The following table data test: Is there a relationship between students living a

Question

The following table data test: Is there a relationship between students living arrangement and exercise status? None Regular Sporadic 28 42 15 32 30 Dormitory On-campus Apt Off-campus Apt At Home 74 110 39 64 25 If the result of this hypothesis test is x2-11 which of the following is the correct conclusion OA.P Accept Ho, no significant association between exercise status and living arrangement. 0 a P>,Reject Ho, no significant correlation between exercise status and living arrangement. O c.Pc 11 . Accept Ho, no significant association between exercise status and living arrangement. ODP11 Reject Ho there is a signficant correlation between exercise status andi ing arrangement

Explanation / Answer

option:D

there is significant relation between exercise status and living arrangement

Given table data is as below MATRIX col1 col2 col3 TOTALS row 1 32 30 28 90 row 2 74 64 42 180 row 3 110 25 15 150 row 4 39 6 5 50 TOTALS 255 125 90 470 ------------------------------------------------------------------

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 row 4 row4*col1/N row4*col2/N row4*col3/N ------------------------------------------------------------------

expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 row 1 48.83 23.936 17.234 row 2 97.66 47.872 34.468 row 3 81.383 39.894 28.723 row 4 27.128 13.298 9.574 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 32 48.83 -16.83 283.249 5.801 30 23.936 6.064 36.772 1.536 28 17.234 10.766 115.907 6.725 74 97.66 -23.66 559.796 5.732 64 47.872 16.128 260.112 5.433 42 34.468 7.532 56.731 1.646 110 81.383 28.617 818.933 10.063 25 39.894 -14.894 221.831 5.561 15 28.723 -13.723 188.321 6.556 39 27.128 11.872 140.944 5.196 6 13.298 -7.298 53.261 4.005 5 9.574 -4.574 20.921 2.185 ^2 o = 60.439 ------------------------------------------------------------------

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, = 0.09
from standard normal table, chi square value at right tailed, ^2 /2 =10.948
since our test is right tailed,reject Ho when ^2 o > 10.948
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 60.439
critical value
the value of |^2 | at los 0.09 with d.f (r-1)(c-1)= ( 4 -1 ) * ( 3 - 1 ) = 3 * 2 = 6 is 10.948
we got | ^2| =60.439 & | ^2 | =10.948
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0


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: 60.439
critical value: 10.948 =11
p-value:0
decision: reject Ho

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