A recent survey asked 1,000 men and 1,000 women how they would rate their credit
ID: 3320194 • Letter: A
Question
A recent survey asked 1,000 men and 1,000 women how they would rate their credit score, and the results are as follows:
Poor
Fair
Good
Excellent
Unsure
Women
100
250
450
125
75
Men
75
225
325
250
125
Is there a difference in the distribution of responses between women and men? (PLEASE SHOW ALL WORK).
A) Is this a test of independence or homogeneity?
B) Write the hypothesis.
C) Check the necessary assumptions and conditions.
D) Find the P-value of your test.
E) State the conclusion and analysis.
Poor
Fair
Good
Excellent
Unsure
Women
100
250
450
125
75
Men
75
225
325
250
125
Explanation / Answer
a.
chi square test of independence
------------------------------------------------------------------
set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =9.488
since our test is right tailed,reject Ho when ^2 o > 9.488
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 79.216
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 5 - 1 ) = 1 * 4 = 4 is 9.488
we got | ^2| =79.216 & | ^2 | =9.488
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0
ANSWERS
---------------
b.
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
c.
test statistic: 79.216
critical value: 9.488
d.
p-value:0
e.
decision: reject Ho
Given table data is as below MATRIX col1 col2 col3 col4 col5 TOTALS row 1 100 250 450 125 75 1000 row 2 75 225 325 250 125 1000 TOTALS 175 475 775 375 200 N = 2000 ------------------------------------------------------------------calculation formula for E table matrix E-TABLE col1 col2 col3 col4 col5 row 1 row1*col1/N row1*col2/N row1*col3/N row1*col4/N row1*col5/N row 2 row2*col1/N row2*col2/N row2*col3/N row2*col4/N row2*col5/N ------------------------------------------------------------------
expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 col4 col5 row 1 87.5 237.5 387.5 187.5 100 row 2 87.5 237.5 387.5 187.5 100 ------------------------------------------------------------------
calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 100 87.5 12.5 156.25 1.786 250 237.5 12.5 156.25 0.658 450 387.5 62.5 3906.25 10.081 125 187.5 -62.5 3906.25 20.833 75 100 -25 625 6.25 75 87.5 -12.5 156.25 1.786 225 237.5 -12.5 156.25 0.658 325 387.5 -62.5 3906.25 10.081 250 187.5 62.5 3906.25 20.833 125 100 25 625 6.25 ^2 o = 79.216
------------------------------------------------------------------
set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =9.488
since our test is right tailed,reject Ho when ^2 o > 9.488
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 79.216
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 5 - 1 ) = 1 * 4 = 4 is 9.488
we got | ^2| =79.216 & | ^2 | =9.488
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0
ANSWERS
---------------
b.
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
c.
test statistic: 79.216
critical value: 9.488
d.
p-value:0
e.
decision: reject Ho
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