Question

A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information. Today (Population 1) Five Years Ago (Population 2) Sample Mean 84 89 Sample Variance 112.5 54 Sample Size 45 36 The 92.50% confidence interval for the difference between the two population means is (Round your answers to 2 decimal places.

Answer 1

Sample #1   ---->   sample 1  
mean of sample 1,    x̅1=   84.000  
standard deviation of sample 1,   s1 =    112.500  
size of sample 1,    n1=   45.000  
          
Sample #2   ---->   sample 2  
mean of sample 2,    x̅2=   89.000  
standard deviation of sample 2,   s2 =    54.000  
size of sample 2,    n2=   36.000  
Degree of freedom, DF=   n1+n2-2 =    79          

Level of Significance ,    α =    0.075
   
t-critical value =    t α/2 =    1.8043   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    91.3288              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    20.4217              
margin of error, E = t*SE =    1.8043   *   20.4217   =   36.8464  
                      
difference of means =    x̅1-x̅2 =    84.0000   -   89.000   =   -5.0000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -5.0000   -   36.8464   =   -41.85
Interval Upper Limit=   (x̅1-x̅2) + E =    -5.0000   +   36.8464   =   31.85
...............

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