PSGY2002/PSGY4025 Autumn revision quiz
Quiz
•
Mathematics
•
University
•
Practice Problem
•
Hard
Megan Barnard
Used 12+ times
FREE Resource
10 questions
Show all answers
1.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
I predict there will be no significant difference in mean speeds between older and younger drivers. This is an example of a:
null statistical hypothesis
null theoretical hypothesis
alternative statistical hypothesis
alternative theoretical hypothesis
2.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
What can we say about the two population distributions shown in this image?
They probably show support for the alternative hypothesis
They probably show support for the null hypothesis
They have similar means and variances
We cannot tell from this image
3.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Which of the following statements is not true about ANOVAs?
The assumption of sphericity must be met for all types of ANOVA.
Smaller sample sizes will result in more uncertainty.
Between-groups deviation reflects treatment effects and experimental error.
Knowledge of the F-ratio and degrees of freedom are required to test significance.
4.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
In a one-way ANOVA where our independent variable has four levels, the between-group degrees of freedom would be:
3
1
2
4
5.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Which of the following is true about the Bonferroni correction?
Its use results in an increased risk of Type II error
It is calculated by dividing the desired alpha value by the number of treatment levels.
It uses the studentised range distribution Q to test for significance levels.
It is the least conservative type of statistical correction.
6.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Partial eta squared is used to estimate effect size in an ______ when there is______:
ANOVA; more than one independent variable
ANOVA; one independent variable
t-test; more than one independent variable
t-test: one independent variable
7.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
In a two-way factorial ANOVA, the degrees of freedom for the interaction term is calculated as:
(a-1)(b-1)
ab (s-1)
a-1
b-1
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