Question

(1 point) Here we investigate whether the register balance at a local retail store is better on days with a manager than days without a manager. This evidence might be used to determine whether or not you should always schedule a manager. The table gives the register balance for a sample of 10 days with a manager and 10 days without a manager. Here, 0 means the register balance is right on, negative means there is less money than there should be, and positive means there is more money than there should be. The number of degrees of freedom (d.f.) that you should use in your calculations is given in the table.

	Register Balance (10 days each)										mean	s2s2	ss
With Manager (x1x1)	-3	-7	-3	-5	-4	2	6	-4	-10	1	-2.7	21.3444444444444	4.62000481000231
Without Manager (x2x2)	-5	-6	-4	-12	-4	-9	-5	1	-8	-9	-6.1	12.9888888888889	3.6040101122068
degrees of freedom: d.f. = 17

Test the claim that the average register balance is better (greater) for all days with a manager than those days without a manager. Use a 0.01 significance level.

(a) Find the test statistic (to 3 decimal places).

(b) Find the P value (to 3 decimal places).

(c) Is there sufficient data to support the claim?
Yes
No

(d) Test the same claim at the 0.05 significance level. Is there sufficient data to support the claim?
Yes
No

Answer 1

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 >   0          
                  
Level of Significance ,    α =    0.01          
                  
Sample #1   ---->   successful          
mean of sample 1,    x̅1=   -2.70          
standard deviation of sample 1,   s1 =    4.62          
size of sample 1,    n1=   10          
                  
Sample #2   ---->   unsuccessful          
mean of sample 2,    x̅2=   -6.100          
standard deviation of sample 2,   s2 =    3.60          
size of sample 2,    n2=   10          
                  
difference in sample means = x̅1-x̅2 =    -2.700   -   -6.1000   =   3.4000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    1.8529

       
a)

t-statistic = ((x̅1-x̅2)-µd)/SE = (   3.4000   /   1.8529   ) =   1.835
                  
Degree of freedom,     df= 17
                  

b)

p-value =        0.043   [excel function: =T.DIST.RT(t stat,df) ]      

Conclusion:     p-value>α , Do not reject null hypothesis  

c)

no , not sufficeint data to support the claim

as p-value>α , Do not reject null hypothesis  

d)

at 0.05 significance level

p-value =        0.043   [excel function: =T.DIST.RT(t stat,df) ]      
Conclusion:     p-value<α , Reject null hypothesis         

   
so yes,sufficient data to support the clain