Virtual Campus 6:09 AM Sun Jun 30 202405221053127570_UNIT_06_CHI_SQUARE_MBA ва Home Insert Draw Page Layout Formulas Data Review View e B IUS O Av Ev General Read Only - Save a copy to edit. fx Enter text or formula here Chi-Square Test of Independence Save a copy H N ° P Q R T U W 3 Observed Frequencies Column variable 5 Row variable Pool No Pool Total Brick 30 40 7 Stucco 18 17 35 8 Wood 14 11 25 Use the YELLOW cells to set up the Chi Square table. The table can handle up to 5 rows and 5 columns of values. 9 0 If fewer rows or columns are needed, leave the excess blank. 10 11 Total 62 38 이 100 The BLUE table computes the expected frequencies needed to compute the chi square statistic. The only values that ultimately matter to you is in the RESULTS table. 12 13 Expected Frequencies 14 Column variable 15 Row variable Pool No Pool ° Total 16 Brick 24.80 15.20 0.00 0.00 0.00 40 Here we are testing if having a pool is independent of what the house is made of. 17 Stucco 21.70 13.30 0.00 0.00 0.00 35 The p-value of .0858 is greater than .05, and so we do not reject the null hypothesis. 18 15.50 9.50 0.00 0.00 0.00 There is insufficient evidence to conclude a relationship between the construction of a house and whether it has a poo 19 0.00 0.00 0.00 0.00 0.00 0 20 0.00 0.00 0.00 0.00 0.00 0 21 Total 38 0 100 22 23 Data 24 Level of Significance 0.05 25 Number of Rows Number of Columns 27 Degrees of Freedom 2 28 29 Results 30 Critical Value 31 Chi-Square Test Statistic 5.991465 4.911472 32 p-Value 0.0858 33 Do not reject the null hypothesis 34 35 36 37 38 This tests the null hypothesis that the row variable and column variable are independent. Rejecting the null implies that the two variables are related (one is dependent on the other). ©2007 Dr JimMirabella.com 39 40 42 43 44 45 47 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 86 88 89 90 Unite 6 equal Unit 6 unequal Unit 6 prop Variable descriptions Age age of student in were MBA GPA overall GPA in the MBA program Works FT 0 (No), 1 (Yes

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 76E
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1. Historically, the MBA program at Whatsamattu U. has about 40% of their students choose a Leadership major, 30% choose a Finance major, 20% choose a Marketing major, and 10% choose no major. Does the most recent class of 200 MBA students fit that same pattern or has there been a shift in the choice of majors. Using the sample of 200 students (in the data file), conduct a Chi Square Goodness of Fit test to determine if the current distribution fits the historical pattern. Use a .05 significance level. 2. While job opportunities for men and women are considerably more balanced than they were 40 years ago, the career aspirations may still differ. Is there a difference in majors chosen by men and women? Using the sample of 200 MBA students (in the data file), conduct a Chi Square Test of Independence to determine if one's choice of major is independent of their gender. Use a .05 significance level.
Virtual Campus 6:09 AM Sun Jun 30
202405221053127570_UNIT_06_CHI_SQUARE_MBA
ва
Home
Insert
Draw
Page Layout
Formulas
Data Review View
e
B
IUS O
Av Ev General
Read Only - Save a copy to edit.
fx Enter text or formula here
Chi-Square Test of Independence
Save a copy
H
N
°
P
Q
R
T
U
W
3
Observed Frequencies
Column variable
5
Row variable
Pool
No Pool
Total
Brick
30
40
7
Stucco
18
17
35
8
Wood
14
11
25
Use the YELLOW cells to set up the Chi Square table.
The table can handle up to 5 rows and 5 columns of values.
9
0
If fewer rows or columns are needed, leave the excess blank.
10
11
Total
62
38
이
100
The BLUE table computes the expected frequencies needed to compute the chi square
statistic. The only values that ultimately matter to you is in the RESULTS table.
12
13
Expected Frequencies
14
Column variable
15
Row variable
Pool
No Pool
°
Total
16
Brick
24.80
15.20
0.00
0.00
0.00
40
Here we are testing if having a pool is independent of what the house is made of.
17
Stucco
21.70
13.30
0.00
0.00
0.00
35
The p-value of .0858 is greater than .05, and so we do not reject the null hypothesis.
18
15.50
9.50
0.00
0.00
0.00
There is insufficient evidence to conclude a relationship between the construction of a house and whether it has a poo
19
0.00
0.00
0.00
0.00
0.00
0
20
0.00
0.00
0.00
0.00
0.00
0
21
Total
38
0
100
22
23
Data
24
Level of Significance
0.05
25
Number of Rows
Number of Columns
27
Degrees of Freedom
2
28
29
Results
30
Critical Value
31
Chi-Square Test Statistic
5.991465
4.911472
32
p-Value
0.0858
33
Do not reject the null hypothesis
34
35
36
37
38
This tests the null hypothesis that the row variable and column variable are independent.
Rejecting the null implies that the two variables are related (one is dependent on the other).
©2007 Dr JimMirabella.com
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Unite 6 equal
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Unit 6 prop
Transcribed Image Text:Virtual Campus 6:09 AM Sun Jun 30 202405221053127570_UNIT_06_CHI_SQUARE_MBA ва Home Insert Draw Page Layout Formulas Data Review View e B IUS O Av Ev General Read Only - Save a copy to edit. fx Enter text or formula here Chi-Square Test of Independence Save a copy H N ° P Q R T U W 3 Observed Frequencies Column variable 5 Row variable Pool No Pool Total Brick 30 40 7 Stucco 18 17 35 8 Wood 14 11 25 Use the YELLOW cells to set up the Chi Square table. The table can handle up to 5 rows and 5 columns of values. 9 0 If fewer rows or columns are needed, leave the excess blank. 10 11 Total 62 38 이 100 The BLUE table computes the expected frequencies needed to compute the chi square statistic. The only values that ultimately matter to you is in the RESULTS table. 12 13 Expected Frequencies 14 Column variable 15 Row variable Pool No Pool ° Total 16 Brick 24.80 15.20 0.00 0.00 0.00 40 Here we are testing if having a pool is independent of what the house is made of. 17 Stucco 21.70 13.30 0.00 0.00 0.00 35 The p-value of .0858 is greater than .05, and so we do not reject the null hypothesis. 18 15.50 9.50 0.00 0.00 0.00 There is insufficient evidence to conclude a relationship between the construction of a house and whether it has a poo 19 0.00 0.00 0.00 0.00 0.00 0 20 0.00 0.00 0.00 0.00 0.00 0 21 Total 38 0 100 22 23 Data 24 Level of Significance 0.05 25 Number of Rows Number of Columns 27 Degrees of Freedom 2 28 29 Results 30 Critical Value 31 Chi-Square Test Statistic 5.991465 4.911472 32 p-Value 0.0858 33 Do not reject the null hypothesis 34 35 36 37 38 This tests the null hypothesis that the row variable and column variable are independent. Rejecting the null implies that the two variables are related (one is dependent on the other). ©2007 Dr JimMirabella.com 39 40 42 43 44 45 47 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 86 88 89 90 Unite 6 equal Unit 6 unequal Unit 6 prop
Variable descriptions
Age age of student in were
MBA GPA overall GPA in the MBA program
Works FT 0 (No), 1 (Yes
Transcribed Image Text:Variable descriptions Age age of student in were MBA GPA overall GPA in the MBA program Works FT 0 (No), 1 (Yes
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