Question

In: Statistics and Probability

Directions Use the Crosstabs option in the Descriptives menu to answer the questions based on the...

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).

Scenario

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

Satisfied

Yes: Freshman (23) Sophmore (21) Junior (15) Senior (8)

No: Freshman (8) Sophmore (4) Junior (15) Senior (24)

Questions:

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.

Solutions

Expert Solution

#1)

Percentages of Freshmen, Sophomore, Junior, and Senior that were satisfied are

Freshmen = 23/67 = 34.3%

Sophomore = 21/67 = 31.3%

Junior = 15/67 = 22.4%

Senior = 8/67 = 11.9%

#2) Null and alternative hypotheses :

H0 : The variables level of satisfaction and student classification is independent.

Ha : The variables level of satisfaction and student classification is not independent.

#3) chi square statistic:

Observed frequency table :

We need to find expected frequency.

Expected frequency =

For example expected frequency for satisfied and Freshmen = 67*31/118 = 17.6017

Expected frequency table :

χ2    =                   ; O is observed and E is expected frequency.

χ2     = 25.1103

#4 )

Degrees of freedom (df) = (r-1)(c-1)           r is level of satisfaction and c is categories of students.

r = 2 and c = 4

df = (2-1)*(4-1) = 1*3 = 3

df = 3

#5) We are given level of significance α = 0.05

Therefore critical value χ2(0.05,3) = 7.815 -------- ( From chi square table )

#6)

Decision : As χ2 = 25.1103is greater than χ2(0.05,3) = 7.815

We reject H0,

Conclusion: We have significant evidence that there is a relationship between the level of satisfaction and student classification.


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