By Dr. Sultan Muhammad Khan
Keywords; Using
SPSS create new categorical variables for Verbal and Quant of level 3 each? Apply
chi square test to new variable and interpret the result? Calculate the
correlation between verbal and quant and interpret the results? Apply chi square
test to Job and Experience of any cell has less than 5 expected frequencies
then combine the adjacent levels and try to find chi square test, repeat until
there is no cell with expected frequency less than 5 and interpret the result?
Statistics Test Explanation
Question
A: Using SPSS create new
categorical variables for Verbal and Quant of level 3 each?
Answer:
To create new
categorical variables use recode into different variables from the transform menu, input the name of
target variable and give new name for variable in the output variable
window and press change button. Then click on old and new values
and enter date by range and assign them a new value e.g. 1,2,3….
Attach are the
sheets which was created by this method with name of Varbnew and Quantnew,
using range 80 thru 100 for value 1 and 101 to 120 for value 2 and 131 to 140
for value 3.
In this way we get
two new categorical variables
Verbnew and Quantnew
Question B: Apply chi square test to new variable and
interpret the result?
Answer:
After creating the new categorical
variable verbnew and quantnew in question A, here we apply chi square test to
these new variables using command syntax Analyze- Descriptive Statistic- Crosstab,
a new window will appear, move verbnew to row window and quantnew to column
window and then press on the Statistic button select chi-square in
new window and press continue button and then press ok. Now Computer will calculate
the chi square of
Verbnew and Quantnew, the
output sheet is attached.
In
the result table we have the degree of freedom 4 in two cases and 1 in linear
by linear association, which is good, the degree of freedom should not be 0
because then it have no freedom it will effect the whole calculation.
In the case processing summary we have
N=100 mean one hundred observation for both new variables.
In the last
table we have the original chi square test table which shows us that the asymp
significance is .000 for all three, which
mean that we can reject this hypothesis
because there is a definite difference between the two. We reject the hypothesis
because the difference is high if the value is less then .5 which in this case
it is .000.
If the value of Asymp sig is
greater than .5 then there is
no difference and we accept the hypothesis which is not in this case.
Question C: Calculate
the correlation between verbal and quant and interpret the results?
Answer:
To find correlation, we will use the
bivariate correlation to determine if the two variables are linearly related to
each other, for this purpose we will use command syntax
Analyze-correlation-bivariate
to calculate the correlation between two variables.
Select two variables, in this case
we select verbal and quant and move it to variable
pane for correlation then click on one
tailed option. After that click on the option button and select mean and slandered deviation then
click ok and check the output in the output viewer.
The
descriptive statistics section gives the mean, slandered deviation and number
of observation (n) for each variable.
The mean for quant is 1.1232E2 and verbal 1.1163E2 and slandered deviation
for variables 12.88071, 12.30926 and observation for each is 75. The
correlation section gives us the values of the specified correlation test.
It is a one 1-tailed test
therefore .000 is
the significance of this correlation, which tells us that the correlation is
significant in verbal and quant, because the value is less than .5
The correlation coefficient
is .872 and the middle number is significance of this correlation which is .000 and 75 is the number of observation in this case.
Question D: Apply chi square test to Job and Experience of
any cell has less than 5 expected frequencies then combine the adjacent levels
and try to find chi square test, repeat until there is no cell with expected
frequency less than 5 and interpret the result?
Answer:
After
creating new variables job1, job2, job3, job4 and exper1, exper2, exper3,
exper4 to achieve the result of no cell that have the expected frequency less
than 5 and calculate chi square five times to research the end result.
In
all five chi square result we have different degree of freedom and every time we
combine the adjacent levels the degree of freedom decrease e.g. in first X2
test it is 30 and in the last test it is 2 with no cell that have expected frequency less than 5 which was our aim
in this case to find the result till there is no cell which have expected
frequency less than 5 and we achieved it on fifth chi square test and
in all chi square test we have the degree of freedom 1 in linear by linear
association,
In the case processing summary we
have N=100 mean one hundred observation for all variables.
In all five
chi square result we have find a high difference because in all chi square test
the Asymp sig is .000 for all which
is less than .5, Means we can reject the hypothesis and
can say that there is a definite difference between them.
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