In: Statistics and Probability
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
Exam Pass 20 18 8 46
Fail 8 6 10 24
28 24 18 N=70
(a) Conduct a chi-square test for independence at a 0.05 level of significance. (Round your answer to two decimal places.)
x^2{obt}=
(b) Compute effect size using Cramer's V. (Round your answer to two decimal places.)
V =
In order to test , whether the loudness of noise during an exam (low, medium, and high) is independent of exam grades (pass, fail) we define the test hypothesis as:
Null hypothesis:
Test grade and noise level are independent.
Against the alternative hypothesis is
Test grade and noise level are not independent.
Observed data (frequency) are as follows:
Noise Level | |||||
Low | Medium | High | |||
exam | Pass | 20 | 18 | 8 | 46 |
Fail | 8 | 6 | 10 | 24 | |
28 | 24 | 18 | 70 |
Now the expected data (frequency) are calculated as:
Expected frequency=(Row total X Colunm total) / Grand total
Noise Level | |||||
Low | Medium | High | Sum | ||
exam | Pass | 18.4 | 15.77143 | 11.82857 | 46 |
Fail | 9.6 | 8.228571 | 6.171429 | 24 | |
28 | 24 | 18 | 70 |
To calculate the Chi square test statistics we define the chi square as:
Noise Level | |||||
Low | Medium | High | Sum | ||
exam | Pass | 21.73913 | 20.54348 | 5.410628 | 47.69324 |
Fail | 6.666667 | 4.375 | 16.2037 | 27.24537 | |
sum | 28.4058 | 24.91848 | 21.61433 | 74.93861 |
=74.93861-70
=4.93861
Now as it is a contingency table view , therefore the degree of freedom of statistics is;
Therefore the critical value of test statistics at level of significance is:
from the standard chi square table .
in order to determined the P value for corresponding Chi square we follow Excell as:
=CHIDIST(4.9386,2)
Therefore ,
and corresponding
We fail to reject the null hypothesis and therefore we retain the null hypothesis.
b) Cramer’s V is measure the inter correlation between two contingency ( discrete variables). It is use to post test of strengths of association after chi-square significance for more than 2x2 rows and columns.
It is denoted as:
r: row and c: column
r: row and c: column
=0.07055
Therefore the effect size by cramer's V is :
V=0.07