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Chi-square goodness of fit: observed 30, 50, 20; expected 25%, 50%…

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

Step by step

The question

Question: we saw the counts 30, 50, 20, and expected 25%, 50%, 25%: at the 5% level, do the data fit?

observed 30, 50, 20; expected 25%, 50%, 25%; 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: 100 times 0.25 is 25.

E(A) = 100 × 0.25 = 25

B: 100 times 0.5 is 50.

E(B) = 100 × 0.5 = 50

C: 100 times 0.25 is 25.

E(C) = 100 × 0.25 = 25

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.

30 minus 25 is 5.

Squared, 25.

Over 25: 1.

50 minus 50 is 0.

Squared, 0.

Over 50: 0.

20 minus 25 is −5.

Squared, 25.

Over 25: 1.

Careful!

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

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

4 Add them

1 plus 0 plus 1 is 2.

χ2 = 1 + 0 + 1 = 2

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.

2 is not past the cut-off.

So we do not reject H zero.

= χ2 = 2 < 5.991 → do not reject H0

7 Say it in words

At the 5% level, there is not enough evidence that

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

the data do not fit.

This does not prove H zero is true.

We just cannot rule it out.

not rejecting H0 does not prove H0

8 Check

Let's check: the expected counts add up to 100, like the observed ones.

E: 25 + 50 + 25 = 100 = n

And the differences O minus E add up to 0.

So it is right!

O − E: 5 + 0 + (−5) = 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 = 2 < 5.991 → do not reject H0

Your turn

Try this one yourself, then check:

observed 100, 40, 60; expected 50%, 25%, 25%; α = 0.05
Show the answer
do not reject H0

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