Question

Consider the hypothesis test with null hypothesis μ1=μ2 and alternative hypothesis μ1 > μ2.Suppose that sample sizes n1 = 10 and n2 =10, that the sample means are 4.9 and 2.8 respectively, and that the sample variances are 2 and 3 respectively. Assume that the population variances are equal and that the data are drawn from normal distributions.

a) Test the hypothesis that at α= 0.05 and provide a conclusion statement

b) Provide an adequate confidence interval with 95% confidence level. c) Do your answers in a) and b) coincide? Explain.

Answer 1

a)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   4.90                  
standard deviation of sample 1,   s1 =    1.41                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   2.80                  
standard deviation of sample 2,   s2 =    1.73                  
size of sample 2,    n2=   10                  
                          
difference in sample means =    x̅1-x̅2 =    4.9000   -   2.8   =   2.10  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.5811                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.7071                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   2.1000   -   0   ) /    0.71   =   2.9698
                          
Degree of freedom, DF=   n1+n2-2 =    18                  
t-critical value , t* =        1.734   (excel function: =t.inv(α,df)              
Decision:   | t-stat | > | critical value |, so, Reject Ho                      
p-value =        0.0041   [excel function: =T.DIST.RT(t stat,df) ]              
Conclusion:     p-value <α , Reject null hypothesis                      

There is enough evidence to that  μ1 > μ2

b)

Degree of freedom, DF=   n1+n2-2 =    18              
t-critical value =    t α/2 =    2.1009   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.5811              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.7071              
margin of error, E = t*SE =    2.1009   *   0.71   =   1.49  
                      
difference of means =    x̅1-x̅2 =    4.9000   -   2.800   =   2.1000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    2.1000   -   1.4856   =   0.6144
Interval Upper Limit=   (x̅1-x̅2) + E =    2.1000   +   1.4856   =   3.5856

since, both ends of CI is positive, so, reject Ho

yes, both answers coincide