Question

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6973 subjects randomly selected from an online group involved with ears. There were 1299 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than​ 20%. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution.

Answer 1

Ho :   p =    0.2                  
H1 :   p <   0.2       (Left tail test)          
                          
Level of Significance,   α =    0.01                  
Number of Items of Interest,   x =   1299                  
Sample Size,   n =    6973                  
                          
Sample Proportion ,    p̂ = x/n =    0.1863                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0048                  
Z Test Statistic = ( p̂-p)/SE = (   0.1863   -   0.2   ) /   0.0048   =   -2.8621
                          

p-Value   =   0.0021   [excel function =NORMSDIST(z)]              
Decision:   p-value<α , reject null hypothesis                       
There is enough evidence to conclude that return rate is less than​ 20%