For each question: A.State the claim. B.State the test used and any necessary as
ID: 3225036 • Letter: F
Question
For each question:
A.State the claim.
B.State the test used and any necessary assumptions to conduct the test.
C.State the hypothesis in words relating to the question and as symbols.
D.State the probability distribution used to find critical values.
E.State the test statistic(Value)
F.State if you reject or fail to reject the null hypothesis, support with p-value or traditional test.
G.State your conclusion in terms of the claim.
H.Determine if there is a chance of a Type I or a Type II error, then state what the possible error may be in terms of the question
E & J College is doing a study on their policies. After randomly gathering data from a sample of 40 graduates, they put the raw data in a table and did not now how to proceed. They are asking for your statistical expertise to provide an analysis. They need you to answer the following questions. Assume all populations are normally distributed and these are random samples.
5) Records indicate that Region A is 15% region B is 60% and Region C is 25%. Is there enough evidence to support this claim at a 5% level of significance?
Here is the table:
Subject
HS GPA
ACT Score
College G.P.A
Number of years took to Graduate
Residential Region
Participated in Support Service Program
GPA Before Support Services
1
2.05
29
2.38
5.5
B
N
2
4.00
22
3.92
5
C
Y
3.61
3
3.00
26
3.51
4
B
Y
3.23
4
3.89
24
3.57
5
C
Y
3.03
5
3.36
20
3.78
5
C
N
6
3.08
25
2.39
3.5
A
Y
2.06
7
3.28
25
3.11
5.5
B
N
8
1.69
16
2.38
4
B
N
9
2.17
23
2.59
3
C
N
10
3.06
27
3.31
5
B
N
11
2.77
22
2.29
4
B
Y
1.65
12
2.14
21
2.78
8
B
Y
2.09
13
2.03
25
2.63
5
B
Y
2.36
14
2.73
24
2.66
6
A
Y
2.43
15
2.82
25
2.44
5
C
N
16
3.77
29
3.30
4
B
N
17
2.51
21
2.08
3
C
N
18
2.88
22
2.86
3.5
C
Y
2.64
19
3.37
25
2.55
4.5
A
Y
2.51
20
2.39
22
2.42
5.5
A
Y
1.97
21
3.15
21
3.40
6
B
N
22
2.31
26
2.28
5
B
N
23
3.37
26
3.29
4
A
N
24
2.30
19
2.52
4.5
A
Y
2.56
25
3.53
22
3.17
5
C
Y
2.67
26
3.43
23
2.57
4
B
N
27
2.58
23
2.46
4
A
Y
2.26
28
4.00
24
3.27
5
C
N
29
2.00
31
2.32
5.5
C
Y
1.85
30
2.80
21
2.72
4.5
A
N
31
2.49
23
2.67
6
B
Y
2.19
32
3.73
23
3.00
5
C
N
33
2.36
24
2.15
4
A
N
34
3.60
31
4.00
4.5
B
N
35
3.43
21
3.30
5
A
N
36
1.46
18
2.25
4
A
Y
1.93
37
4.00
28
3.41
4
A
N
38
3.57
31
2.79
5
B
Y
2.22
39
2.74
21
2.99
5.5
C
N
40
2.37
22
2.48
4.5
C
Y
1.82
Subject
HS GPA
ACT Score
College G.P.A
Number of years took to Graduate
Residential Region
Participated in Support Service Program
GPA Before Support Services
1
2.05
29
2.38
5.5
B
N
2
4.00
22
3.92
5
C
Y
3.61
3
3.00
26
3.51
4
B
Y
3.23
4
3.89
24
3.57
5
C
Y
3.03
5
3.36
20
3.78
5
C
N
6
3.08
25
2.39
3.5
A
Y
2.06
7
3.28
25
3.11
5.5
B
N
8
1.69
16
2.38
4
B
N
9
2.17
23
2.59
3
C
N
10
3.06
27
3.31
5
B
N
11
2.77
22
2.29
4
B
Y
1.65
12
2.14
21
2.78
8
B
Y
2.09
13
2.03
25
2.63
5
B
Y
2.36
14
2.73
24
2.66
6
A
Y
2.43
15
2.82
25
2.44
5
C
N
16
3.77
29
3.30
4
B
N
17
2.51
21
2.08
3
C
N
18
2.88
22
2.86
3.5
C
Y
2.64
19
3.37
25
2.55
4.5
A
Y
2.51
20
2.39
22
2.42
5.5
A
Y
1.97
21
3.15
21
3.40
6
B
N
22
2.31
26
2.28
5
B
N
23
3.37
26
3.29
4
A
N
24
2.30
19
2.52
4.5
A
Y
2.56
25
3.53
22
3.17
5
C
Y
2.67
26
3.43
23
2.57
4
B
N
27
2.58
23
2.46
4
A
Y
2.26
28
4.00
24
3.27
5
C
N
29
2.00
31
2.32
5.5
C
Y
1.85
30
2.80
21
2.72
4.5
A
N
31
2.49
23
2.67
6
B
Y
2.19
32
3.73
23
3.00
5
C
N
33
2.36
24
2.15
4
A
N
34
3.60
31
4.00
4.5
B
N
35
3.43
21
3.30
5
A
N
36
1.46
18
2.25
4
A
Y
1.93
37
4.00
28
3.41
4
A
N
38
3.57
31
2.79
5
B
Y
2.22
39
2.74
21
2.99
5.5
C
N
40
2.37
22
2.48
4.5
C
Y
1.82
Explanation / Answer
From the given data, first we find the number of students from each of the 3 regions as below:
Region
A
B
C
Total
Number of Students
12
15
13
50
Let p1, p2 and p3 be the proportion of students from region A, B and C respectively.
A.Claim: Region A is 15% region B is 60% and Region C is 25%.
i.e., p1 = 0.15, p2 = 0.6 and p3 = 0.25
B.Test used: Chi-square goodness of fit test.
C.Hypothesis:
In words: the distribution of students among the 3 regions is as per the claim.
As symbols: H0 : p1 = 0.15, p2 = 0.6 and p3 = 0.25 Vs HA : H0 is false
D. Probability distribution used to find critical values: Chi-square with 2 degrees of freedom [rule is: number of cells – 1 = 3 - 1 ]
E.Test statistic(Value): 2 = [1,3]{(Oi - Ei)2/Ei}, where Oi = observed (given) number of students from region i; Ei = expected number of students from region I = n x pi, n being the total number of students = 40. (i = 1 for A. 2 for B, 3 for C)
Calculation: 2 = {(12 - 6)2/6} + {(15 - 24)2/24} +{(13 - 10)2/10} = 10.267
F.Decision, i.e., reject or fail to reject the null hypothesis,
Under H0, 2 ~ 22
Given 5% level of significance, 22, 0.05 = 5.99.
Since calculated value of 2 > 5.99, H0 is rejected.
G.Conclusion: There is not enough statistical evidence to support the claim that Region A is 15% region B is 60% and Region C is 25%.
DONE
Region
A
B
C
Total
Number of Students
12
15
13
50
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