Passed Failed White candidates Minority candidates Results from a civil servant
ID: 3318542 • Letter: P
Question
Passed Failed White candidates Minority candidates Results from a civil servant exam are shown in the table to the right. Is there sufficient evidence to support the claim that the results from the test are discriminatory? Use a 0.05 significance level. 14 18 Determine the null and alternative hypotheses A. Ho: White and minority candidates have the same chance of passing the test. O B. Ho: A white candidate is not more likely to pass the test than a minority candidate H1. White and minority candidates do not have the same chance of passing the test. H1: A white candidate is more likely to pass the test than a minority candidate Ho: White and minority candidates do not have the same chance of passing the test. H1. White and minority candidates have the same chance of passing the test. Ho: A white candidate is more likely to pass the test than a minority candidate. H1: A white candidate is not more likely to pass the test than a minority candidate C. 0 D. Determine the test statistic x2-D | (Round to three decimal places as needed.) Determine the P-value of the test statistic. P-value =L (Round to four decimal places as needed.) Is there sufficient evidence to support the claim that the results from the test are discriminatory? ( A. 0 B. ° C. 0 D. There is sufficient evidence to reject the claim that a white candidate is more likely to pass the test than a minority candidate. There is not sufficient evidence to support the claim that the results are discriminatory. There is not sufficient evidence to reject the claim that a white candidate is more likely to pass the test than a minority candidate There is sufficient evidence to support the claim that the results are discriminatory.Explanation / Answer
null and alternative hypothesis:
option:D
there is not sufficient to evidence to support the claim that white candidates is more likely to pass the test than a minority candidates
Given table data is as below MATRIX col1 col2 TOTALS row 1 14 18 32 row 2 7 24 31 TOTALS 21 42 N = 63 ------------------------------------------------------------------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 ------------------------------------------------------------------
expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 row 1 10.667 21.333 row 2 10.333 20.667 ------------------------------------------------------------------
calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 14 10.667 3.333 11.109 1.041 18 21.333 -3.333 11.109 0.521 7 10.333 -3.333 11.109 1.075 24 20.667 3.333 11.109 0.538 ^2 o = 3.175 ------------------------------------------------------------------
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 =3.841
since our test is right tailed,reject Ho when ^2 o > 3.841
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 3.175
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.841
we got | ^2| =3.175 & | ^2 | =3.841
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.075
ANSWERS
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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: 3.175
critical value: 3.841
p-value:0.075
decision: do not reject Ho