Enrique teaches five sections of Introductory Psychology. He wants to know whether or not the average score on the second exam is the same for all five sections. Enrique's student's exam scores are listed below. Use an ANOVA test with a level of significance of 6% to test whether the average score is the same for all five sections. Section 1: 80, 80, 84, 88, 85, 89, 80, 79, 85, 80, 70, 82, 76, 86, 72, 84, 80, 79, 87, 74, 81, 86, 74, 79, 81, 77, 75, 76, 82, 76, 84, 80, 88, 92 Section 2: 78, 90, 77, 83, 79, 82, 82, 89, 78, 82, 81, 81, 80, 85, 81, 84, 86, 84, 75, 81, 82, 80, 83, 83, 81, 86, 85, 82, 79, 81, 82 Section 3: 82, 80, 78, 81, 82, 83, 83, 81, 76, 81, 84, 78, 75, 86, 83, 89, 79, 84, 80, 80, 77, 76, 80, 81, 79, 76, 79, 87, 77, 85, 82, 85, 81, 85, 75, 83, 87, 84, 79 Section 4: 84, 88, 77, 87, 89, 83, 88, 83, 84, 70, 85, 82, 82, 84, 83, 82, 78, 83, 82, 85, 78, 81, 78, 86, 87 Section 5: 83, 85, 78, 82, 82, 87, 82, 90, 80, 84, 87, 83, 89, 81, 80, 89, 82, 83, 84, 92, 85, 79, 83, 91, 84, 84, 81, 82, 84, 89, 88, 73, 86, 78, 77 Step 1: State the null and alternative hypotheses. Ho: P1 = H2 = P3 = Ha = Hs Ha: At least one mean isn't equal to the other means v Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of test statistics. v distribution with numerator degrees of freedom We will use a(n) F 159 dfbetween and denominator degrees of freedom dfwithin %3D 4 Part 3 of 4
Enrique teaches five sections of Introductory Psychology. He wants to know whether or not the average score on the second exam is the same for all five sections. Enrique's student's exam scores are listed below. Use an ANOVA test with a level of significance of 6% to test whether the average score is the same for all five sections. Section 1: 80, 80, 84, 88, 85, 89, 80, 79, 85, 80, 70, 82, 76, 86, 72, 84, 80, 79, 87, 74, 81, 86, 74, 79, 81, 77, 75, 76, 82, 76, 84, 80, 88, 92 Section 2: 78, 90, 77, 83, 79, 82, 82, 89, 78, 82, 81, 81, 80, 85, 81, 84, 86, 84, 75, 81, 82, 80, 83, 83, 81, 86, 85, 82, 79, 81, 82 Section 3: 82, 80, 78, 81, 82, 83, 83, 81, 76, 81, 84, 78, 75, 86, 83, 89, 79, 84, 80, 80, 77, 76, 80, 81, 79, 76, 79, 87, 77, 85, 82, 85, 81, 85, 75, 83, 87, 84, 79 Section 4: 84, 88, 77, 87, 89, 83, 88, 83, 84, 70, 85, 82, 82, 84, 83, 82, 78, 83, 82, 85, 78, 81, 78, 86, 87 Section 5: 83, 85, 78, 82, 82, 87, 82, 90, 80, 84, 87, 83, 89, 81, 80, 89, 82, 83, 84, 92, 85, 79, 83, 91, 84, 84, 81, 82, 84, 89, 88, 73, 86, 78, 77 Step 1: State the null and alternative hypotheses. Ho: P1 = H2 = P3 = Ha = Hs Ha: At least one mean isn't equal to the other means v Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of test statistics. v distribution with numerator degrees of freedom We will use a(n) F 159 dfbetween and denominator degrees of freedom dfwithin %3D 4 Part 3 of 4
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Concept explainers
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
Topic Video
Question
Transcribed Image Text:test statistics.
We will use a(n) F
vv distribution with numerator degrees of freedom
v and denominator degrees of freedom dfwithin =
d fpetween
4
159
Part 3 of 4
Step 3: Find the p-value of the test statistic.
I1 = 80.91176471/
I2 =
82
81
T4 =
82.76
83.82352941X
T = 82.10494118 X
ANOVA Table
SS
df
MS
F
49.7567796 ×
2.986335399 X
Between
199.0271184 X
4
153
16.66148404 X
Within
2549.207059 X
P(F2♥ 0.02079
Transcribed Image Text:Enrique teaches five sections of Introductory Psychology. He wants to know whether or not the
average score on the second exam is the same for all five sections. Enrique's student's exam scores
are listed below. Use an ANOVA test with a level of significance of 6% to test whether the average
Score is the same for all five sections.
Section 1:
80, 80, 84, 88, 85, 89, 80, 79, 85, 80, 70, 82, 76, 86, 72, 84, 80, 79, 87, 74, 81, 86, 74, 79, 81, 77, 75, 76,
82, 76, 84, 80, 88, 92
Section 2:
78, 90, 77, 83, 79, 82, 82, 89, 78, 82, 81, 81, 80, 85, 81, 84, 86, 84, 75, 81, 82, 80, 83, 83, 81, 86, 85, 82,
79, 81, 82
Section 3:
82, 80, 78, 81, 82, 83, 83, 81, 76, 81, 84, 78, 75, 86, 83, 89, 79, 84, 80, 80, 77, 76, 80, 81, 79, 76, 79, 87,
77, 85, 82, 85, 81, 85, 75, 83, 87, 84, 79
Section 4:
84, 88, 77, 87, 89, 83, 88, 83, 84, 70, 85, 82, 82, 84, 83, 82, 78, 83, 82, 85, 78, 81, 78, 86, 87
Section 5:
83, 85, 78, 82, 82, 87, 82, 90, 80, 84, 87, 83, 89, 81, 80, 89, 82, 83, 84, 92, 85, 79, 83, 91, 84, 84, 81, 82,
84, 89, 88, 73, 86, 78, 77
Step 1: State the null and alternative hypotheses.
Ho: P1 = P2 = P3 = P4 = Ps
Ha: At least one mean isn't equal to the other means v
Part 2 of 4
Step 2: Assuming the null hypothesis is true, determine the features of the distribution of
test statistics.
vv distribution with numerator degrees of freedom
159
We will use a(n) F
dfpetween =
v and denominator degrees of freedom d fwithin =
4
Part 3 of 4
homualue of the test statistic.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 16 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.Recommended textbooks for you
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
Elementary Statistics: Picturing the World (7th E…
Statistics
ISBN:
9780134683416
Author:
Ron Larson, Betsy Farber
Publisher:
PEARSON
The Basic Practice of Statistics
Statistics
ISBN:
9781319042578
Author:
David S. Moore, William I. Notz, Michael A. Fligner
Publisher:
W. H. Freeman
Introduction to the Practice of Statistics
Statistics
ISBN:
9781319013387
Author:
David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:
W. H. Freeman