Question

A new nail salon business just opened, in their first week of business they decided that thy would conduct a promotion in which a customer's bill can be randomly selected to receive a discount. When a customer's bill is printed, a program in the cash register randomly determines whether the customer will receive a discount on the bill. The program was written to generate a discount with a probability of 0.2, that is, giving a discount to 20 percent of the bills in the long run. However, the owner is concerned that the program has a mistake that results in the program not generating the intended long-run proportion of 0.2.

(a) The owner selected a random sample of 100 bills and found that only 16 percent of them received discounts. The conditions for inference are met. Using the sample data collected by the owner, calculate a 95% confidence interval for the true proportion of bills that will receive a discount in the long run.

(b) Observing the value that you received in part a. Do you believe that the confidence interval provide convincing statistical evidence to indicate that the program is not working as intended?

Answer 1

Level of Significance,   α =    0.05          
Sample Size,   n =    100          
                  
Sample Proportion ,    p̂ = 0.160          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0367          
margin of error , E = Z*SE =    1.960   *   0.0367   =   0.0719
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.160   -   0.0719   =   0.0881
Interval Upper Limit = p̂ + E =   0.160   +   0.0719   =   0.2319
                  
95%   confidence interval is (   0.088   < p <    0.232   )

b)

Ho: p=0.20

H1: p╪0.20

the confidence interval do not provide convincing statistical evidence to indicate that the program is not working as intended because proportion 0.20 lies in the confidence interval