Kowalski and Taylor (2009) observed that students in an introductory psychology course often enter the course with

Question:

Kowalski and Taylor (2009) observed that students in an introductory psychology course often enter the course with misconceptions and complete the course with the misconceptions intact. One example of a misconception is that “most people use only 10% of their brain power” (p. 153). Kowalski and Taylor designed a study to find out whether refutational lectures and readings would be effective in dispelling misconceptions of introductory psychology students. The investigators identified 45 misconceptions and embedded them in a Psychological Information Questionnaire that was administered as a pretest during the first class. Nine misconceptions were then assigned to one of five different combinations of coverage in a within subjects design where all students received all five combinations. The five combinations were:

(a) Refutational lecture and refutational reading,

(b) Refutational lecture and no reading,

(c) Standard lecture and standard reading,

(d) No lecture and refutational reading, and

(e) No lecture and no reading.

At the end of the course, the students again completed the Psychological Information Questionnaire. A normalized gain score was calculated for each student and the mean gain score was computed for each combination of coverage.

1. Which statistical test seems most appropriate for the statistical analyses?

a. t test for one sample

b. t test for two independent samples

c. t test for two related samples

d. t test for a correlation coefficient

e. One-factor F-test

f. Two-factor F-test

g. Repeated-measures F test

h. One-variable χ2 test

i. Two-variable χ2 test

A paragraph from Kowalski and Taylor’s results section is shown below. Use the information presented in this paragraph to answer questions 2 to 5.

The initial analysis involved a one-way repeated measures ANOVA assessing the effect of the five different coverage methods on . This overall ANOVA was significant, F(4,61) = 110.42, p < .001, η2 = .88. (p. 156)

2. Numerator degrees of freedom were equal to ___.

3. Denominator degrees of freedom were equal to ___.

4. With a significance level of .05, critical F is equal to ___.

5. The proportion of explained variance was equal to ___.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question

Statistics

ISBN: 9781119254515

11th Edition

Authors: Robert S. Witte, John S. Witte

Question Posted: