Question

A local TV station claims that 60% of people support candidates A, 30% support candidate B, and 10% support candidate C. A survey of 500 registered voters is taken. The accompanying table indicates how they are likely to vote.

Candidate A	Candidate B	Candidate C
350	125	25

1. Specify the competing hypotheses to test whether the TV station's claim can be rejected by the data.
2. Test the hypothesis a the 1% significance level.
3. Conclusion? Your opinion supported the above questions

Answer 1

Solution:

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

Null hypothesis: H₀: A local TV stations claim is correct.

Alternative hypothesis: H_a: A local TV stations claim is not correct.

We are given level of significance = α = 0.01

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

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

We are given

N = 3

Degrees of freedom = df = N – 1 = 3 – 1 = 2

α = 0.01

Critical value = 9.210340372

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Candidate	P	O	E	(O - E)^2/E
A	0.6	350	300	8.333333333
B	0.3	125	150	4.166666667
C	0.1	25	50	12.5
Total	1	500	500	25

Chi square = ∑[(O – E)^2/E] = 25

P-value = 0.0000

(By using Chi square table or excel)

P-value < α = 0.01

So, we reject the null hypothesis

There is not sufficient evidence to conclude that a local TV stations claim is correct.