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(9) A study looking at the education level and diabetes status of patients was completed and the following contingency table
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Answer #1

Given table data is as below

MATRIX col1 col2 TOTALS
row 1 33 218 251
row 2 25 389 414
row 3 20 393 413
row 4 17 178 195
TOTALS 95 1178 1273

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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 18.731 232.269
row 2 30.896 383.104
row 3 30.821 382.179
row 4 14.552 180.448

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calculate chi-square test statistic using given observed frequencies, calculated expected frequencies from above

Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei
33 18.731 14.269 203.604 10.87
218 232.269 -14.269 203.604 0.877
25 30.896 -5.896 34.763 1.125
389 383.104 5.896 34.763 0.091
20 30.821 -10.821 117.094 3.799
393 382.179 10.821 117.094 0.306
17 14.552 2.448 5.993 0.412
178 180.448 -2.448 5.993 0.033
ᴪ^2 o = 17.513

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set up null vs alternative as
null, Ho: diabetes status is independent of education level
alternative, H1: diabetes status is dependent of education level
level of significance, alpha = 0.05
from standard normal table, chi square value at right tailed, ᴪ^2 alpha/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 = 17.513
critical value
the value of |ᴪ^2 alpha| 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| =17.513 & | ᴪ^2 alpha | =7.815
make decision
hence value of | ᴪ^2 o | > | ᴪ^2 alpha| and here we reject Ho
ᴪ^2 p_value =0.001
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[ANSWERS]
null, Ho: diabetes status is independent of education level
alternative, H1: diabetes status is dependent of education level
test statistic: 17.513
critical value: 7.815
p-value:0.001
decision: reject Ho                                                  
evidence to support that diabetes status is dependent of education level

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