For each question: A.State the claim. B.State the test used and any necessary as
ID: 3225434 • 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.
Is there evidence to support that on average student GPA’s increased due to use of support services 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
Let 1 denote the mean for the gpa before and 2 denote the mean for gpa after program.
Hypothesis:
Null hypothesis: There mean differnce of GPA after and before support services is zero, H0:12=0
Alternate hypothesis: There mean differnece of GPA after and before support services is less than zero, Ha:12<0
Step 2. Significance level:
=0.05
Stpep3:
hence t statistic obtained < t statistic at 0.05 level of significnce. we can say that the data doesnt provide sufficient evidence to reject the null hypothesis
A.State the claim- There is significant difference in the group mean before and after the support services
B.State the test used and any necessary assumptions to conduct the test.- Two sample t-test with pooled variance
Assumptions: The variance for population 1 is about the same as that of population 2, we can estimate the common variance by pooling information from samples from population 1 and population 2. An informal check for this is to compare the ratio of the two sample standard deviations. Here 0.50/0.46 =1.08
When the sample sizes are nearly equal (admittedly "nearly equal" is somewhat ambiguous so often if sample sizes are small one requires they be equal), then a good Rule of Thumb to use is to see if this ratio falls from 0.5 to 2 (that is neither sample standard deviation is more than twice the other). If this rule of thumb is satisfied we can assume the variances are equal.
C.State the hypothesis in words relating to the question and as symbols.
Null hypothesis: The mean differnce of GPA before and after support services is zero, H0:12=0
Alternate hypothesis: The mean differnece of GPA before and after support services is less than zero, Ha:12<0
D.State the probability distribution used to find critical values.
Normal Distribution is used to find the critical values, since data assumed to be normally distributed
E.State the test statistic(Value) : t= -0.45
F.State if you reject or fail to reject the null hypothesis, support with p-value or traditional test.: t statistic obtained < t statistic at 0.05 level of significnce. we can say that the data doesnt provide sufficient evidence to reject the null hypothesis
G.State your conclusion in terms of the claim: There is no significant difference in the change of GPA before and after support services
subject GPA Before Support Services College G.P.A after support services 2 3.61 3.92 3 3.23 3.51 4 3.03 3.57 6 2.06 2.39 11 1.65 2.29 12 2.09 2.78 13 2.36 2.63 14 2.43 2.66 18 2.64 2.86 19 2.51 2.55 20 1.97 2.42 24 2.56 2.52 25 2.67 3.17 27 2.26 2.46 29 1.85 2.32 31 2.19 2.67 36 1.93 2.25 38 2.22 2.79 40 1.82 2.48 mean 2.372631579 2.749473684 standard deviation 0.506697829 0.469851911 n 19 19 common standard deviation 2.567974508 SQRT(((39*K21^2)+(39*L21^2))/78) t statistic -0.452304091 (K21-L21)/(N24*SQRT((1/19)+(1/19))) t at 0.05 significance -1.685 from t tableRelated Questions
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