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Perform a goodness of fit test to determine if DSCC Introductory Statistics students match the reported distribution for age OR employment.

Category: Accounting & Finance Paper Type: Report Writing Reference: N/A Words: 819

According to the National Center for Education Statistics, the National Postsecondary Student Aid Study (NPSAS), 2012 revealed the following about students attending community college.

Student Age

Under 20                           21%

Between 20 and 24         34%

Between 25 and 29          16%

30 or older                          28%

Job While Enrolled in School

No Job                                 31%

Part-Time                            36%

Full-Time                             33

Construct an appropriate table for observed and expected frequency counts. 

Student Age

Observed Frequency

Expected Frequency

Under 20                          

21%

58%

Between 20 and 24        

34%

12%

Between 25 and 29         

16%

8%

30 or older                         

28%

14%

 State the null and alternative hypothesis

 H0: DSCC Introductory Statistics students match the reported distribution for age

H1: DSCC Introductory Statistics students does not match the reported distribution for age

 Compute the appropriate test statistic. Explain what method you used to calculate the test statistic including anything you put into your calculator, Excel, or formulas. Also include output.

The method of t-test is used on excel for appropriate test statistics. The data analysis tab is used and one sample t-test is applied using 5% level of significance or 0.05 alpha. The excel has given following output.

t-Test

 

Age

Mean

23.1

Variance

61.92857

Observations

50

Hypothesized Mean Difference

0

df

49

t Stat

20.75637

P(T<=t) one-tail

3.16E-26

t Critical one-tail

1.676551

P(T<=t) two-tail

6.31E-26

t Critical two-tail

2.009575


Determine if the null hypothesis can be rejected.  Fully explain your decision

The t statistics value in above table is 20.76 and t critical value is 1.68. The p-value is

3.16E-26 which is greater than the level of significance i.e. 5% so we are unable to reject null hypothesis.

State a conclusion.

 In the nutshell of above values, we cannot reject the null hypothesis in the favor of alternative hypothesis and we conclude that DSCC Introductory Statistics students does not match the reported distribution for age and the values for both results are different.

  What is a possible explanation for the results you have found?

The results have shown that the reported distribution for age is different than DSCC introductory statistics and possible explanation for the results found is that the data for both statistics could be different or biasness might exist in data distribution analysis.

  Conduct ONE of the following tests of independence.

 In our survey, we collected data about employment. We also collected data on whether a student had dependents. Perform a test of independence to determine if these two variables are independent.  


Part time

Full time

Unemployed

Responsible for dependent child

2

7

7

Not responsible for dependent child

21

4

9

In our survey we collected data regarding employment area. We also collected data on whether an Introductory Statistics student had dependents. Perform a test of independence to determine if these two variables are independent.

 

Health Care

Food

Office

Retail

Other

Unemployed

Responsible for dependent child

6

1

0

1

1

7

Not responsible for dependent child

2

11

1

5

6

9

 State the null and alternative hypothesis

H0: These two variables are not independent

H1: These two variables are independent

 Compute the appropriate test statistic. Explain what method you used to calculate the test statistic including anything you put into your calculator, Excel, or formulas. Also include output.

A first step, employment category is given codes i.e. food = 1, unemployed = 2, education = 3, retail = 4, healthcare = 5, office support = 6, factory = 7, and other =8.  Then the method of t-test is used on excel for appropriate test statistics. The data analysis tab is used and one sample t-test is applied using 5% level of significance or 0.05 alpha. The excel has given following output:

t-Test

 

Employment Category

Mean

3.14

Variance

4.694286

Observations

50

Hypothesized Mean Difference

0

df

49

t Stat

10.24778

P(T<=t) one-tail

4.44E-14

t Critical one-tail

1.676551

P(T<=t) two-tail

8.88E-14

t Critical two-tail

2.009575


Determine if the null hypothesis can be rejected. Fully explain your decision

The t statistics value in above table is 10.24778 and t critical value is 1.676551. The p-value is 4.44E-14which is greater than the level of significance i.e. 5% so we are unable to reject null hypothesis.

State a conclusion.

 

In the nutshell of above values, we cannot reject the null hypothesis in the favor of alternative hypothesis and we conclude that these two variables are not independent.

What is a possible explanation for the results you have found?

 

The results have shown that these two variables are not independent and possible explanation for the results found the employment category of student depends on various factors such as financial status of student family and need of students, the study area of students, and availability of opportunity etc.

 

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