Question

Given the following Hypotheses:

Null: An equal number of fruits is distributed per consumer box

Alt: An unequal number of fruits is distributed in each box

Coconut	21
Raspberry	34
Banana	33
Avocado	0
Orange	30
Starfruit	14

Total: 132

Mean: 22

SD: 2.6

Find the degrees of freedom and chi Square analysis. What is the value of chi-square needed for statistical significance at a 0.05 level for the data?

I am struggling to find the correct chi square for this, I am unsure if I am using the right expected value in the chi-square test.

Answer 1

Category	Observed Frequency (O)	Proportion, p	Expected Frequency (E)	(O-E)²/E
Coconut	21	1/6	132 * 1/6 = 22	(21 - 22)²/22 = 0.0455
Raspberry	34	1/6	132 * 1/6 = 22	(34 - 22)²/22 = 6.5455
Banana	33	1/6	132 * 1/6 = 22	(33 - 22)²/22 = 5.5
Avocado	0	1/6	132 * 1/6 = 22	(0 - 22)²/22 = 22
Orange	30	1/6	132 * 1/6 = 22	(30 - 22)²/22 = 2.9091
Starfruit	14	1/6	132 * 1/6 = 22	(14 - 22)²/22 = 2.9091
Total	132	0.00	132	39.9091

Test statistic:

χ² = ∑ ((O-E)²/E) = 39.909

df = n-1 = 5

Critical value:

χ²α = CHISQ.INV.RT(0.05, 5) = 11.070

Decision:

Reject the null hypothesis