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Pages:
4 pages/≈1100 words
Sources:
Level:
MLA
Subject:
Business & Marketing
Type:
Other (Not Listed)
Language:
English (U.S.)
Document:
MS Word
Date:
Total cost:
$ 17.28
Topic:

Survey Project Paper (Other (Not Listed) Sample)

Instructions:

It was a project of Bussiness Statistic class,It conntained 42 questionnaires and data set about the projects.The project was in children Technology education.

source..
Content:
Name
Tutor
Subject
Date
DESCRIPTIVE STATISTICS
1. Categorical variables: Children's gender(D8)
i. & ii. Frequency Distribution and Percentages
Children’s Gender

Freq

%

No Kids

10

23.809524

Male

17

40.476190

Female

3

7.142857

Male and Female

12

28.571429

Total

42

100

iii.) Bar Chart
lefttop
iv.) Pie Chart
2. Numerical Variable: Monthly Income (Q5)
Monthly Income



Mean

2.341463

Standard Error

0.225263

Median

2

Mode

1

Standard Deviation

1.44239

Sample Variance

2.080488

Kurtosis

-0.74437

Skewness

0.727962

Range

4

Minimum

1

Maximum

5

Sum

96

Count

41

B. Inferential Statistics
1. Monthly Income VsNumbers of kids
The correlation is positive and with a magnitude of 0.41 statistically significant (F0.00567<0.05)
Therefore Increase in one monthly income leads more children but not at a very high rate.
SUMMARY OUTPUT

















Regression Statistics








Multiple R

0.419610875








R Square

0.176073286








Adjusted R Square

0.155475118








Standard Error

0.673171369








Observations

42

















ANOVA









 

df

SS

MS

F

Significance F




Regression

1

3.873612297

3.873612

8.54801

0.005670046




Residual

40

18.1263877

0.45316






Total

41

22

 

 

 













 

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept

1.492741247

0.202216315

7.381903

5.5E-09

1.084046828

1.901435666

1.084046828

1.901435666

X Variable 1

0.215200683

0.073605659

2.923698

0.00567

0.066438097

0.36396327

0.066438097

0.36396327

2. Construct a 95% confidence interval for "Monthly Income”
95% CI = Mean +/- Standard Error
T-test is used since we are to estimate the population variance.
df=42-1=41 at (α/2) level of significance gives t=2.015
Mean

2.341463

Standard Error

0.225263

t= 2.015
95% CI
Upper limit 2.7952
Lower limit 1.8876
Here we are 95% confident that similarly constructed intervals will contain the true population. i.e. We are 95% confident that the true population will be between 1.8876 and 2.7952.
Further translation: We are 95%confident that in the population, the average Monthly Income tends to be between 2001 to 6000.
Extra Credit
E.C 1.1 Monthly Income Vs Parent’s Age
The correlation is positive and with a magnitude of 0.44 statistically significant (F0.002826<0.05)
Therefore Increase in one’s monthly income results from increase in his/her age but not at a high rate.
SUMMARY OUTPUT

















Regression Statistics








Multiple R

0.449461








R Square

0.202015








Adjusted R Square

0.182066








Standard Error

0.746116








Observations

42

















ANOVA









 

df

SS

MS

F

Significance F




Regression

1

5.637184

5.637184

10.12626

0.002826




Residual

40

22.26758

0.556689






Total

41

27.90476

 

 

 













 

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept

2.34045

0.224128

10.44245

5.46E-13

1.887469

2.79343

1.887469

2.79343

X Variable 1

0.259607

0.081582

3.182179

0.002826

0.094725

0.42449

0.094725

0.42449

E.C 1.2 Parent’s Age Vs Number of Kids
1. Monthly Income Vs Numbers of kids
The correlation is positive and with a magnitude of 0.44 statistically significant (F0.0032<0.05)
Therefore Increase in one will have more kids as he/ she grows older but not at a very high rate....
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