Question

(Solve by Rstudio)

Farmers would like to know the amount of sunshine in given locations they may buy land. They hypothesize that a place near the coast may have different cloud cover than a place further from the coast at the same latitude so that the sun angles are the same.   In the data sheet you have daily noontime reports of the fraction of cloud cover near the coast and inland. Justify your acceptance or rejection of the hypothesis using these numbers, and state the confidence level you would use.

(Data)

Cld cover coast: ,71,26,100,16,100,83,61,32,84,15

Cld cover inland: ,60,9,6,29,12,19,21,62,82,0,4,84,51,4,19

Answer 1

confidence level used = 0.05

coast	inland
71	60
26	9
100	6
16	29
100	12
83	19
61	21
32	62
84	82
15	0
	4
	84
	51
	4
	19

Sample #1   ---->   1                  
mean of sample 1,    x̅1=   58.80                  
standard deviation of sample 1,   s1 =    33.85                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   30.80                  
standard deviation of sample 2,   s2 =    29.15                  
size of sample 2,    n2=   15                  
                          
difference in sample means =    x̅1-x̅2 =    58.8000   -   30.8   =   28.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    31.0721                  
std error , SE =    Sp*√(1/n1+1/n2) =    12.6852                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   28.0000   -   0   ) /    12.69   =   2.2073
                          
Degree of freedom, DF=   n1+n2-2 =    23                  
  
p-value =        0.037542   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis                      
                          
There is enough evidence of significant difference