Answer:
a. 15
b.
Sr.no Samples Sample mean
1 (79,64) 71.5
2 (79,84) 81.5
3 (79,82) 80.5
4 (79,92) 85.5
5 (79,77) 78
6 (64,84) 74
7 (64,82) 73
8 (64,92) 78
9 (64,77) 70.5
10 (84,82) 83
11 (84,92) 88
12 (84,77) 80.5
13 (82,92) 87
14 (82,77) 79.5
15 (92,77) 84.5
c.
mean of sample mean=population mean=79.67
Step-by-step explanation:
a.
The different samples of two test grade are nCr, where n=6 and r=2.
nCr=6C2=6!/2!(6-2)!=6*5*4!/2!4!=30/2=15.
Thus, there are 15 different samples of two test grade.
b.
All possible samples are listed below:
Sr.no Samples
1 (79,64)
2 (79,84)
3 (79,82)
4 (79,92)
5 (79,77)
6 (64,84)
7 (64,82)
8 (64,92)
9 (64,77)
10 (84,82)
11 (84,92)
12 (84,77)
13 (82,92)
14 (82,77)
15 (92,77)
The sample means for each sample can be calculated as
Sr.no Samples Sample mean
1 (79,64) (79+64)/2=71.5
2 (79,84) (79+84)/2=81.5
3 (79,82) (79+82)/2=80.5
4 (79,92) (79+92)/2=85.5
5 (79,77) (79+77)/2=78
6 (64,84) (64+84)/2=74
7 (64,82) (64+82)/2=73
8 (64,92) (64+92)/2=78
9 (64,77) (64+77)/2=70.5
10 (84,82) (84+82)/2=83
11 (84,92) (84+92)/2=88
12 (84,77) (84+77)/2=80.5
13 (82,92) (82+92)/2=87
14 (82,77) (82+77)/2=79.5
15 (92,77) (92+77)/2=84.5
c.
The sample means of sample mean μxbar will calculated by taking average of sample means
μxbar=(71.5+ 81.5+ 80.5+ 85.5+ 78+ 74+ 73+ 78+ 70.5+ 83+ 88+ 80.5+ 87+ 79.5+ 84.5)/15
μxbar=1195/15=79.67
Population mean=μ=(79+64+84+82+92+77)/6
μ=478/6=79.67
Sample means of sample mean μxbar=Population mean μ.
