Question

the following tabel contains the number of successes and failures for three categories of a variable. test whether the proportions are equal for each category at the a= .01 level of significance

Failures Successes Category 1 Category 2 Category 3 74 56 56 80 81 25

Answer 1

Solution:

Here, we have to use chi square test. The null and alternative hypotheses for this test are given as below:

H₀: p₁ = p₂ = p₃

H₁: At least one of the proportions is different from the others.

We are given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

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

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 1*2 = 2

α = 0.01

Critical value = 9.21034

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Category
Row variable	1	2	3	Total
Failures	74	56	56	186
Successes	80	81	25	186
Total	154	137	81	372

Expected Frequencies
	Category
Row variable	1	2	3	Total
Failures	77	68.5	40.5	186
Successes	77	68.5	40.5	186
Total	154	137	81	372

Calculations
(O - E)
-3	-12.5	15.5
3	12.5	-15.5

(O - E)^2/E
0.116883	2.281022	5.932099
0.116883	2.281022	5.932099

Test Statistic = Chi square = ∑[(O – E)^2/E] = 16.66001

χ² ₀ = 16.660

P-value = 0.000241

(By using Chi square table or excel)

The P-value is less than .01.

P-value < α = 0.01

So, we reject the null hypothesis

B. The P-value is less than α, so reject H₀. There is sufficient evidence that the proportions are different from each other.