In: Statistics and Probability
Assignment 6: Testing Correlations
Directions
Use the Bivatiate Correlation function and the Options submenu to answer each of the questions based on the scenario.
Scenario
The superintendent has continued the examination of data by examining the relationship between attendance rate and percent of students eligible for free or reduced priced lunch. The district data used for the analysis are contained below.
School |
% Free or Reduced |
Attendance Rate |
1 |
47.2 |
94.8 |
2 |
31.1 |
96.1 |
3 |
58.9 |
94.9 |
4 |
44.9 |
94.2 |
5 |
24.1 |
95.7 |
6 |
52.4 |
94.8 |
7 |
54.7 |
93.8 |
8 |
68.1 |
92.6 |
9 |
49.6 |
93.6 |
10 |
42.9 |
93.8 |
11 |
38.1 |
92.8 |
12 |
27.2 |
95.7 |
13 |
58.4 |
93.4 |
14 |
52.4 |
93.6 |
15 |
58.4 |
94.6 |
16 |
64.9 |
93.0 |
17 |
75.5 |
92.9 |
18 |
35.6 |
95.1 |
19 |
79.4 |
92.8 |
20 |
67.3 |
92.6 |
21 |
54.7 |
95.5 |
22 |
74.7 |
91.8 |
Questions
Note: The table must be created using your word processing program. Tables that are copied and pasted from SPSS are not acceptable.
A) Mean can be calculated as follows:
Mean = Sum of all values/Number of values
The mean percent of students receiving free or reduced lunch = 52.75
The mean attendance rate = 94.0045
b) The standard deviation can be calculated using the following command (=stdev.s)
C) Null and Alternate Hypothesis
Null Hypothesis: Ho - There is no significant relationship between attendance rate and % free or reduced lunch
Alternate Hypothesis: Ha - There is a significant relationship between attendance rate and % free or reduced lunch
D) Correlation Coefficient
Following formula can be used to calculate the correlation coefficient:
The correlation can be calculated as follows:
Since the value of correlation coefficient is nearly approaching zero, there is weak relationship between the two variables
Degree of freedom = (n1 + n2 -2) = (22+22-2) = 42
The level of significance can be assumed as 0.05.
Therefore, it can be concluded that, there is no significant relationship between the two variables.