Hi there, I hope you're well. I have attached the results of a chi-square goodness of fit test for a set of data. We are required to visually represent this data either using a bar graph, pie chart, distribution graph etc. We are required to use two different visual representations of the data. I have tried to youtube various tutorials on how it's done in excel, but have had absolutely no luck. Would you be able to assist me by any chance please?
Hi there, I hope you're well. I have attached the results of a chi-square goodness of fit test for a set of data. We are required to visually represent this data either using a bar graph, pie chart, distribution graph etc. We are required to use two different visual representations of the data. I have tried to youtube various tutorials on how it's done in excel, but have had absolutely no luck. Would you be able to assist me by any chance please?
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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Hi there,
I hope you're well. I have attached the results of a chi-square goodness of fit test for a set of data. We are required to visually represent this data either using a bar graph, pie chart, distribution graph etc. We are required to use two different visual representations of the data. I have tried to youtube various tutorials on how it's done in excel, but have had absolutely no luck.
Would you be able to assist me by any chance please?
I would really appreciate your help.
Thank you kindly.
Kind Regards,
Carmen Loureiro
Transcribed Image Text:2. Chi-square goodness of fit test
Difference in Academic Term Averages
(%)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
D.F = 15 - 1
Sports
Participation
Treatment x
x2= [(Or,c-Er,c)2Er,c
= 31.83756224
≈ 31.84
DF = k-1
D.F = 14
85
86
88
82
88
90
92
90
90
77
80
89
79
83
92
à = 0.05
P value = 0.0042
No Sports
Participation
Treatment y
P (x2 > 31.84) = 0.0042
86
67
65
68
68
79
79
83
72
84
82
75
74
52
75
Sum
1
23
4
5
6
7
8
9
10
11
12
13
34
14
15
Null hypothesis (Ho): The distribution of Sports Participation Treatment(x) does not differ from No
Sports Participation Treatment(y).
Alternate hypothesis (H₁): The distribution of Sports Participation Treatment(x) does differ from No
Sports Participation Treatment(y).
Observed
frequency
85
86
88
82
88
90
92
90
90
77
80
89
79
83
92
Expected
frequency
E(x) = n*P(x)
86
0.011627907
67 5.388059701
65
8.138461538
68
2.882352941
68
5.882352941
79
1.53164557
79 2.139240506
82 0.780487805
72
84
0.583333333
82
0.048780488
75 2.613333333
74
0.337837838
52
18.48076923
75
3.853333333
31.83756224
(Or,c-
Er,c)2Er,c
Since the P-value (0.0042) is less than the significance level (0.05), we have to reject the null
hypothesis.
We have to Reject Ho as there is convincing evidence to conclude that the distribution of Sports
Participation Treatment(x) does differ from No Sports Participation Treatment(y).
4.5
Expert Solution
Step 1
Given information:
| Sports Participation Treatment x | No Sports Participation Treatment y | |
| 1 | 85 | 86 |
| 2 | 86 | 67 |
| 3 | 88 | 65 |
| 4 | 82 | 68 |
| 5 | 88 | 68 |
| 6 | 90 | 79 |
| 7 | 92 | 79 |
| 8 | 90 | 82 |
| 9 | 90 | 72 |
| 10 | 77 | 84 |
| 11 | 80 | 82 |
| 12 | 89 | 75 |
| 13 | 79 | 74 |
| 14 | 83 | 52 |
| 15 | 92 | 75 |
Step by step
Solved in 3 steps with 2 images
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