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A researcher wants to study the relationship between salary and gender. She randomly selects 215 individuals and determines t
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Answer #1
Given table data is as below
MATRIX col1 col2 TOTALS
row 1 22 20 42
row 2 23 20 43
row 3 31 34 65
row 4 34 31 65
TOTALS 110 105 215
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calculation formula for E table matrix
E-TABLE col1 col2
row 1 row1*col1/N row1*col2/N
row 2 row2*col1/N row2*col2/N
row 3 row3*col1/N row3*col2/N
row 4 row4*col1/N row4*col2/N
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expected frequencies calculated by applying E - table matrix formula
E-TABLE col1 col2
row 1 21.488 20.512
row 2 22 21
row 3 33.256 31.744
row 4 33.256 31.744
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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above
Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei
22 21.488 0.512 0.262 0.012
20 20.512 -0.512 0.262 0.013
23 22 1 1 0.045
20 21 -1 1 0.048
31 33.256 -2.256 5.09 0.153
34 31.744 2.256 5.09 0.16
34 33.256 0.744 0.554 0.017
31 31.744 -0.744 0.554 0.017
ᴪ^2 o = 0.465
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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, α = 0.05
from standard normal table, chi square value at right tailed, ᴪ^2 α/2 =7.815
since our test is right tailed,reject Ho when ᴪ^2 o > 7.815
we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei
from the table , ᴪ^2 o = 0.465
critical value
the value of |ᴪ^2 α| at los 0.05 with d.f (r-1)(c-1)= ( 4 -1 ) * ( 2 - 1 ) = 3 * 1 = 3 is 7.815
we got | ᴪ^2| =0.465 & | ᴪ^2 α | =7.815
make decision
hence value of | ᴪ^2 o | < | ᴪ^2 α | and here we do not reject Ho
ᴪ^2 p_value =0.927


ANSWERS
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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
test statistic: 0.465
critical value: 7.815
p-value:0.927
decision: do not reject Ho

we do not have enough evidence to support the claim that there is a relationship between salary and gender.

Expected value for the number of men with an income below 25000$ is 21.5

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