8. -/3 points PriviteraStats2 17.E.023 My Notes Ask Your Teacher A professor tes

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8. -/3 points PriviteraStats2 17.E.023 My Notes Ask Your Teacher A professor tests whether the loudness of noise during an exam (low, medium, and high) is independent of exam grades (pass, fail). The following table shows the observed frequencies for this test Noise Level Low Medium High 17 6 23 21 47 26 N=73 Exam Pass Fail 30 20 (a) Conduct a chi-square test for independence at a 0.05 level of significance. (Round your answer to two decimal places.) Decide whether to retain or reject the null hypothesis Retain the null hypothesis Reject the null hypothesis (b) Compute effect size using Cramer's V. (Round your answer to two decimal places.) You may need to use the appropriate table in Appendix B to answer this question Additional Materials Section 17.1

Explanation / Answer

b.

Effective size for cramers V rule
V =sqrt(chisquare/n*df)
degree of freedom df =min(r-1,c-1)
df = min(2-1,3-1) = min (1,2) = 1*2 =2
df =0.21 (medium)
V = sqrt(4.6/73*0.21)
V = 0.54

Given table data is as below MATRIX col1 col2 col3 TOTALS row 1 21 17 9 47 row 2 9 6 11 26 TOTALS 30 23 20 N = 73 ------------------------------------------------------------------

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 19.32 14.81 12.88 row 2 10.68 8.19 7.12 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 21 19.32 1.68 2.82 0.15 17 14.81 2.19 4.8 0.32 9 12.88 -3.88 15.05 1.17 9 10.68 -1.68 2.82 0.26 6 8.19 -2.19 4.8 0.59 11 7.12 3.88 15.05 2.11 ^2 o = 4.6 ------------------------------------------------------------------

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 =5.99
since our test is right tailed,reject Ho when ^2 o > 5.99
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 4.6
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 5.99
we got | ^2| =4.6 & | ^2 | =5.99
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.1


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: 4.6
critical value: 5.99
p-value:0.1
decision: do not reject Ho

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