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

Satisfied	Freshmen	Sophomore	Junior	Senior
Yes	21	18	14	19
No	9	13	21	25

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? The value of chi-square is 7.12.

4. What are the reported degrees of freedom? The df is 3.

5. What is the reported level of significance? The reported level of significance is .052.

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

P(Freshmen | Satisfied) = P(Freshmen and Satisfied) / P(Satisfied)

= (21/140) / (72/140) = 0.292 = 29.2%

P(Sophomore | Satisfied) = P(Sophomore and Satisfied) / P(Satisfied)

= (18/140) / (72/140) = 0.25 = 25%

P(Junior | Satisfied) = P(Junior and Satisfied) / P(Satisfied)

= (14/140) / (72/140) = 0.194 = 19.4%

P(Senior | Satisfied) = P(Senior and Satisfied) / P(Satisfied)

= (19/140) / (72/140) = 0.264 = 26.4%

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Observed Frequencies
	Freshmen	Sophomore	Junior	Senior	Total
Yes	21	18	14	19	72
No	9	13	21	25	68
Total	30	31	35	44	140
Expected Frequencies
	Freshmen	Sophomore	Junior	Senior	Total
Yes	30 * 72 / 140 = 15.4286	31 * 72 / 140 = 15.9429	35 * 72 / 140 = 18	44 * 72 / 140 = 22.6286	72
No	30 * 68 / 140 = 14.5714	31 * 68 / 140 = 15.0571	35 * 68 / 140 = 17	44 * 68 / 140 = 21.3714	68
Total	30	31	35	44	140
	(fo-fe)²/fe
Yes	(21-15.4286)²/15.4286=2.0119	(18-15.9429)²/15.9429=0.2654	(14-18)²/18=0.8889	(19-22.6286)²/22.6286=0.5819
No	(9-14.5714)²/14.5714=2.1303	(13-15.0571)²/15.0571=0.2811	(21-17)²/17=0.9412	(25-21.3714)²/21.3714=0.6161

Null and Alternative hypothesis:

Ho: there is no relationship between the level of satisfaction and student classification

H1: there is a relationship between the level of satisfaction and student classification

Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 7.7166

df = (r-1)(c-1) = 3

p-value = CHISQ.DIST.RT(7.7166, 3) = 0.0522

Decision:

p-value > α, Do not reject the null hypothesis.

There is not enough evidence to conclude that there is an association between e-textbook satisfaction and academic classification.