Math Core

In Unit 6 you tested a claim about one proportion. But many claims involve a whole distribution across several categories at once: "Our bags are 30% cherry, 25% lime, 25% orange and 20% grape," or "Library visits are spread evenly across the school week." A chi-square goodness-of-fit test checks whether sample counts across several categories are consistent with a claimed distribution.

Observed versus expected counts

Suppose a school librarian believes that checkouts are uniformly distributed across the five weekdays. She takes a random sample of $170$ checkouts from the past year and records the day of each.

Day	Mon	Tue	Wed	Thu	Fri	Total
Observed	$40$	$28$	$30$	$25$	$47$	$170$

If the claim is true, each day should get $20\%$ of the checkouts. The expected count for each day is

\text{expected} = n \times p_i = 170 \times 0.20 = 34.

The observed counts will never match the expected counts exactly, because of sampling variability. The real question is the same one you have asked all year: are the differences bigger than chance alone would reasonably produce?

To answer that, you need one number that measures the total distance between the observed and expected counts.

Definition

Chi-square statistic

For a table of counts,

\chi^2 = \sum \frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}},

where the sum is over all categories. Each term $\dfrac{(O - E)^2}{E}$ is called that category's contribution (or component) to $\chi^2$ .

Squaring makes every difference count positively, and dividing by the expected count scales each difference: being off by $6$ matters much more when you expected $10$ than when you expected $1{,}000$ . A value of $\chi^2$ near $0$ means the data fit the claim well. A large value means the data fit poorly, which is evidence against the claim.

Worked example: Computing the statistic for the library data

Find $\chi^2$ for the librarian's sample.

Every expected count is $34$ . Compute each contribution:

Day	$O$	$E$	$(O - E)^2 / E$
Mon	$40$	$34$	$36/34 \approx 1.059$
Tue	$28$	$34$	$36/34 \approx 1.059$
Wed	$30$	$34$	$16/34 \approx 0.471$
Thu	$25$	$34$	$81/34 \approx 2.382$
Fri	$47$	$34$	$169/34 \approx 4.971$

Add the contributions: $\chi^2 \approx 1.059 + 1.059 + 0.471 + 2.382 + 4.971 \approx 9.94$ . (Adding the unrounded fractions, $\dfrac{338}{34}$ , gives $9.941$ .)

Friday contributes the most by far: many more checkouts happened on Fridays than the uniform claim predicts.

The chi-square distributions

When the null hypothesis is true, the sampling distribution of the $\chi^2$ statistic is approximately a chi-square distribution. There is a different chi-square distribution for each number of degrees of freedom. For a goodness-of-fit test,

\text{df} = (\text{number of categories}) - 1.

Why minus one? Once you know $n$ and all but one of the counts, the last count is forced.

Chi-square distributions have three features worth remembering:

They take only values $\chi^2 \ge 0$ and are skewed right.
The mean of a chi-square distribution equals its df.
As df increases, the peak moves right and the curve becomes more symmetric.

Chi-square density curves with df = 3 (highest peak), df = 5, and df = 8 (lowest, farthest right). Each is skewed right, and the center moves right as df grows.Open in grapher →

Because only large values of $\chi^2$ count as evidence against $H_0$ , the P-value is always the area in the right tail: $P\text{-value} = P(\chi^2_{\text{df}} \ge \text{observed } \chi^2)$ .

Finding P-values: technology and the table

On a calculator, $\chi^2\text{cdf}(\text{lower}=\chi^2, \text{upper}=\infty, \text{df})$ gives the P-value directly (many calculators also run the whole test with $\chi^2\text{GOF-Test}$ ). On the AP exam you may also use a table of critical values. Here is an excerpt. Each entry is the value with the given right-tail area.

df	$0.10$	$0.05$	$0.025$	$0.01$	$0.005$	$0.001$
$1$	$2.706$	$3.841$	$5.024$	$6.635$	$7.879$	$10.828$
$2$	$4.605$	$5.991$	$7.378$	$9.210$	$10.597$	$13.816$
$3$	$6.251$	$7.815$	$9.348$	$11.345$	$12.838$	$16.266$
$4$	$7.779$	$9.488$	$11.143$	$13.277$	$14.860$	$18.467$
$5$	$9.236$	$11.070$	$12.833$	$15.086$	$16.750$	$20.515$
$6$	$10.645$	$12.592$	$14.449$	$16.812$	$18.548$	$22.458$

For the library data, df $= 5 - 1 = 4$ and $\chi^2 = 9.941$ . In the df $= 4$ row, $9.941$ falls between $9.488$ (tail area $0.05$ ) and $11.143$ (tail area $0.025$ ). So $0.025 < P\text{-value} < 0.05$ . Technology gives $P \approx 0.041$ .

