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Chi-square goodness of fit: observed 102, 48, 50; expected 60%, 20%…

Step by step, with the reason for every step · Chi-square · University Statistics

Step by step

The question

Question: we saw the counts 102, 48, 50, and expected 60%, 20%, 20%: at the 5% level, do the data fit?

observed 102, 48, 50; expected 60%, 20%, 20%; α = 0.05

1 Write the hypotheses

H₀ is the "nothing new" idea.

Here: the data fit the expected parts.

H0 : the data fit the expected parts

Hₐ is what we want to show.

Here: the data do not fit.

Ha : the data do not fit

2 Expected counts

If H zero is true, each cell expects n times its part.

E = n × p (if H0 is true)

A: 200 times 0.6 is 120.

E(A) = 200 × 0.6 = 120

B: 200 times 0.2 is 40.

E(B) = 200 × 0.2 = 40

C: 200 times 0.2 is 40.

E(C) = 200 × 0.2 = 40

Every expected count is at least 5, so the test works.

all E at least 5: the test works

3 Each part

Chi-square adds up, for each cell, the observed minus the expected, squared, over the expected.

χ2 = add up all (O − E)2E

A table: each cell, what we saw, O, and what we expected, E.

102 minus 120 is −18.

Squared, 324.

Over 120: 2.7.

48 minus 40 is 8.

Squared, 64.

Over 40: 1.6.

50 minus 40 is 10.

Squared, 100.

Over 40: 2.5.

Careful!

We always divide by the expected count, not the observed one.

Careful: divide by E (expected), not by O

4 Add them

2.7 plus 1.6 plus 2.5 is 6.8.

χ2 = 2.7 + 1.6 + 2.5 = 6.8

5 The cut-off

df is the number of cells minus 1: 3 minus 1 is 2.

df = k − 1 = 3 − 1 = 2

Chi-square is always a right-tail test: big means far from what we expected.

Row 2, column 0.05.

Row 2, column .05: the cut-off is 5.991.

6 Decide

The chi-square curve for df 2.

Past the cut-off, 5.991, is the reject part.

6.8 is past the cut-off.

So we reject H zero.

= χ2 = 6.8 > 5.991 → reject H0

7 Say it in words

At the 5% level, there is enough evidence that

In words: at the 5% level, there is enough evidence that the data do not fit.

the data do not fit.

8 Check

Let's check: the expected counts add up to 200, like the observed ones.

E: 120 + 40 + 40 = 200 = n

And the differences O minus E add up to 0.

So it is right!

O − E: (−18) + 8 + 10 = 0

Let's look back.

First: write the hypotheses.

Then: expected counts.

Then: each part.

Then: add them.

Then: the cut-off.

Then: decide.

Then: say it in words.

Last: check.

Answer
= χ2 = 6.8 > 5.991 → reject H0

Your turn

Try this one yourself, then check:

observed 102, 50, 48; expected 60%, 20%, 20%; α = 0.05
Show the answer
reject H0

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