Question

The data below are yields for two different types of corn seed that were used on adjacent plots of land. Assume that the data are simple random samples and that the differences have a distribution that is approximately normal. Construct a​ 95% confidence interval estimate of the difference between type 1 and type 2 yields. What does the confidence interval suggest about farmer​ Joe's claim that type 1 seed is better than type 2​ seed?

Type 1	21402140	18991899	20492049	23962396	22032203	19911991	21712171	14411441
Type 2	20712071	19171917	20842084	24212421	21122112	19061906	21772177	14491449

Answer 1

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
2140	2071	69	2487.5156
1899	1917	-18	1378.2656
2049	2084	-35	2929.5156
2396	2421	-25	1947.0156
2203	2112	91	5166.0156
1991	1906	85	4339.5156
2171	2177	-6	631.2656
1441	1449	-8.0000	735.7656

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	16290	16137	153	19614.875

mean of difference ,    D̅ =ΣDi / n =   19.1250000
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    52.9351018

sample size ,    n =    8          
Degree of freedom, DF=   n - 1 =    7   and α =    0.05  
t-critical value =    t α/2,df =    2.3646   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    52.9351          
                  
std error , SE = Sd / √n =    52.9351   / √   8   =   18.7154
margin of error, E = t*SE =    2.3646   *   18.7154   =   44.2549
                  
mean of difference ,    D̅ =   19.125          
confidence interval is                   
Interval Lower Limit= D̅ - E =   19.125   -   44.2549   =   -25.1299
Interval Upper Limit= D̅ + E =   19.125   +   44.2549   =   63.3799
                  
so, confidence interval is (   -25.1299   < Dbar <   63.3799   )  

since, confidence interval contains 0, so, test is not significant

hence, confidence interval does not have enough evidence to claim that  type 1 seed is better than type 2​ seed