In: Statistics and Probability
please answer both of the following:
1.
Exhibit 10-3
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. Final examination scores from a random
sample of students enrolled today and from a random sample of
students enrolled five years ago were selected. You are given the
following information.
| 
 Today  | 
 Five Years Ago  | 
|
| 
 82  | 
 88  | 
|
| σ2 | 
 112.5  | 
 54  | 
| n | 
 45  | 
 36  | 
Refer to Exhibit 10-3. The standard error of is _____.
A: 12.9
B: 9.3
C: 4
D: 2
2.
Exhibit 10-3
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. Final examination scores from a random
sample of students enrolled today and from a random sample of
students enrolled five years ago were selected. You are given the
following information.
| 
 Today  | 
 Five Years Ago  | 
|
| 
 82  | 
 88  | 
|
| σ2 | 
 112.5  | 
 54  | 
| n | 
 45  | 
 36  | 
Refer to Exhibit 10-3. The 95% confidence interval for the difference between the two population means is _____.
A: -9.92 to -2.08
B: -3.92 to 3.92
C: -13.84 to 1.84
D: -24.228 to 12.23
10.3)
Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(112.49996356/45 + 54.00045225/36)
sp = 2
standard error = 2
2)
Given CI level is 0.95, hence α = 1 - 0.95 = 0.05  
           
   
α/2 = 0.05/2 = 0.025, zc = z(α/2, df) = 1.96  
           
   
          
       
Margin of Error          
       
ME = zc * sp          
       
ME = 1.96 * 2          
       
ME = 3.92          
       
          
       
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc *
sp)          
       
CI = (82 - 88 - 1.96 * 2 , 82 - 88 - 1.96 * 2  
           
   
CI = (-9.92 , -2.08)      
           
           A: -9.92 to
-2.08