Question

7. A recent Iowa straw poll for the Democratic Primary found the following number of supporters for these candidates (Observed Frequencies).

Elizabeth Warren	Bernie Sanders	Joe Biden	Pete Buttigieg
20	25	10	25

a) Specify the null and alternative hypotheses for a chi-square Goodness of Fit test of candidate preference.

b) Fill in the table below with the expected frequencies.

Elizabeth Warren	Bernie Sanders	Joe Biden	Pete Buttigieg

c) Use the observed and expected frequencies to calculate a chi-square Goodness of Fit test to decide if people in Iowa have a preference for any of the candidates, using alpha = .05. Report the critical value and your decision.

Answer 1

Solution:

Given:

A recent Iowa straw poll for the Democratic Primary found the following number of supporters for these candidates (Observed Frequencies).

Elizabeth Warren	Bernie Sanders	Joe Biden	Pete Buttigieg	Total : N
20	25	10	25	80

Part a) Specify the null and alternative hypotheses for a chi-square Goodness of Fit test of candidate preference

H0: People in Iowa do not have preference for any of the candidates, That is all candidates are equal preference.

Vs

H1: People in Iowa have a preference for any of the candidates.

Part b) Find expected frequencies.

Ei = Expected frequencies = N / k = 80 / 4 = 20

Thus

Elizabeth Warren	Bernie Sanders	Joe Biden	Pete Buttigieg
20	20	20	20

Part c) Use the observed and expected frequencies to calculate a chi-square Goodness of Fit test to decide if people in Iowa have a preference for any of the candidates, using alpha = .05. Report the critical value and your decision.

Find test statistic:

Chi square test statistic for goodness of fit

Where

O_i = Observed Counts

E_i =Expected Counts

Thus we need to make following table

	Oi : Observed Frequencies	Ei : Expected Frequencies	Oi^2/Ei
Elizabeth Warren	20	20	20
Bernie Sanders	25	20	31.25
Joe Biden	10	20	5
Pete Buttigieg	25	20	31.25
	N = 80

Thus

Find Chi-square critical value:

df = k - 1 = 4 - 1 = 3

Level of significance = 0.05

Chi-square critical value = 7.815

Decision Rule:

Reject null hypothesis H0, if Chi square test statistic > Chi-square critical value = 7.815 , otherwise we fail to reject H0.

Since Chi square test statistic = < Chi-square critical value = 7.815, we fail to reject null hypothesis H0.

Thus there is not sufficient evidence to conclude that: people in Iowa have a preference for any of the candidates.

Thus: People in Iowa do not have preference for any of the candidates, That is all candidates are equal preference.