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Problem 8 (15 marks) Five people have declared their intentions to run for a trustee seat in the next local election. Some po
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Answer #1

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H0: There are no any preferences among the voters.   

Alternative hypothesis: Ha: There are preferences among the voters.   

We are given level of significance = α = 0.05

We are given

Number of categories = N = 5

Degrees of freedom = df = N - 1 = 4

α = 0.05

Critical value = 9.48772904

(by using Chi square table or excel)

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Calculation tables for test statistic are given as below:

Candidate

O

E

(O - E)^2/E

A

209

204

0.12254902

B

269

204

20.7107843

C

229

204

3.06372549

D

159

204

9.92647059

E

154

204

12.254902

Total

1020

1020

46.0784314

Test Statistic = Chi square = ∑[(O – E)^2/E] = 46.0784314

χ2 = 46.0784314

Critical value = 9.48772904

Test statistic value is greater than critical value.

So, we reject the null hypothesis

There is sufficient evidence to conclude that there is clear preference among the voters.

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