The chi-square distribution with df = 3. The shaded right tail beyond 7.815 has area 0.05.Open in grapher →

Conditions and the full test

Like every significance test, a goodness-of-fit test needs conditions before the chi-square distribution is a trustworthy model.

Chi-square goodness-of-fit test

Hypotheses. $H_0$ : the claimed distribution is correct (list the proportions). $H_a$ : at least one of the claimed proportions is incorrect.

Conditions.

Random: the data come from a random sample (or a randomized experiment).
10%: when sampling without replacement, $n \le 10\%$ of the population.
Large counts: all expected counts are at least $5$ .

Statistic. $\chi^2 = \sum \dfrac{(O - E)^2}{E}$ with df $=$ number of categories $- 1$ .

P-value. The right-tail area under the chi-square curve.

Common mistake

The Large Counts condition is about expected counts, not observed counts. An observed count of $3$ is fine as long as the matching expected count is at least $5$ . Also, always use counts, never percents or proportions, in the $\chi^2$ formula. Plugging in percents changes the value of the statistic and gives a wrong P-value.

Worked example: A full four-step test: fruit chew flavors

A candy company claims its fruit chews are $30\%$ cherry, $25\%$ lime, $25\%$ orange and $20\%$ grape. A student buys a random sample of $200$ chews from many stores and counts $50$ cherry, $58$ lime, $42$ orange and $50$ grape. Is there convincing evidence at $\alpha = 0.05$ that the company's claim is wrong?

State. $H_0$ : $p_{\text{cherry}} = 0.30$ , $p_{\text{lime}} = 0.25$ , $p_{\text{orange}} = 0.25$ , $p_{\text{grape}} = 0.20$ . $H_a$ : at least one of these proportions is wrong.

Plan. Chi-square goodness-of-fit test. Random: random sample of chews. $10\%$ : $200$ is less than $10\%$ of all the company's chews. Large counts: the expected counts are $200(0.30) = 60$ , $200(0.25) = 50$ , $50$ , and $200(0.20) = 40$ , all at least $5$ .

Do.

\chi^2 = \frac{(50-60)^2}{60} + \frac{(58-50)^2}{50} + \frac{(42-50)^2}{50} + \frac{(50-40)^2}{40} = 1.667 + 1.28 + 1.28 + 2.5 \approx 6.727

with df $= 4 - 1 = 3$ . In the table, $6.727$ is between $6.251$ and $7.815$ , so $0.05 < P < 0.10$ . Technology: $P \approx 0.081$ .

Conclude. Because $0.081 > 0.05$ , fail to reject $H_0$ . There is not convincing evidence that the company's claimed flavor distribution is wrong.

Notice what the conclusion does not say. Failing to reject $H_0$ does not prove the claim is true; the data are simply consistent with it.

Worked example: Following up a significant result

For the library data, $\chi^2 \approx 9.941$ , df $= 4$ , and $P \approx 0.041$ . (Conditions: the sample is random, $170$ is less than $10\%$ of a year's checkouts, and all expected counts are $34$ .) What should the librarian conclude at $\alpha = 0.05$ , and where is the claim failing?

Since $0.041 < 0.05$ , reject $H_0$ . There is convincing evidence that checkouts are not uniformly distributed across the weekdays.

To see why, look at the contributions. Friday's contribution ( $4.971$ ) is half of the total, and Friday had more checkouts than expected ( $47$ versus $34$ ). Thursday is next ( $2.382$ ), with fewer checkouts than expected ( $25$ versus $34$ ). The biggest departure from the claim is a surplus of Friday checkouts.

Tip

A quick sanity check: the expected counts must add up to the same total $n$ as the observed counts. If they don't, you made an arithmetic slip or used proportions that don't sum to $1$ .

Practice

Practice 1

A textbook claims that blood types in a region are $42\%$ type O, $40\%$ type A, $11\%$ type B and $7\%$ type AB. In a random sample of $200$ blood donors, what is the expected count of type B donors if the claim is true?