Question

1. More than $70 billion is spent each year in the drive-thru lanes of America’s fast-food restaurants. Having quick, accurate, and friendly service at a drive-thru window translates directly into revenue for the restaurant. According to Jack Greenberg, former CEO of McDonald’s, sales increase 1% for every six seconds saved at the drive-thru. So industry executives, stockholders, and analysts closely follow the ratings of fast-food drive-thru lanes that appear annually in QSR, a publication that reports on the quick-service restaurant industry.

The 2012 QSR magazine drive-thru study involved visits to a random sample of restaurants in the 20 largest fast-food chains in all 50 states. During each visit, the researcher ordered a modified main item (for example, a hamburger with no pickles), a side item, and a drink. If any item was not received as ordered, or if the restaurant failed to give the correct change or supply a straw and a napkin, then the order was considered “inaccurate.” Service time, which is the time from when the car stopped at the speaker to when the entire order was received, was measured each visit. Researchers also recorded whether or not
each restaurant had an order-confirmation board in its drive-thru.

Here are some results from the 2012 QSR study:

• For restaurants with order-confirmation boards, 1169 of 1327 visits (88.1%) resulted in accurate orders. For restaurants with no order-confirmation board, 655 of 726 visits (90.2%) resulted in accurate orders.
• McDonald’s average service time for 362 drive-thru visits was 188.83 seconds with a standard
  deviation of 17.38 seconds. Burger King’s service time for 318 drive-thru visits had a mean of 201.33
  seconds and a standard deviation of 18.85 seconds.
• a. Is there a significant difference in order accuracy between restaurants with and without order-confirmation boards? Carry out an appropriate test at the a = 0.05 level to help answer this question.
• b. A 95% confidence interval for the difference in the population proportions of accurate orders at restaurants with and without order-confirmation boards is (–0.049, 0.00649). Interpret the confidence interval and explain how the confidence interval is consistent with your conclusion from Question 1.
• c. Construct and interpret a 99% confidence interval for the difference in the mean service times at McDonald’s and Burger King drive-thru's.

Answer 1

a)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0          
                  
sample #1   ----->              
first sample size,     n1=   1327          
number of successes, sample 1 =     x1=   1169          
proportion success of sample 1 , p̂1=   x1/n1=   0.8809          
                  
sample #2   ----->              
second sample size,     n2 =    726          
number of successes, sample 2 =     x2 =    655          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.9022          
                  
difference in sample proportions, p̂1 - p̂2 =     0.8809   -   0.9022   =   -0.0213
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.888455918          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.01453          
Z-statistic = (p̂1 - p̂2)/SE = (   -0.021   /   0.0145   ) =   -1.46
                        
p-value =        0.1433   [excel formula =2*NORMSDIST(z)]      
decision :    p-value>α,Don't reject null hypothesis               
                  
Conclusion:   There is not enough evidence to conclued that there is a significant difference in order accuracy between restaurants with and without order-confirmation boards

b)

level of significance, α =   0.05              
Z critical value =   Z α/2 =    1.960   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.01416          
margin of error , E = Z*SE =    1.960   *   0.0142   =   0.02776
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.021   -   0.0278   =   -0.0490272
upper limit = (p̂1 - p̂2) + E =    -0.021   +   0.0278   =   0.0064884
                  
so, confidence interval is (   -0.049   < p1 - p2 <   0.006   )  

As we can see 0 lies in the interval and hence the results are not significant

c)

Degree of freedom, DF=   n1+n2-2 =    678              
t-critical value =    t α/2 =    2.5831   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    18.0822              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    1.3898              
margin of error, E = t*SE =    2.5831   *   1.39   =   3.59  
                      
difference of means =    x̅1-x̅2 =    188.8300   -   201.330   =   -12.5000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -12.5000   -   3.5899   =   -16.0899
Interval Upper Limit=   (x̅1-x̅2) + E =    -12.5000   +   3.5899   =   -8.9101

Please let me know in case of any doubt.

Thanks in advance!

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