In: Math
Some Alzheimer’s caregivers were asked to respond to the following statement: “Caregiving enabled me to develop a more positive attitude toward life.” The responses are reflected in the following table.
| 
 Strongly Disagree  | 
 Somewhat Disagree  | 
 No Opinion  | 
 Somewhat Agree  | 
 Strongly Agree  | 
| 
 166  | 
 116  | 
 171  | 
 234  | 
 542  | 
a)

from graph, data indicate the true distribution is not uniform
b)
Ho: data indicate the true distribution is uniform
H1: data indicate the true distribution is not uniform
Chi square test for Goodness of fit  
  
expected frequncy,E = expected proportions*total
frequency  
total frequency=   1229
| observed frequencey, O | expected proportion | expected frequency,E | (O-E) | (O-E)² | (O-E)²/E | 
| 166 | 0.2000 | 245.800 | -79.80 | 6368.04 | 25.907 | 
| 116 | 0.2000 | 245.800 | -129.80 | 16848.04 | 68.544 | 
| 171 | 0.2000 | 245.800 | -74.80 | 5595.04 | 22.763 | 
| 234 | 0.2000 | 245.800 | -11.80 | 139.24 | 0.566 | 
| 542 | 0.200 | 245.800 | 296.20 | 87734.44 | 356.934 | 
chi square test statistic,X² = Σ(O-E)²/E =  
474.7144          
   
          
       
level of significance, α=   0.05  
           
Degree of freedom=k-1=   5   -  
1   =   4
          
       
P value =   0.0000   [ excel function:
=chisq.dist.rt(test-stat,df) ]      
   
Decision: P value < α, Reject Ho      
           
conclusion: there is enough evidence to conclude that data indicate the true distribution is not uniform