In: Statistics and Probability
Satisified | Freshmen | Sophomore | Junior | Senior |
Yes | 21 | 19 | 10 | 20 |
No | 14 | 9 | 17 | 11 |
Directions: Use the crosstabe option in the Descriptives menu to answer the questions based on the following scenario. ( Be sure to select Chi-square from the sStatistics submenu and Observed, Expected, Row, and Column in the Cell submenu. Assume a level of significance of .05)
The school district recently adopted the use of e-textbooks, and superintendent is interested in determining the level of satisfaction with e-textbooks among students classification. The superintendent selected a sample of students from one high school and asked them how satisified they were with the use of e-textbooks. The data that were collected are presented in the following table.
1. Of the students that were satisfied, what percent were Freshmen, Sophomore, Junior, and Senior? (Round your final answer to 1 decimal place.
2. State an appropriate null hypothesis for this analysis.
3. What is the value of the chi-square statistic?
4. What are the reported degrees of freedom?
5. What is the reported level of significance?
6. Based on the results of the chi-square test of independence is there an association between e-textbook satisfaction and academic classification?
7 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 the analysis.
Note: The table must be created using your word processing program. Tables that are copied and pasted from SPSS are not acceptable.
Soln
Satisfied |
Freshmen |
Sophomore |
Junior |
Senior |
Total |
Yes |
21 |
19 |
10 |
20 |
70 |
No |
14 |
9 |
17 |
11 |
51 |
Total |
35 |
28 |
27 |
31 |
121 |
1)
Of Satisfied Students,
% Freshmen = 21/70 = 0.30 or 30%
% Sophomore = 19/70 = 0.271 or 27.1%
% Junior = 10/70 = 0.143 or 14.3%
% Senior = 20/70 = 0.286 or 28.6%
2)
Null and Alternate Hypothesis
H0: The two variables (Level of Satisfaction and Students) are independent.
Ha: The two variables are associated.
3)
Given,
Observed Values:
Satisfied |
Freshmen |
Sophomore |
Junior |
Senior |
Total |
Yes |
21 |
19.00 |
10 |
20 |
70 |
No |
14 |
9 |
17 |
11 |
51 |
Total |
35 |
28 |
27 |
31 |
121 |
Expected Values:
Expected Values are calculated as:
Eij = (Ti * Tj)/N
where,
Ti = Total in ith row
Tj = Total in jth column
N = table grand total
Satisfied |
Freshmen |
Sophomore |
Junior |
Senior |
Yes |
20.25 |
16.20 |
15.62 |
17.93 |
No |
14.75 |
11.80 |
11.38 |
13.07 |
Test Statistic:
Chi Square = ∑(Oij – Eij)2/Eij = (21-20.25)2/20.25 + ……………….. + (11 – 13.07)2/13.07 = 6.58
4)
df = (r-1)*(c-1) = (2-1)*(4-1) = 1*3 = 3
5)
Alpha = 0.05
6)
Chi Square Critical = 7.81
Decision Rule:
If Chi Square> Chi Square Critical reject the null hypothesis
Result:
Since, Chi Square< Chi Square Critical fail to reject the null hypothesis.
7)
Conclusion:
Level of Satisfaction and Students are independent