Complete parts A through D Parts A and B are shown C.find the critical value Thi
ID: 3324669 • Letter: C
Question
Complete parts A through D Parts A and B are shownC.find the critical value This Question: 1 pl 3 or5 (0 complete) This Test: 5 pts possibls You want to determine whother the reasons given by workers for continuing the data shown in the contingency table. Al a-005, can you conckude that the reason and type of workriComplele parts (a) their education is related to job s+ ody you rarndornly collect ? Complete parts (a through (d) 9 134 (a) Identify the claim and state the null and alternative hypotheses Ho Reasons are H, Reasons are The (b) Determine the degrees of freédom, find the critical value, and identily the rejection region the type of worker the type of worker What are the degrees of freedom Find the ee al val Click to select your answer(s) 8 9
Explanation / Answer
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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, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =5.991
since our test is right tailed,reject Ho when ^2 o > 5.991
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 11.116
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991
we got | ^2| =11.116 & | ^2 | =5.991
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.004
ANSWERS
---------------
a.
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
b.
d.f =(r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991
c.
test statistic: 11.116
critical value: 5.991
p-value:0.004
d.
decision: reject Ho
we have enough evidence to support the claim
Given table data is as below MATRIX col1 col2 col3 TOTALS row 1 30 39 34 103 row 2 50 20 35 105 TOTALS 80 59 69 N = 208 ------------------------------------------------------------------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 ------------------------------------------------------------------
expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 row 1 39.615 29.216 34.168 row 2 40.385 29.784 34.832 ------------------------------------------------------------------
calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 30 39.615 -9.615 92.448 2.334 39 29.216 9.784 95.727 3.277 34 34.168 -0.168 0.028 0.001 50 40.385 9.615 92.448 2.289 20 29.784 -9.784 95.727 3.214 35 34.832 0.168 0.028 0.001 ^2 o = 11.116
------------------------------------------------------------------
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, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =5.991
since our test is right tailed,reject Ho when ^2 o > 5.991
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 11.116
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991
we got | ^2| =11.116 & | ^2 | =5.991
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.004
ANSWERS
---------------
a.
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
b.
d.f =(r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991
c.
test statistic: 11.116
critical value: 5.991
p-value:0.004
d.
decision: reject Ho
we have enough evidence to support the claim
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