Question

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.

Answer 1

`p_v = P(\chi^2_(3,0.05) >8.180)=0.0424`

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.

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
`\alpha=0.05`

The statistic to check the hypothesis is given by:

`\sum_(i=1)^n ((O_i -E_i)^2)/(E_i)`

Now we just need to calculate the expected values with the following formula
`E_i = \% * total`

And the calculations are given by:

`E_(Democratic) =0.48*1200=576`

`E_(Republican) =0.38*1200=456`

`E_(Independent) =0.04*1200=48`

`E_(Undecided) =0.1*1200=120`

And now we can calculate the statistic:

`\chi^2 = ((540-576)^2)/(576)+((480-456)^2)/(456)+((40-48)^2)/(48)+((140-120)^2)/(120) =2.25+1.263158+1.33333+3.33333=8.180`

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

`df=(categories-1)=(4-1)=3`

And we can calculate the p value given by:

`p_v = P(\chi^2_(3,0.05) >8.180)=0.0424`

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.