Question

Hypothesis testing regarding the equality of 2 population means: DATA SET: [4 55 73 4 4 6 4 615 6 6 4 16 8 646 655556 3 78] 1) Take the average of the first three numbers in the first part of your dataset and let's call it 'c'. Test the hypothesis that the difference between 2 population means is equal to 'd'. Choose 5% level of significance and perform the test using critical value approach. Moreover, you have to perform the same test in left-tailed, right-tailed and two-tailed version but keep a = 0.05 in all cases. 1.1. Divide your dataset in two equal parts with 15 data values in each part. Assume the first part is a random sample from the daily expenditures of ADA students(Population 1), while the second part is a random sample from the daily expenditures of BSU students(Population 2). What is estimator and estimate of the difference between 2 population means? 2) Build 90%, 95% and 99% confidence intervals for the difference between population means. 3) How many observations do I need to have the margin of error equal to 2 at 95% confidence level?

Answer 1

1)

Ho :   µ1 - µ2 =   4.67                  
Ha :   µ1-µ2 ╪   4.67               
                          
Level of Significance ,    α =    0.05   
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   4.667                  
standard deviation of sample 1,   s1 =    1.496                  
size of sample 1,    n1=   15                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   5.400                  
standard deviation of sample 2,   s2 =    1.805                  
size of sample 2,    n2=   15                  
                          
difference in sample means =    x̅1-x̅2 =    4.6667   -   5.4   =   -0.73  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.6576                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.6053                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.7333   -   4.666666667   ) /    0.61   =   -8.9217
                          
Degree of freedom, DF=   n1+n2-2 =    28                  
  
p-value =        0.000000   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis      

2)

90%

--------

Degree of freedom, DF=   n1+n2-2 =    28              
t-critical value =    t α/2 =    1.7011   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.6576              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.6053              
margin of error, E = t*SE =    1.7011   *   0.61   =   1.03  
                      
difference of means =    x̅1-x̅2 =    4.6667   -   5.400   =   -0.7333
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -0.7333   -   1.0296   =   -1.7630
Interval Upper Limit=   (x̅1-x̅2) + E =    -0.7333   +   1.0296   =   0.2963

95%

------------

Degree of freedom, DF=   n1+n2-2 =    28              
t-critical value =    t α/2 =    2.0484   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.6576              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.6053              
margin of error, E = t*SE =    2.0484   *   0.61   =   1.24  
                      
difference of means =    x̅1-x̅2 =    4.6667   -   5.400   =   -0.7333
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -0.7333   -   1.2398   =   -1.9732
Interval Upper Limit=   (x̅1-x̅2) + E =    -0.7333   +   1.2398   =   0.5065

99%

------

Degree of freedom, DF=   n1+n2-2 =    28              
t-critical value =    t α/2 =    2.7633   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.6576              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.6053              
margin of error, E = t*SE =    2.7633   *   0.61   =   1.67  
                      
difference of means =    x̅1-x̅2 =    4.6667   -   5.400   =   -0.7333
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -0.7333   -   1.6725   =   -2.4058
Interval Upper Limit=   (x̅1-x̅2) + E =    -0.7333   +   1.6725   =   0.9392

Please let me know in case of any doubt.

Thanks in advance!

Please upvote!