Assignment #6: Testing Correlations Directions: Use the Bivatiate Correlation fu
ID: 3264819 • Letter: A
Question
Assignment #6: Testing Correlations
Directions: Use the Bivatiate Correlation function and the Options submenu to answer each of the questions based on the above scenario.
The superintendent has continued the examination of data by examining the relationship between school performance scores and percent of students receiving free or reduced lunch. The district data used for the analysis is contained below.
School
Performance score
% Free or Reduced
School
Performance score
% Free or Reduced
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
124
119
109
96
86
107
115
126
90
91
101
112
82
89
118
97
106
109
102
101
51
62
29
67
83
62
40
33
85
42
44
48
93
87
71
67
72
56
64
66
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
125
124
107
132
116
122
90
113
121
86
112
115
92
113
114
101
131
132
104
120
36
50
71
20
69
47
70
30
44
93
59
38
80
28
42
71
35
47
61
57
1. State an appropriate null hypothesis for this analysis.
2. What is the value of the correlation coefficient? how would you classify the strength of this relationship?
3. Based on the information from the scenario, what is the appropriate value for the degrees of freedom?
4. Present the results as they might appear in an article. This must include a table and narrative statement that reports and interprets the results of your analysis.
PLEASE COULD U HELP ME BECAUSE THE DUE DATE IS TODAY? Please
School
Performance score
% Free or Reduced
School
Performance score
% Free or Reduced
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
124
119
109
96
86
107
115
126
90
91
101
112
82
89
118
97
106
109
102
101
51
62
29
67
83
62
40
33
85
42
44
48
93
87
71
67
72
56
64
66
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
125
124
107
132
116
122
90
113
121
86
112
115
92
113
114
101
131
132
104
120
36
50
71
20
69
47
70
30
44
93
59
38
80
28
42
71
35
47
61
57
Explanation / Answer
1. Null hypothesis H0: rho=0
Alternative Ha: rho not equals 0
2. Correlation coefficient r=-0.7278
As this is greater than 0.70 in magnitude so the relationship is strongly negative.
3. Degree of freedom =n-2=40-2=38
4. Test statistic t= r*sqrt(n-2)/sqrt(1-r2) = (-0.7278)*sqrt(38)/sqrt(1-0.7278*0.7278)=-6.54
critical t= tinv(0.05,28)=2.024
As calculated |t|>2.024, we reject the nullh ypothesis and concldue there is significant correlation between thevariables