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.