Question

Assignment #8: Chi-Square Test of Independence

Directions: Use the Crosstabs option in the Descriptives menu to answer the questions based on the following scenario. (Be sure to select Chi-square from the Statistics submenu and Observed, Expected, Row, and Column in the Cells submenu. Assume a level of significance of .05)

The school district recently adopted the use of e-textbooks, and the superintendent is interested in determining the level of satisfaction with e-textbooks among students and if there is a relationship between the level of satisfaction and student classification. The superintendent selected a sample of students from one high school and asked them how satisfied they were with the use of e-textbooks. The data that were collected are presented in the following table.

Student Classification
Sophomore Junior Senior

Satisfied Freshmen
Yes 18 19 13 12

No 13 14 15 17

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.

Answer 1

Solution :

Freshman   Sophomore   Junior   Senior   Total
Yes 18 19 13 12 62
No 13 14 15 17 59

Total   31   33 28 29 121

a) Percent of satisfied from each group:

Freshman= 18/31= 0.5806=58.06%.

Sophomore= 18/33= 0.57= 57%.

Junior= 13/28= 0.4842= 48.42%.

Senior= 12/29= 0.413= 41.3%.

2) Chi-square association test:

H0: There is no relationship between student classification and satisfied

Ha: There is relationship between student classification and satisfied

3) Expected value:

E(X)= (Sum of Row * Sum of column)/ Total

E(Freshman* Yes)= (62*31)/121= 15. 88

   Freshman       Sophomore   Junior   Senior       Total
Yes   15.88 16.91 14.35 14.86 66
No 15.12 16. 09 13.65 14.24 59  

X^2=SUMMATION(Oi-Ei)**2/Ei.

=2.4961

4) The degree of freedom: (c-1)(r-1)= (4-1)(2-1)= 3

c= no of columns and r= No of rows

5) Level of significant= 0.05

6) P-value:0.4760

The test statistic is significant and rejects H0. There is sufficient evidence to support the claim that there is relationship between student classification and satisfied.

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