A hypothesis test is to be conducted to determine if the mean social marginality scores are not the same for all four age groups. That is, the following hypotheses will be tested. Ho: Myouths young Adults Adults Seniors H: at least two of the four ,'s are different Since we are told to assume that it is reasonable to regard the four samples as representative of the U.S. population in the corresponding age groups and that the distributions of social marginality scores for these four groups are approximately normal with the same standard deviation, the assumptions for the ANOVA test have been met. When performing an ANOVA test, a table is helpful to organize values in calculation of the test statistic and corresponding P-value. Recall the format of a partial ANOVA table, given below. Source of Variation Treatments Error Degrees of Freedom The given summary statistics are below. Age Group Youths k-1 N-k Young Sum of Squares Mean Square 55Tr SSE Adults Seniors SSTr k-1 = MST SSE NE MSE N-K = F MSTr MSE F ==
A hypothesis test is to be conducted to determine if the mean social marginality scores are not the same for all four age groups. That is, the following hypotheses will be tested. Ho: Myouths young Adults Adults Seniors H: at least two of the four ,'s are different Since we are told to assume that it is reasonable to regard the four samples as representative of the U.S. population in the corresponding age groups and that the distributions of social marginality scores for these four groups are approximately normal with the same standard deviation, the assumptions for the ANOVA test have been met. When performing an ANOVA test, a table is helpful to organize values in calculation of the test statistic and corresponding P-value. Recall the format of a partial ANOVA table, given below. Source of Variation Treatments Error Degrees of Freedom The given summary statistics are below. Age Group Youths k-1 N-k Young Sum of Squares Mean Square 55Tr SSE Adults Seniors SSTr k-1 = MST SSE NE MSE N-K = F MSTr MSE F ==
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![←
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All Black - Full Dark Them X
WLE3 - MATH142-2_E01_20 X
webassign.net/web/Student/Assignment-Responses/tutorial?pos=7&dep=31365371&tags=autosave#question4526942_7
with the same standard deviation.
Is there evidence that the mean social marginality scores are not the same for all four age groups? Test the relevant hypotheses using a = 0.01.
Step 1
A hypothesis test is to be conducted to determine if the mean social marginality scores are not the same for all four age groups. That is, the following hypotheses will be tested.
Ho Myouths young Adults Adults Seniors
Source of Variation
Treatments
83 F
Cloudy
H₂: at least two of the four ,'s are different
Since we are told to assume that it is reasonable to regard the four samples as representative of the U.S. population in the corresponding age groups and that the distributions of social marginality scores for these four groups are approximately normal with the
same standard deviation, the assumptions for the ANOVA test have been met.
When performing an ANOVA test, a table is helpful to organize values in calculation of the test statistic and corresponding P-value. Recall the format of a partial ANOVA table, given below.
Error
The given summary statistics are below.
Age Group Youths
Sample Size
x
5
LTI Launch
328935761_48874....jpg
117
2.00
1.46
Degrees
of Freedom
k-1
N-k
Young
Adults
242
3.50
1.76
Sum of Squares
302
Adults Seniors
3.18
55Tr
1.59
55E
42
2.73
1.97
Mean Square
SST=MSTR
k-1
SSE
N-K
= MSC
▬▬▬▬
F=
Facebook
F
We will first focus on the column for Degrees of Freedom. The value of k is the number of populations being compared. Here, we are comparing the means of four age groups, so we have k = 4 and k - 1 = ||
The value of N is the total number of observations in the data set, so we have N = 117 +242 + 302 +42 =
Thus, the degrees of freedom for Error is N-k=
Search
MSTr
MSE
[Solved] Engineering Data X
e =
=JO
61
ES
degrees of freedom for Treatments.
Show all
5:38 PM
2/2/2023
x
X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe24267a4-0d99-41ad-9e59-48078801137f%2F87314b8f-a76a-4bdf-bcb2-62fdbf179a64%2Fqrhgrge_processed.jpeg&w=3840&q=75)
Transcribed Image Text:←
✔
All Black - Full Dark Them X
WLE3 - MATH142-2_E01_20 X
webassign.net/web/Student/Assignment-Responses/tutorial?pos=7&dep=31365371&tags=autosave#question4526942_7
with the same standard deviation.
Is there evidence that the mean social marginality scores are not the same for all four age groups? Test the relevant hypotheses using a = 0.01.
Step 1
A hypothesis test is to be conducted to determine if the mean social marginality scores are not the same for all four age groups. That is, the following hypotheses will be tested.
Ho Myouths young Adults Adults Seniors
Source of Variation
Treatments
83 F
Cloudy
H₂: at least two of the four ,'s are different
Since we are told to assume that it is reasonable to regard the four samples as representative of the U.S. population in the corresponding age groups and that the distributions of social marginality scores for these four groups are approximately normal with the
same standard deviation, the assumptions for the ANOVA test have been met.
When performing an ANOVA test, a table is helpful to organize values in calculation of the test statistic and corresponding P-value. Recall the format of a partial ANOVA table, given below.
Error
The given summary statistics are below.
Age Group Youths
Sample Size
x
5
LTI Launch
328935761_48874....jpg
117
2.00
1.46
Degrees
of Freedom
k-1
N-k
Young
Adults
242
3.50
1.76
Sum of Squares
302
Adults Seniors
3.18
55Tr
1.59
55E
42
2.73
1.97
Mean Square
SST=MSTR
k-1
SSE
N-K
= MSC
▬▬▬▬
F=
Facebook
F
We will first focus on the column for Degrees of Freedom. The value of k is the number of populations being compared. Here, we are comparing the means of four age groups, so we have k = 4 and k - 1 = ||
The value of N is the total number of observations in the data set, so we have N = 117 +242 + 302 +42 =
Thus, the degrees of freedom for Error is N-k=
Search
MSTr
MSE
[Solved] Engineering Data X
e =
=JO
61
ES
degrees of freedom for Treatments.
Show all
5:38 PM
2/2/2023
x
X
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