Before the presidential debates, it was expected that the percentages of registered voters in favor of various candidates to be as follows. Percent Democrats 48% Republicans 38% Independent 04% Undecided 10% After the presidential debates, a random sample of 1200 voters showed that 540 favored the Democratic candidate; 480 were in favor of the Republican candidate; 40 were in favor of the Independent candidate, and 140 were undecided. At 95% confidence, test to see if the proportion of voters has changed.

Respuesta :

Answer:

[tex]p_v = P(\chi^2_{3,0.05} >8.180)=0.0424[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(8.180,3,TRUE)"

Since the p value is lower than the significance level we can to reject the null hypothesis at 5% of significance, and we can conclude that we have significant differences in the proportions assumed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

The observed values are given by:

Democratic 540

Republican  480

Independent 40

Undecided   140

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the proportions for the political party

H1: There is a difference in the proportions for the political party

The level os significance assumed for this case is [tex]\alpha=0.05[/tex]

The statistic to check the hypothesis is given by:

[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

Now we just need to calculate the expected values with the following formula [tex]E_i = \% * total[/tex]

And the calculations are given by:

[tex]E_{Democratic} =0.48*1200=576[/tex]

[tex]E_{Republican} =0.38*1200=456[/tex]

[tex]E_{Independent} =0.04*1200=48[/tex]

[tex]E_{Undecided} =0.1*1200=120[/tex]

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(540-576)^2}{576}+\frac{(480-456)^2}{456}+\frac{(40-48)^2}{48}+\frac{(140-120)^2}{120} =2.25+1.263158+1.33333+3.33333=8.180[/tex]

Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(categories-1)=(4-1)=3[/tex]

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{3,0.05} >8.180)=0.0424[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(8.180,3,TRUE)"

Since the p value is lower than the significance level we can to reject the null hypothesis at 5% of significance, and we can conclude that we have significant differences in the proportions assumed.