Question

A manufacturer wants to compare the number of defects on the day shift with the number on the evening shift. A sample of production from recent shifts showed the following defects: Day Shift 5 8 7 6 9 7

Evening Shift 8 10 7 11 9 12 14 9

The objective is to determine whether the mean number of defects on the night shift is greater than the mean number on the day shift at the 95% confidence level.

State the null and alternate hypotheses.

What is the level of significance? What is the test statistic? What is the decision rule? Use the Excel Data Analysis pack to analyze the problem. Include the output with your answer. (Note: You may calculate by hand if you prefer). What is your conclusion? Explain. Does the decision change at the 99% confidence level?

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->  DAY
mean of sample 1,    x̅1=   7.00                  
standard deviation of sample 1,   s1 =    1.41                  
size of sample 1,    n1=   6                  
                          
Sample #2   ---->   NIGHT
mean of sample 2,    x̅2=   10.00                  
standard deviation of sample 2,   s2 =    2.27                  
size of sample 2,    n2=   8                  
                          
difference in sample means =    x̅1-x̅2 =    7.0000   -   10.0   =   -3.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.9579                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.0574                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -3.0000   -   0   ) /    1.06   =   -2.837
                          
Degree of freedom, DF=   n1+n2-2 =    12                  

p-value =        0.007487   [ excel function: =T.DIST(t stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis      

mean number of defects on the night shift is greater than the mean number on the day shift

............

AT 0.01

P VALUE < 0.01,

REJECTHo

result does not changes

..............

Please revert in case of any doubt.

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