Question

Select two data sets and decide which hypothesis testing discussed in class should be used and matched with your data set. Construct a 99% confidence interval for a mean difference and Test at a 1% significance level. Use both the P-value and the critical-value approaches to make this test and interpret your results. If you cannot reject the null hypothesis, please adjust to 5% or 10% significance level.

East Asia and Pacific	Europe and Central Asia	Difference
9.67104	7.51427	2.15677
7.58733	7.15481	.43252
5.37422	6.64138	-1.26716
5.14952	7.23006	-2.08054
4.78177	6.59366	-1.81189
4.81419	5.52866	-7.1447
4.84156	3.70016	1.1414
4.73152	3.48066	1.25086
4.81446	4.71345	.10101
5.41959 4.28956	5.051 4.85657	.36859 -.56701

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   East Asia and Pacific                
mean of sample 1,    x̅1=   5.59                  
standard deviation of sample 1,   s1 =    1.60                  
size of sample 1,    n1=   11                  
                          
Sample #2   ---->      Europe and Central Asia    
mean of sample 2,    x̅2=   5.68                  
standard deviation of sample 2,   s2 =    1.43                  
size of sample 2,    n2=   11                  
                          
difference in sample means =    x̅1-x̅2 =    5.5886   -   5.7   =   -0.09  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.5207                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.6484                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.0900   -   0   ) /    0.65   =   -0.139
                          
Degree of freedom, DF=   n1+n2-2 =    20                  
t-critical value , t* =        2.0860   (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | < | critical value |, so, Do not Reject Ho             

       
p-value =        0.891004   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis                      
                          

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Degree of freedom, DF=   n1+n2-2 =    20              
t-critical value =    t α/2 =    2.8453   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.5207              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.6484              
margin of error, E = t*SE =    2.8453   *   0.6484   =   1.8449  
                      
difference of means =    x̅1-x̅2 =    5.5886   -   5.679   =   -0.0900
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -0.0900   -   1.8449   =   -1.9349
Interval Upper Limit=   (x̅1-x̅2) + E =    -0.0900   +   1.8449   =   1.7549

THANKS

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