Question

# Given the following Hypotheses: Null: An equal number of fruits is distributed per consumer box Alt:...

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.

 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

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