stath math

Name   :

NIM    :

HOMEWORK

Problem:

Let A is a population that contains five score, there are  4   6    7    9    10

X₃ be sample that contain 3 observation from population A

·      List all samples that can be formed

·      Find the mean from each sample

·      Find the mean of sample mean #

·      Compare the result at # and

·      Make the conclusion from this result

·      Why does the condition be happened

·      Find the variance and expectation from each sample

Answer:

·      All samples that can be formed are:

1.    4          6          7                                  6. 4      9          10

2.    4          6          9                                  7. 6      7          9

3.    4          6          10                                8. 6      7          10

4.    4          7          9                                  9. 6      9          10

5.    4          7          10                                10. 7    9          10

·      The mean from each sample

1.       =  =  = 5.67                    6.  =  =  = 7.67

2.       =  =  = 6.33                    7.  =  =  = 7.33

3.       =  =  = 6.67                  8.  =  =  = 7.67

4.       =  =  = 6.67                    9.  =  =  = 8.33

5.       =  =  = 7                       10.  =  =  = 8.67

·      The mean of sample mean

 =  =  = 7.2

·      Comparison of # and

From the problem above

we have, =  =  = 7.2

So, the comparison is  =

·     Conclusion

Mean of population is equal to mean of sample mean

·      The condition # equals , since the expectation value of the sample mean is the population mean

·      The variance of each sample

1.      4  6  7  à

S² =   =   =

     = (2.7889 + 2.7889 + 1.7689)/2 = 7.3467/2 = 3.67335

2.      4  6  9  à

S² =   =   =

     = (5.4289 + 0.1089 + 7.1289)/2 = 12.6667/2 = 6.33335

3.      4  6  10  à

S² =   =   =

     = (7.1289 + 0.4489 + 11.0889)/2 = 18.6667/2 = 9.33335

4.      4  6  9  à

S² =   =   =

     = (7.1289 + 0.4489 + 5.4289)/2 = 13.0067/2 = 6.50335

5.      4  7  10  à

S² =   =   =

     = (9 + 0 + 9)/2 = 18/2 = 9

6.      4  9  10  à

S² =   =   =

     = (13.4689 + 1.7689 + 5.4289)/2 = 20.6667/2 = 10.3335

7.      6  7  9  à

S² =   =   =

     = (1.7689 + 0.1089 + 2.7889)/2 = 4.6667/2 = 2.33335

8.      6  7  10  à

S² =   =   =

     = (2.7889 + 0.4489 + 5.4289)/2 = 8.6667/2 = 4.33335

9.      6  9  10  à

S² =   =   =

     = (5.4289 + 0.4489 + 2.7889)/2 = 8.6667/2 = 4.33335

10.  7  9  10  à

S² =   =   =

     = (2.7889 + 0.1089 + 1.7689)/2 = 4.6667/2 = 2.33335

·      The expectation of each sample is same with the variance of each sample, since the sample variance S² is an unbiased estimator of the population variance.