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 (N = 126)

Satisfied Freshmen Sophomore Junior Senior

Yes 13 13 21 20

No 22 17 8 12

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

Following table shows the row total and column total:

Satisfied	Freshmen	Sophomore	Junior	Senior	Total
Yes	13	13	21	20	67
No	22	17	8	12	59
Total	35	30	29	32	126

(1)

The percent of Freshman:

(13 /67) *100% = 19.4%

The percent of Sophomore:

(13 /67) *100% = 19.4%

The percent of Junior:

(21/67) *100% = 31.34%

The percent of Senior:

(20/67) *100% = 29.85%

(2)

Hypotheses are:

H0: There is no association between e-textbook satisfaction and academic classification.

Ha: There is an association between e-textbook satisfaction and academic classification.

3:

Expected frequencies will be calculated as follows:

Following table shows the expected frequency:

Satisfied	Freshmen	Sophomore	Junior	Senior	Total
Yes	18.611	15.952	15.421	17.016	67
No	16.389	14.048	13.579	14.984	59
Total	35	30	29	32	126

Following table shows the calculations for chi square test staistics:

O	E	(O-E)^2/E
13	18.611	1.691651228
13	15.952	0.546282849
21	15.421	2.018367227
20	17.016	0.523287259
22	16.389	1.921003173
17	14.048	0.620323462
8	13.579	2.292160027
12	14.984	0.594250934
Total		10.20732616

So,

4:

Degree of freedom: df =( number of rows -1)*(number of columns-1) = (2-1)*(4-1)=3

5:

The level of signficance:

6:

The p-value: 0.0169

Excel function used for p-value: "=CHIDIST(10.21,3)"

Since p-value is less than 0.05 so we reject the null hypothesis.

That is we can conclude that there is an association between e-textbook satisfaction and academic classification.