is the same? Explain

Answer 1

a)

Sample #1   ----> Wool weight with low nitrogen   
mean of sample 1,    x̅1=   10.900                  
standard deviation of sample 1,   s1 =    2.700                  
size of sample 1,    n1=   17                  
                          
Sample #2   ---->   Wool weight with high nitrogen
mean of sample 2,    x̅2=   15.400                  
standard deviation of sample 2,   s2 =    3.500                  
size of sample 2,    n2=   14                  
                          
difference in sample means =    x̅1-x̅2 =    10.9000   -   15.4   =   -4.50  

Degree of freedom, DF=   n1+n2-2 =    29              
t-critical value =    t α/2 =    1.3114   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    3.0844              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    1.1132              
margin of error, E = t*SE =    1.3114   *   1.11   =   1.46  
                      
difference of means =    x̅1-x̅2 =    10.9000   -   15.400   =   -4.5000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -4.5000   -   1.4598   =   -5.9598
Interval Upper Limit=   (x̅1-x̅2) + E =    -4.5000   +   1.4598   =   -3.0402

b)

From the confidence interval it is clear not value of zero is there. It means wool weight produce by high nitrogen is higher than low wool weight.

Please let me know in case of any doubt.

Thanks in advance!

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