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Question I. A random sample of 75 cities has been classified as small or large by population and as high or low on crime rate

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Answer #1

1.

Given table data is as below
MATRIX col1 col2 TOTALS
row 1 45 10 55
row 2 10 10 20
TOTALS 55 20 N = 75

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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

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expected frequencies calculated by applying E - table matrix formulae
E-TABLE col1 col2
row 1 40.333 14.667
row 2 14.667 5.333

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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
45 40.333 4.667 21.781 0.54
10 14.667 -4.667 21.781 1.485
10 14.667 -4.667 21.781 1.485
10 5.333 4.667 21.781 4.084
ᴪ^2 o = 7.594

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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 =3.841
since our test is right tailed,reject Ho when ᴪ^2 o > 3.841
we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei
from the table , ᴪ^2 o = 7.594
critical value
the value of |ᴪ^2 α| at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.841
we got | ᴪ^2| =7.594 & | ᴪ^2 α | =3.841
make decision
hence value of | ᴪ^2 o | > | ᴪ^2 α| and here we reject Ho
ᴪ^2 p_value =0.006


ANSWERS
---------------
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: 7.594
critical value: 3.841
p-value:0.006
decision: reject Ho

we have enough evidence to support the claim that there is association between dependent and independent variable.

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