Chi-square: a die is rolled 60 times: 8, 12, 9, 11, 6, 14; is it… Easy

The question

Question: a die is rolled 60 times, and the faces 1 to 6 come up 8, 12, 9, 11, 6, 14 times: at the 5% level, is the die fair?

1 Write the hypotheses

H₀ is the "nothing new" idea.

Here: the die is fair.

Hₐ is what we want to show.

Here: the die is not fair.

2 Expected counts

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

Each face: 60 times 1/6 is 10.

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

3 Each part

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

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

8 minus 10 is −2.

Squared, 4.

Over 10: 0.4.

12 minus 10 is 2.

Squared, 4.

Over 10: 0.4.

9 minus 10 is −1.

Squared, 1.

Over 10: 0.1.

11 minus 10 is 1.

Squared, 1.

Over 10: 0.1.

6 minus 10 is −4.

Squared, 16.

Over 10: 1.6.

14 minus 10 is 4.

Squared, 16.

Over 10: 1.6.

Careful!

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

4 Add them

0.4 plus 0.4 plus 0.1 plus 0.1 plus 1.6 plus 1.6 is 4.2.

5 The cut-off

df is the number of cells minus 1: 6 minus 1 is 5.

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

Row 5, column 0.05.

Row 5, column .05: the cut-off is 11.070.

6 Decide

The chi-square curve for df 5.

Past the cut-off, 11.070, is the reject part.

4.2 is not past the cut-off.

So we do not reject H zero.

7 Say it in words

In words: at the 5% level, there is not enough evidence that the die is not fair.

This does not prove H zero is true.

We just cannot rule it out.

8 Check

Let's check: the expected counts add up to 60, like the observed ones.

And the differences O minus E add up to 0.

So it is right!

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.

▶ Watch it solved

Chi-square goodness of fit: observed 60, 30, 10; expected 50%, 30%… Medium

The question

Question: we saw the counts 60, 30, 10, and expected 50%, 30%, 20%: at the 5% level, do the data fit?

1 Write the hypotheses

H₀ is the "nothing new" idea.

Here: the data fit the expected parts.

Hₐ is what we want to show.

Here: the data do not fit.

2 Expected counts

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

A: 100 times 0.5 is 50.

B: 100 times 0.3 is 30.

C: 100 times 0.2 is 20.

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

3 Each part

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

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

60 minus 50 is 10.

Squared, 100.

Over 50: 2.

30 minus 30 is 0.

Squared, 0.

Over 30: 0.

10 minus 20 is −10.

Squared, 100.

Over 20: 5.

Careful!

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

4 Add them

2 plus 0 plus 5 is 7.

5 The cut-off

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

7 is past the cut-off.

So we reject H zero.

7 Say it in words

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

8 Check

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

And the differences O minus E add up to 0.

So it is right!

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.

▶ Watch it solved

Chi-square goodness of fit: observed 90, 60, 50; expected 50%, 25%… Harder

The question

Question: we saw the counts 90, 60, 50, and expected 50%, 25%, 25%: at the 1% level, do the data fit?

1 Write the hypotheses

H₀ is the "nothing new" idea.

Here: the data fit the expected parts.

Hₐ is what we want to show.

Here: the data do not fit.

2 Expected counts

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

A: 200 times 0.5 is 100.

B: 200 times 0.25 is 50.

C: 200 times 0.25 is 50.

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

3 Each part

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

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

90 minus 100 is −10.

Squared, 100.

Over 100: 1.

60 minus 50 is 10.

Squared, 100.

Over 50: 2.

50 minus 50 is 0.

Squared, 0.

Over 50: 0.

Careful!

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

4 Add them

1 plus 2 plus 0 is 3.

5 The cut-off

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

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

Row 2, column 0.01.

Row 2, column .01: the cut-off is 9.210.

6 Decide

The chi-square curve for df 2.

Past the cut-off, 9.210, is the reject part.

3 is not past the cut-off.

So we do not reject H zero.

7 Say it in words

In words: at the 1% level, there is not enough evidence that the data do not fit.

This does not prove H zero is true.

We just cannot rule it out.

8 Check

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

And the differences O minus E add up to 0.

So it is right!

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.

▶ Watch it solved