Question

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

1. What is the mean percent of students receiving free or reduced lunch? What is the mean attendance rate?
2. What are the standard deviations for the percent of students receiving free or reduced lunch and attendance rate?
3. State an appropriate null hypothesis for this analysis.
4. What is the value of the correlation coefficient?
5. Based on the value of the correlation coefficient, how would you classify the strength of this relationship?
6. Based on the information from the scenario, what is the appropriate value for the degrees of freedom?
7. What is the reported level of significance?
8. 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.

Note: The table must be created using your word processing program. Tables that are copied and pasted from SPSS are not acceptable.

Answer 1

1. The mean percent of students receiving free or reduced lunch is 52.750
2. The mean attendance rate is 94.005
3. The standard deviations for the percent of students receiving free or reduced lunch is 15.530
4. Attendance rate is 1.210
5. Correlation coefficient= -0.7426

6] The correlation coefficient, the strength of this relationship is -0.7426 . This is negative correlation.

7] Degrees of freedom=n-1= 22-1=21

8] Reported level of significance =0.05

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha: (TWO TAILED TEST)

test statistic t= r*Sqrt(n-2)/sqrt(1-r^2)

Degrees of freedom= n-1=22-1=21

P value= 0.0000

Since P value SMALLER THAN 0.05 LEVEL SIGNIFICANCE

DECISION: REJECT NULL HYPOTHESIS H0.

CONCLUSION: WE HAVE SUFFICIENT EVIDENCE TO CONCLUDE THAT THERE IS CORRELATION BETWEEN ATTENDACE RATE AND % FREE OR REDUCED LUNCH.