Example: Test at the 5% significance level whether the number of bedrooms in a home and the number of children in the home are statistically independent.

Number of Bedrooms	0 Children	1 Child	2 Children	3 or More Children	Total
2 or fewer	12	18	6	4	40
3	22	18	0	0	40
4 or more	2	16	19	3	40
Total	36	52	25	7	120

---

Step 1: State the null and alternate hypotheses, assuming independence as the null hypothesis:

$H_0$: Number of bedrooms and number of children are statistically independent

$H_1$: Number of bedrooms and number of children are not statistically independent

---

Step 2: Based on the assumption that the events are independent, calculate from the totals the expected frequencies.

$\frac {36}{120} \times \frac{40}{\cancel{120}} \times \cancel{120} = 12$

$\frac {36}{120} \quad \text {P(0 Children})$

$\frac {40}{120} \quad P(\le 2 \ Rooms)$

$120 \ (Total)$

---

Step 3: This method leads to large errors if any entries in the table of expectations are less than 5. If this is the case, columns should be combined to eliminate these values.

Ex	0	1	2+	3+
≤ 2	12	17.3	8.3	2.3
3	12	17.3	8.3	2.3
≥ 4	12	17.3	8.3	2.3

Since there are expectations less than $5$, we should combine the "$3+$" column with a neighboring column. Try to combine with a column that has similar characteristics.

Ex	0	1	2+
≤ 2	12	17.3	10.6
3	12	17.3	10.6
≥ 4	12	17.3	10.6

Now we have all entries greater than or equal to 5.

---

Step 4: Perform the same column combination to the table of observations.

Obs	0	1	2+
≤ 2	12	18	10
3	22	18	0
≥ 4	2	16	22

---

Step 5: Using the formula:

$$\text{Contributions} = \frac{(\text{Observations} - \text{Expectations})^2}{\text{Expected}}$$

For the contributions table:

Cont	0	1	2+
≤ 2	0	0.02567	0.04159
3	8.333	0.02567	10.667
≥ 4	8.333	0.10251	12.04384

$$\frac{(12 - 12)^2}{12} = 0$$$$\frac{(18 - 17.333)^2}{17.333} = 0.02567$$

---

Step 6: The test statistic is the sum of all those contributions. It is called:

$$\chi^2_{calc} = :highlight[39.571]$$

To get $\chi^2$ from the graphical calculator, follow these steps:

---

This shows us that we need to combine columns $3$ and $4$. Do this, then re-input the combined observations matrix.

---

Step 7**: Calculate the number of "degrees of freedom"** and compare your $\chi^2_{\text{calc}}$ to the $\chi^2_{\text{crit}}$ in the table.

Degrees of freedom are variables whose values are left to advance. If we can deduce the value of a variable, then it is not "free."

The number of degrees of freedom, represented by the Greek letter $\nu$ ("nu"), is:

$(\text{no of rows} - 1) \times (\text{no of columns} - 1)$

Here $\nu = 2 \times 2 = :highlight[4]$ (as our combined observations matrix was $3 \times 3$).

This is the acceptance level but sig level is the rejection level

---

Step 8: Conclude

$$:highlight[39.571 > 9.488]$$$$\Rightarrow \text {:success[Reject]} \ H_0$$

Sufficient evidence to suggest that the number of bedrooms and number of children are not independent.

Example: The following contingency table shows the results of $100$ people in a driving test along with their gender:

Result	Male	Female
Pass	34	24
Fail	24	18

Test at the 10% significance level if the outcome of the test is independent of the gender.

• Null Hypothesis ($H_0$): Result and gender are independent.
• Alternative Hypothesis ($H_1$): Result and gender are not independent. Since $r = 1$, we must perform a Yates' correction.

Observed Table:

Expected Table:

Note: Do not use the calculator's test statistic $\chi^2_{\text{calc}}$ as Yates' correction is not performed by the calculator.

Contributions:

$$\frac{(|34 - 33.64| - 0.5)^2}{33.64} = \frac{49}{84100}$$$$\frac{(|24 - 24.36| - 0.5)^2}{24.36} = \frac{7}{8300} \quad \times 2$$$$\frac{(|18 - 17.64| - 0.5)^2}{17.64} = \frac{1}{900}$$$$\chi^2_{\text{calc}} = \frac{49}{84100} + 2\frac{7}{8700} + \frac{1}{900} = \frac{25}{7359} \approx :highlight[3.303 \times 10^{-3}]$$$$:highlight[3.303 \times 10^{-3} < 2.706]$$

Conclusion:

• Do not reject $H_0$
• Insufficient evidence to suggest that gender and results are dependent.