Question

The superintendent of a local maintenance company claims that, after their technicians completed a new training program, less than 20 percent of customers were dissatisfied with the services rendered. A customer service consultant evaluated this claim by randomly surveying 80 customers and found that only 10 were dissatisfied. a. Test the GM’s claim at the 5 percent level of significance. b. Construct the 95 percent confidence interval for the proportion of customers that are dissatisfied with the service.

Answer 1

n =   80      
x =    10      
Sample proportion, p̄ =   0.125      
α =    0.05      
  
a) Null and Alternative hypothesis:          
Ho : p =   0.2      
H1 : p <   0.2      
          
Critical value :          
At α = 0.05, left tailed critical value, z crit = NORM.S.INV(0.05) = -1.645  
          
Test statistic:          
z =(p̄ -p)/(√(p*(1-p)/n))          
= -1.6771     
          
Decision:          

As z = -1.6771 < z crit = -1.645, we reject the null hypothesis     

Conclusion:

There is enough evidence to claim that less than 20 percent of customers were dissatisfied with the services rendered at 0.05 significance level.

b) At α = 0.05, two tailed critical value, z crit = NORM.S.INV(0.05/ 2) = 1.960

95% Confidence interval :      
                
Lower Bound = p̄ - z-crit*(√( p̄ *(1- p̄ )/n)) = 0.0525  
          
Upper Bound = p̄ + z-crit*(√( p̄ *(1- p̄ )/n)) = 0.1975