During a recent trip to a modern art museum,

I was absolutely delighted to find that the collection contains

Robert Rauschenberg’s “White Paintings” from 1951.

To my dismay, however, I found that

one of the pictures has been hung upside down.

The security guard that yelled “You’re too close!”

couldn’t have cared less, saying “You got a screw loose buddy —

orientation means nothing with respect to the abstract.

Now get a move-on before I taze you.”

So I decided to passive-aggressively conduct a poll to prove that

one of the paintings is incorrectly hung.

Over the course of a few hours, I asked 50 people to pick which one of three paintings is hung incorrectly,

and whether it is top-left (TL), top-right (TR), or top-down (TD).

My data is shown below.

Painting | TD | TL | TR | TOTAL |

A | 13 | 5 | 6 | 24 |

B | 5 | 0 | 5 | 10 |

C | 4 | 7 | 5 | 16 |

50 |

The guard will only judge something statistically significant if \(p < 0.05\).

Says so on his badge.

So I set to work by testing the null hypothesis that people are choosing one of the paintings and orientations at random.

For this, I can use the \(\chi^2\)-test, and assume that the expected frequency of each of the 9 pairs of painting and orientation will be \(50/9\).

The \(\chi^2\)-distributed test statistic has 8 degrees of freedom, and has a value

$$

W_8 := \mathop{\sum_{p \in \{A, B, C\}}}_{o \in \{TD, TL, TR\}} \frac{(N_{p,o} – 50/9)^2}{(50/9)} = 16.6.

$$

And so, thumbing my nose at the guard, I show him

$$

p = P\{W_8 \ge 16.6\} < 0.035 < 0.05

$$

"And so I reject your naive null hypothesis.

Now bring me the curator; we have work to do."

“No, you’re all wrong, see?” he says, pausing in between each word to deliver brief but crippling shocks.

“Let’s test the hypothesis that people are picking a painting at random, regardless of the orientation.

This \(\chi^2\)-distributed test statistic has 2 degrees of freedom

$$

W_2 := \sum_{p \in \{A, B, C\}} \frac{( \sum_{o \in \{TD, TL, TR\}} N_{p,o} – 50/3)^2}{(50/3)} = 5.92

$$

and thus

$$

p = P\{W_2 \ge 5.92\} > 0.051.

$$

So we cannot reject that null hypothesis.” ZAP.

“And now let’s test whether those that selected painting A

are choosing the orientation at random.

This \(\chi^2\)-distributed test statistic has 2 degrees of freedom

$$

A_2 := \sum_{o \in \{TD, TL, TR\}} \frac{( N_{A,o} – 24/3)^2}{(24/3)} = 4.75

$$

and thus

$$

p = P\{A_2 \ge 4.75\} > 0.09.

$$

So we cannot reject that null hypothesis.” ZAP.

“So you see, sir,

your statistics are stupid. Scram!”

Dazed and confused, I walk away defeated.

What went wrong?

Below is the MATLAB to reproduce the tests:

N = [13 5 6; 5 0 5; 4 7 5]; W1 = (N-50/9).^2./(50/9); pop = 1-chi2cdf(sum(W1(:)),length(N(:))-1) W2 = (sum(N')'-50/3).^2/(50/3); pp = 1-chi2cdf(sum(W2(:)),2) W3 = (N(1,:)- sum(N(1,:))/3).^2./(sum(N(1,:))/3); po = 1-chi2cdf(sum(W3(:)),2)