In: Statistics and Probability
A manufacturer's association claims that 92% of adults own cell phones. A technology researcher thinks the proportion is lower. In a random sample of 300 adults, 267 said they had a cell phone. Using these results and a 0.05 significance level, formulate and test the researcher's hypothesis.
Ho=
Ha=
Sample Statistic =
null value =
SE =
Z statistic =
Test type =
p-value =
Decision =
Ho :   p =    0.92  
           
   
H1 :   p <   0.92  
    (Left tail test)      
   
          
           
   
Level of Significance,   α =   
0.05          
       
Number of Items of Interest,   x =  
267          
       
Sample Size,   n =    300  
           
   
          
           
   
Sample Proportion ,    p̂ = x/n =   
0.89          
       
          
           
   
Standard Error ,    SE = √( p(1-p)/n ) =   
0.0157          
       
Z Test Statistic = ( p̂-p)/SE = (   0.8900  
-   0.92   ) /   0.0157  
=   -1.9153
          
           
         
          
           
   
p-Value   =   0.027725395   [excel
function =NORMSDIST(z)]      
       
Decision:   p-value<α , reject null hypothesis
          
           
There is enough evidence to support the claim
that  proportion is lower
Please revert back in case of any doubt.
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