Question

according to a government website, 42% of us citizens are democrats, 34% are republicans, and 24% are independents. a local municipality would like to know if the distribution of political party affiliation among its citizens differs from the nationwide percentages. a random sample of 500 citizens of the municipality is selected. in the sample, 200 were democrats, 187 were republicans, and 113 were independents. what is the value of the chi - square test statistic and p - value? find the chi - square table here. ( chi^{2}=2.48 ), p - value is between 0.10 and 0.15 ( chi^{2}=2.48 ), p - value is greater than 0.25 ( chi^{2}=2.58 ), p - value is between 0.10 and 0.15 ( chi^{2}=2.58 ), p - value is greater than 0.25

Explanation:

Step1: Calculate expected values

For Democrats: \(E_1 = 500\times0.42=210\)
For Republicans: \(E_2 = 500\times0.34 = 170\)
For Independents: \(E_3=500\times0.24 = 120\)

Step2: Calculate chi - square statistic

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Step3: Determine degrees of freedom

Degrees of freedom \(df=k - 1\), where \(k = 3\) (number of categories), so \(df=3-1 = 2\)

Step4: Find P - value

Using chi - square table with \(df = 2\), \(\chi^{2}=2.58\). The value \(2.58\) is less than \(\chi_{0.10,2}^{2}=4.605\) and greater than \(\chi_{0.25,2}^{2}=2.773\) (wait, no, actually for \(df = 2\), \(\chi_{0.25,2}^{2}=2.773\) and \(\chi_{0.10,2}^{2}=4.605\), but our \(\chi^{2}=2.58\) is less than \(\chi_{0.25,2}^{2}\). Wait, no, check the table again. For \(df=2\):
\(\chi_{0.25,2}^{2}=2.773\), \(\chi_{0.50,2}^{2}=1.386\). Since \(1.386<2.58<2.773\), the \(P -\)value is greater than \(0.25\)

Answer:

\(\chi^{2}=2.58\), \(P -\)value is greater than \(0.25\) (the fourth option: \(\chi^{2}=2.58\), \(P -\)value is greater than \(0.25\))