A professor tests whether the loudness of noise during a test (low, medium, and

Question

A professor tests whether the loudness of noise during a test (low, medium, and high) is independent of test grades (pass, fail). The following table shows the observed frequencies for this test.

(a) Conduct a chi-square test for independence at a 0.05 level of significance. (Round your answer to two decimal places.)

X^2obt =

Decide whether to retain or reject the null hypothesis.

(b) Compute effect size using Cramer's V. (Round your answer to two decimal places.)

V =

Noise Level Low Medium High Pass 21 16 9 46 Fail 9 6 11 26 30 22 20 N = 72

Explanation / Answer

a.

b.

effective size for cramers V
formula for this V =sqrt(chi-square value/n*df)
degree of freedom (df) = (row *colomn)=(r-1*c-1) =(2-1 *4-1) =(1*3) =3
df of 3 = 0.17(medium value)
V =sqrt((4.32/(0.17*72))) = 0.59 medium effect

Given table data is as below MATRIX col1 col2 col3 col4 TOTALS row 1 21 16 9 46 92 row 2 9 6 11 26 52 TOTALS 30 22 20 72 N = 144 ------------------------------------------------------------------

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 19.17 14.06 12.78 46 row 2 10.83 7.94 7.22 26 ------------------------------------------------------------------

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 19.17 1.83 3.35 0.17 16 14.06 1.94 3.76 0.27 9 12.78 -3.78 14.29 1.12 46 46 0 0 0 9 10.83 -1.83 3.35 0.31 6 7.94 -1.94 3.76 0.47 11 7.22 3.78 14.29 1.98 26 26 0 0 0 ^2 o = 4.32 ------------------------------------------------------------------

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 =7.81
since our test is right tailed,reject Ho when ^2 o > 7.81
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 4.32
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 7.81
we got | ^2| =4.32 & | ^2 | =7.81
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.23


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: 4.32
critical value: 7.81
p-value:0.23
decision: do not reject Ho

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