Question

2. State the Null and Alternate hypotheses. Calculate y and interpret the p-value. Candy Skittles Reeses Milk Duds Boys 30 17 Child Girls 9 20

Please show work! thanks!

Answer 1

Solution:

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

Null hypothesis: H₀: The type of candy and gender of child are independent.

Alternative hypothesis: H_a: The type of candy and gender of child are not independent.

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 = 3

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

α = 0.05

Critical value = 5.991465

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Candy
Child	Skittles	Reeses	Milk Duds	Total
Boys	30	17	3	50
Girls	9	20	1	30
Total	39	37	4	80

Expected Frequencies
	Candy
Child	Skittles	Reeses	Milk Duds	Total
Boys	24.375	23.125	2.5	50
Girls	14.625	13.875	1.5	30
Total	39	37	4	80

Calculations
(O - E)
5.625	-6.125	0.5
-5.625	6.125	-0.5

(O - E)^2/E
1.298077	1.622297	0.1
2.163462	2.703829	0.166667

Chi square = ∑[(O – E)^2/E] = 8.054331

P-value = 0.017825

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the type of candy and gender of child are independent.