In the following table, likely voters’ preferences of two candidates are cross-classified by gender.

tefoldaavv

tefoldaavv

Answered question

2022-04-24

In the following table, likely voters’ preferences of two candidates are cross-classified by gender.
 Male  Female  Candidate A 150130 Candidate B 100120
For the chi-square test of independence, the value of the test statistic is:
a. 2.34
b. 1.62
c. 3.25
d. 4

Answer & Explanation

ubafumene42h

ubafumene42h

Beginner2022-04-25Added 13 answers

It is needed to test, whether the there is a relationship between voters’ preferences of two candidates and gender.
The null and alternative hypotheses for the test are:
H0: There is there is no relationship between voters’ preferences of two candidates and gender.
H1: There is there is a significant relationship between voters’ preferences of two candidates and gender.
A chi-squared test of independence is suitable to be used in this situation.
The expected frequency formula for a particular cell is,
Expected frequency=Row Total×Column TotalOverall frequency
Consider the following table for necessary calculation:
 Male  Female  Row Total  Candidate A 150130280 Candidate B 100120220 Column Total 250250500
Thus, the expected frequencies are obtained as follows:
 Expected value  Male  Female  Candidate A 280×250500140280×250500140 Candidate B 220×250500110220×250500110
Consider that Oi is the observed frequency for ith cell and Ei is the expected frequency for ith cell. There are total 2 rows (r) and 2 columns (c).
The chi-square test statistic is defined as,
χ2=(OiEi)2Ei, which follows chi-square distribition with degrees of freedom of (e-1)(c-1).
Consider the following table for necessary calculation.
 Observed (Oi) Expected (Ei)(OiEi)2(OiEi)2Ei1501401000.7142861301401000.7142861001101000.9090911201101000.909091 Total 3.25
Thus, the test statistic is χ2=3.25.
The fegrees of freedom is,
df=(2-1)(2-1)
=1
For the chi-square test of independence, the value of the test statistic is 3.25.

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