Using a chi-square goodness of fit test real world example, how can observed category counts be tested against a claimed distribution using \( \chi^2=\sum (O-E)^2/E \) and a p-value decision?

A chi-square goodness of fit test compares observed counts to expected counts from the claimed proportions, computes \( \chi^2=\sum (O-E)^2/E \) with \(df=k-1\), and rejects the claimed distribution when the p-value is small.

Chi-Square Goodness of Fit Test: Real World Example With Full Calculation

Accepted answer Answer included

The chi-square goodness of fit test real world example below shows how to test whether observed categorical data follow a claimed distribution. The method compares observed counts \(O_i\) to expected counts \(E_i\) and uses the chi-square distribution to quantify how unusual the discrepancies are under the null hypothesis.

Real-world scenario

A coffee shop expects the following long-run order proportions during morning hours: Espresso 25%, Latte 30%, Cappuccino 25%, Tea 20%. A random sample of \(n=200\) morning orders yields the observed counts: Espresso 40, Latte 70, Cappuccino 60, Tea 30. Determine whether the observed distribution matches the claimed proportions at \(\alpha=0.05\).

Step 1: State hypotheses

Let \(p_1,p_2,p_3,p_4\) denote the true category probabilities (Espresso, Latte, Cappuccino, Tea).

\[ H_0:\ (p_1,p_2,p_3,p_4)=(0.25,0.30,0.25,0.20) \qquad\text{vs}\qquad H_A:\ \text{at least one } p_i \text{ differs.} \]

Step 2: Compute expected counts

For each category, \(E_i = n \cdot p_i\).

\[ E_{\text{Espresso}} = 200 \cdot 0.25 = 50,\quad E_{\text{Latte}} = 200 \cdot 0.30 = 60,\quad E_{\text{Cappuccino}} = 200 \cdot 0.25 = 50,\quad E_{\text{Tea}} = 200 \cdot 0.20 = 40 \]

Conditions check

Independence: the sample should be random and each order counted once.
Expected counts: each \(E_i\) should be at least 5 (here: 50, 60, 50, 40).
Categories: mutually exclusive and collectively exhaustive.

Step 3: Compute the chi-square statistic

The test statistic for a goodness-of-fit test is:

\[ \chi^2=\sum_{i=1}^{k}\frac{(O_i-E_i)^2}{E_i} \]

Category	Observed \(O_i\)	Expected \(E_i\)	\((O_i-E_i)\)	\(\dfrac{(O_i-E_i)^2}{E_i}\)
Espresso	40	50	\(-10\)	\(\dfrac{(-10)^2}{50}=\dfrac{100}{50}=2.0000\)
Latte	70	60	\(10\)	\(\dfrac{(10)^2}{60}=\dfrac{100}{60}=1.6667\)
Cappuccino	60	50	\(10\)	\(\dfrac{(10)^2}{50}=\dfrac{100}{50}=2.0000\)
Tea	30	40	\(-10\)	\(\dfrac{(-10)^2}{40}=\dfrac{100}{40}=2.5000\)

\[ \chi^2 = 2.0000 + 1.6667 + 2.0000 + 2.5000 = 8.1667 \]

Step 4: Degrees of freedom

For a chi-square goodness-of-fit test with \(k\) categories and no parameters estimated from the data:

\[ df = k - 1 \]

Here \(k=4\), so \(df=3\).

Step 5: p-value (or critical value) and conclusion

The p-value is the right-tail probability:

\[ p = P\!\left(\chi^2_{(3)} \ge 8.1667\right) \approx 0.0427 \]

Since \(p \approx 0.0427 \le 0.05\), reject \(H_0\). The observed order pattern is inconsistent with the claimed distribution at the 5% significance level.

(Equivalently, the 0.05 critical value for \(df=3\) is approximately \(7.815\), and \(8.1667 > 7.815\).)

Step 6: Practical magnitude (effect size)

A common effect size for goodness-of-fit is Cramer's \(V\):

\[ V=\sqrt{\frac{\chi^2}{n \cdot (k-1)}} = \sqrt{\frac{8.1667}{200 \cdot 3}} = \sqrt{\frac{8.1667}{600}} \approx \sqrt{0.013611} \approx 0.1166 \]

This indicates a small-to-moderate departure from the expected proportions in this sample (interpretation depends on context and domain standards).

Visualization: observed vs expected counts

The largest discrepancies are visible where the observed bar diverges from the expected outline, which drives the statistic \( \chi^2 = 8.1667 \) for \(df=3\).

Common pitfalls (quick checks)

Using proportions instead of counts: the chi-square formula requires counts \(O_i\) and \(E_i\), not percentages.
Small expected counts: if some \(E_i\) are below 5, combine categories when scientifically reasonable.
Wrong test type: goodness-of-fit uses one categorical variable; independence/homogeneity use a two-way table.
Interpretation: rejecting \(H_0\) indicates mismatch with the claimed distribution, not which specific causes produced the mismatch.

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