Question

A) Can we assume the difference between means of population 1 and 3 is equal to 2 with 97% confidence level? (Answer it using hypothesis testing with confidence interval).

Population 1: N(μ1,δ12)

Sample from Population 1: 2, 4, 8, 3, 9, 10, 11, 15, 19

Population 3: N(μ3,δ32)

Sample from Population 3: 11, 13, 7, 14, 10, 30, 13, 3, 20

Answer 1

Ho :   µ1 - µ2 =   2
Ha :   µ1-µ2 ╪   2
      
Level of Significance ,    α =    0.03
      
Sample #1   ---->   1
mean of sample 1,    x̅1=   9.000
standard deviation of sample 1,   s1 =    5.612
size of sample 1,    n1=   9
      
Sample #2   ---->   2
mean of sample 2,    x̅2=   13.444
standard deviation of sample 2,   s2 =    7.796
size of sample 2,    n2=   9

Degree of freedom, DF=   n1+n2-2 =    16              
t-critical value =    t α/2 =    2.3815   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    6.7926              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    3.2020              
margin of error, E = t*SE =    2.3815   *   3.20   =   7.63  
                      
difference of means =    x̅1-x̅2 =    9.0000   -   13.444   =   -4.4444
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -4.4444   -   7.6258   =   -12.0703
Interval Upper Limit=   (x̅1-x̅2) + E =    -4.4444   +   7.6258   =   3.1814

CI is (-12.07 , 3.18 )

Since, CI do contain null hypotheis value 2, so, we do not reject Ho

hence, we can assume the difference between means of population 1 and 3 is equal to 2 with 97% confidence level