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.
Problem 8 (15 marks) Five people have declared their intentions to run for a trustee seat...
Problem 2 (15 marks) Five people have declared their intentions to run for a trustee seat in the next local election. Some political analysts claim that there is no clear preference among the voters. Hence, it is reported that each of the candidate would obtain 20% of the vote on his name. A political poll is conducted during the campaign among 1,020 randomly sampled voters to determine if there is any clear preference among the voters. The responses are shown...
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Problem 2 (15 marks) Five people have declared their intentions to run for a trustee seat in the next local election. Some political analysts claim that there is no clear preference among the voters. Hence, it is reported that each of the candidate would obtain 20% of the vote on his name. A political poll is conducted during the campaign among 1,020 randomly sampled voters to determine if there...