Given table data is as below
MATRIX | col1 | col2 | col3 | TOTALS |
row 1 | 26 | 13 | 18 | 57 |
row 2 | 24 | 32 | 6 | 62 |
row 3 | 10 | 16 | 5 | 31 |
TOTALS | 60 | 61 | 29 | N = 150 |
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calculation formula for E table matrix
E-TABLE | col1 | col2 | col3 |
row 1 | row1*col1/N | row1*col2/N | row1*col3/N |
row 2 | row2*col1/N | row2*col2/N | row2*col3/N |
row 3 | row3*col1/N | row3*col2/N | row3*col3/N |
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expected frequecies calculated by applying E - table matrix formulae
E-TABLE | col1 | col2 | col3 |
row 1 | 22.8 | 23.18 | 11.02 |
row 2 | 24.8 | 25.213 | 11.987 |
row 3 | 12.4 | 12.607 | 5.993 |
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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 |
26 | 22.8 | 3.2 | 10.24 | 0.449 |
13 | 23.18 | -10.18 | 103.632 | 4.471 |
18 | 11.02 | 6.98 | 48.72 | 4.421 |
24 | 24.8 | -0.8 | 0.64 | 0.026 |
32 | 25.213 | 6.787 | 46.063 | 1.827 |
6 | 11.987 | -5.987 | 35.844 | 2.99 |
10 | 12.4 | -2.4 | 5.76 | 0.465 |
16 | 12.607 | 3.393 | 11.512 | 0.913 |
5 | 5.993 | -0.993 | 0.986 | 0.165 |
chi^2 o = 15.727 |
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set up null vs alternative as
null, Ho: no relation b/w symptoms and the work shift
alternative, H1: exists a relation b/w symptoms and the work shift
level of significance, ? = 0.05
from standard normal table, chi square value at right tailed, chi^2 ?/2 =9.488
since our test is right tailed,reject Ho when chi^2 o > 9.488
we use test statistic chi^2 o = ?(Oi-Ei)^2/Ei
from the table , chi^2 o = 15.727
critical value
the value of |chi^2 ?| at los 0.05 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 3 - 1 ) = 2 * 2 = 4 is 9.488
we got | chi^2| =15.727 & | chi^2 ? | =9.488
make decision
hence value of | chi^2 o | > | chi^2 ?| and here we reject Ho
chi^2 p_value =0.003
ANSWERS
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null, Ho: no relation b/w symptoms and the work shift
alternative, H1: exists a relation b/w symptoms and the work shift
test statistic: 15.727
critical value: 9.488
p-value:0.003
decision: reject Ho
conclude that exists a relation b/w symptoms and the work shift
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