100 candidates are classified by whether they were coached and whether they passed an examination, as shown. Compute the expected frequencies under independence and hence the chi-square statistic.

| | Passed | Failed | Total |
|---|---|---|---|
| **Coached** | 40 | 10 | 50 |
| *

Question

100 candidates are classified by whether they were coached and whether they passed an examination, as shown. Compute the expected frequencies under independence and hence the chi-square statistic.

| | Passed | Failed | Total |
|---|---|---|---|
| **Coached** | 40 | 10 | 50 |
| **Not coached** | 30 | 20 | 50 |
| **Total** | 70 | 30 | 100 |

Accepted Answer

**Step 1 — Expected frequencies, $E_{ij} = R_i C_j / N$, $N = 100$.** Row totals both 50; columns 70 (Passed), 30 (Failed).
$$E(	ext{Coached, Passed}) = \dfrac{50 	imes 70}{100} = 35, \qquad E(	ext{Coached, Failed}) = \dfrac{50 	imes 30}{100} = 15$$
$$E(	ext{Not coached, Passed}) = \dfrac{50 	imes 70}{100} = 35, \qquad E(	ext{Not coached, Failed}) = \dfrac{50 	imes 30}{100} = 15$$

**Step 2 — Tabulate $(O-E)^2 / E$.**

| Cell | $O$ | $E$ | $O-E$ | $(O-E)^2$ | $(O-E)^2/E$ |
|---|---|---|---|---|---|
| Coached, Passed | 40 | 35 | 5 | 25 | $25/35 \approx 0.714$ |
| Coached, Failed | 10 | 15 | −5 | 25 | $25/15 \approx 1.667$ |
| Not coached, Passed | 30 | 35 | −5 | 25 | $25/35 \approx 0.714$ |
| Not coached, Failed | 20 | 15 | 5 | 25 | $25/15 \approx 1.667$ |
| **Total** | | | | | **$\approx 4.762$** |

**Step 3 — Sum.**
$$\chi^2 = \dfrac{25}{35} + \dfrac{25}{15} + \dfrac{25}{35} + \dfrac{25}{15} = \dfrac{50}{35} + \dfrac{50}{15} = 1.4286 + 3.3333 = 4.7619 \approx 4.76$$ …

Cell	$O$	$E$	$O-E$	$(O-E)^2$	$(O-E)^2/E$
Coached, Passed	40	35	5	25	$25/35 \approx 0.714$
Coached, Failed	10	15	−5	25	$25/15 \approx 1.667$
Not coached, Passed	30	35	−5	25	$25/35 \approx 0.714$
Not coached, Failed	20	15	5	25	$25/15 \approx 1.667$
Total					$\approx 4.762$

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