I need help asap with this statistic question, is a test. Then be sure that you

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I need help asap with this statistic question, is a test. Then be sure that you know the right answer. Thanks!

Examine and comment on this table of the standardized residuals for the chi-square test of who survived from a sinking ship. Crew -441 3.02 First 9.41 - 6.54 Second 2.75 1.88 Third - 3.31 2.28 Aliv Dead Select the best conclusion about the standardized residuals for this chi-squared test. O A. No conclusions can be drawn from these residuals. B. Third class passengers and crew were most likely to survive, while first class passengers were underrepresented among the survivors. C. All groups were equally represented among survivors. O D. First class passengers were most likely to survive, while third class passengers and crew were underrepresented among the survivors.

Explanation / Answer

option:c

Given table data is as below MATRIX col1 col2 col3 col4 TOTALS row 1 -4.41 9.41 2.75 -3.31 4.44 row 2 3.02 -6.54 -1.88 2.28 -3.12 TOTALS -1.39 2.87 0.87 -1.03 N = 1.32 ------------------------------------------------------------------

calculation formula for E table matrix E-TABLE col1 col2 col3 col4 row 1 row1*col1/N row1*col2/N row1*col3/N row1*col4/N row 2 row2*col1/N row2*col2/N row2*col3/N row2*col4/N ------------------------------------------------------------------

expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 col4 row 1 -4.675 9.654 2.926 -3.465 row 2 3.285 -6.784 -2.056 2.435 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei -4.41 -4.675 0.265 0.07 -0.015 9.41 9.654 -0.244 0.06 0.006 2.75 2.926 -0.176 0.031 0.011 -3.31 -3.465 0.155 0.024 -0.007 3.02 3.285 -0.265 0.07 0.021 -6.54 -6.784 0.244 0.06 -0.009 -1.88 -2.056 0.176 0.031 -0.015 2.28 2.435 -0.155 0.024 0.01 ^2 o = 0.002 ------------------------------------------------------------------

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 =7.815
since our test is right tailed,reject Ho when ^2 o > 7.815
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 0.002
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 4 - 1 ) = 1 * 3 = 3 is 7.815
we got | ^2| =0.002 & | ^2 | =7.815
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =1


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: 0.002
critical value: 7.815
p-value:1
decision: do not reject Ho

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