Question

The following data was collected from a random sample of wood, used as decking for high-end luxury boats. The wood was classified as strong or weak, or dense versus less dense.

	Strong	Weak	Total
Dense	15	5	20
Less Dense	10	20	40
Total	25	25	50

Perform the appropriate hypothesis test and compute the test statistic.

Answer 1

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H0: Two variables strong/weakness of wood and density of wood are independent.

Alternative hypothesis: Ha: Two variables strong/weakness of wood and density of wood are dependent.

We assume 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 = 2

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

α = 0.05

Critical value = 3.841459

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	Strong	Weak	Total
Dense	15	5	20
Less Dense	10	20	30
Total	25	25	50

Expected Frequencies
	Column variable
Row variable	Strong	Weak	Total
Dense	10	10	20
Less Dense	15	15	30
Total	25	25	50

Calculations
(O - E)
5	-5
-5	5

(O - E)^2/E
2.5	2.5
1.666667	1.666667

Test Statistic = Chi square = ∑[(O – E)^2/E] = 8.333333

χ2 statistic = 8.333333

P-value = 0.003892

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that two variables strong/weakness of wood and density of wood are dependent.