Question

A grocery store wants to know which system is better for decreasing the amount of time that their customers wait in checkout lines. Is there a significant difference in the average time customers wait in line using the cashier operated checkout and the self-checkout? The data was collected by recording the amount of time it takes to check out 20 customers that enter the cashier checkout line and 20 customers that enter the self-checkout line. The number of minutes from when the customer enters the checkout line completes the check out process and exits the line.

Cashier Electronic
4.21   9.66
5.55   5.9
3.02   8.02
5.13   5.79
4.77   8.73
2.34   3.82
3.54   8.01
3.2 8.35
4.5 10.49
6.1 6.68
0.38   5.64
5.12   4.08
4.26   6.71
6.19   9.91
3.79   5.47
4.25   4.06
6.32   8.65
4.45   3.5
4.26   5.58
3.46   7.41

1. Create an appropriate graphical display (e.g., bar chart, scatter plot, line chart, or histogram) of the data you collected.

This display should be a summary or representation of your data, not raw data.

2. Report how you analyzed the data using a two-sample t-test analysis technique

3. Include the output and any calculations of the analysis you performed on Excel.

4.. Discuss the results of your data analysis.

5. Discuss the limitation(s) of your data analysis.

6. Recommend a course of action based on your results.

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   4.24                  
standard deviation of sample 1,   s1 =    1.41                  
size of sample 1,    n1=   20                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   6.82                  
standard deviation of sample 2,   s2 =    2.12                  
size of sample 2,    n2=   20                  
                          
difference in sample means =    x̅1-x̅2 =    4.2420   -   6.8   =   -2.58  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.7999                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.5692                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -2.5810   -   0   ) /    0.57   =   -4.535
                          
Degree of freedom, DF=   n1+n2-2 =    38                  
  
p-value =        0.000056   (excel function: =T.DIST.2T(t stat,df) )

              
Conclusion:     p-value <α , Reject null hypothesis      

              
it is conlcuded that  there is a significant difference in the average time customers wait in line using the cashier operated checkout and the self-checkout

pleas erevert back fro doubt