Question

​​​​​​A sample survey of 36 randomly selected households shows that the median household income of the eastern NC residents is $51,000, with a standard deviation of $1075. Test the (null) hypothesis that the actual (real) median household income is $52,752 at 0.05% confidence level.

Answer 1

Ho :   µ =   52752                  
Ha :   µ ╪   52752       (Two tail test)          
                          
Level of Significance ,    α =    0.050                  
sample std dev ,    s =    1075.0000                  
Sample Size ,   n =    36   1155625.0000              
Sample Mean,    x̅ =   51000.0000                  
                          
degree of freedom=   DF=n-1=   35                  
                          
Standard Error , SE = s/√n =   1075.0000   / √    36   =   179.1667      
t-test statistic= (x̅ - µ )/SE = (   51000.000   -   52752   ) /    179.1667   =   -9.779
                          
  
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
Conclusion: There is enough evidence to conclude that actual (real) median household income is not $52,752

at 0.05% confidence level.