Chi-Square Test worksheet

1. p-value: 1.186789e-8

1. Add up each row and each column, since the expected counts are built from these totals. Row totals 33, 98; column totals 60, 71; grand total N

   $N$$=131$

2. If the two things weren't related, each cell would expect its row total times its column total, divided by the grand total N

   $\text{E = (row total × column total) ÷ N}$

3. For example, the expected count (E) for the top-left cell

   $\mathrm{E₁₁}$$=\frac{33\times 60}{131}$$=15.114504$

4. For every cell, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(1-15.114504\right)}^2}{15.114504}+\frac{{\left(32-17.885496\right)}^2}{17.885496}+\frac{{\left(59-44.885496\right)}^2}{44.885496}+\frac{{\left(39-53.114504\right)}^2}{53.114504}$$=32.508393$

5. Count the degrees of freedom (df), how many cells are free to vary once the totals are fixed: (rows − 1) × (columns − 1)

   $\mathrm{df}$$=\left(2-1\right)\times \left(2-1\right)$$=1$

6. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 1 df)

   $p$$=\text{P(χ² >}32.508393\text{) =}$$=0.0000000119$

7. p is at or below α, so the two variables appear to be linked (not independent)

   $\text{Significant: reject H₀ (p ≤ 0.05)}$

8. Measure how strong the link is, from 0 to 1, with Cramér's V = √(χ² ÷ (N × (min(rows, columns) − 1)))

   $V$$=\sqrt{\frac{32.508393}{\mathrm{N(k\; -\; 1)}}}$$=0.498152$

2. p-value: 0.722978

1. Add up each row and each column, since the expected counts are built from these totals. Row totals 49, 26; column totals 28, 47; grand total N

   $N$$=75$

2. If the two things weren't related, each cell would expect its row total times its column total, divided by the grand total N

   $\text{E = (row total × column total) ÷ N}$

3. For example, the expected count (E) for the top-left cell

   $\mathrm{E₁₁}$$=\frac{49\times 28}{75}$$=18.293333$

4. For every cell, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(19-18.293333\right)}^2}{18.293333}+\frac{{\left(30-30.706667\right)}^2}{30.706667}+\frac{{\left(9-9.706667\right)}^2}{9.706667}+\frac{{\left(17-16.293333\right)}^2}{16.293333}$$=0.125657$

5. Count the degrees of freedom (df), how many cells are free to vary once the totals are fixed: (rows − 1) × (columns − 1)

   $\mathrm{df}$$=\left(2-1\right)\times \left(2-1\right)$$=1$

6. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 1 df)

   $p$$=\text{P(χ² >}0.125657\text{) =}$$=0.7229779021$

7. p is above α, so there's no significant evidence of a link

   $\text{Not significant: fail to reject H₀ (p > 0.04)}$

8. Measure how strong the link is, from 0 to 1, with Cramér's V = √(χ² ÷ (N × (min(rows, columns) − 1)))

   $V$$=\sqrt{\frac{0.125657}{\mathrm{N(k\; -\; 1)}}}$$=0.040932$

3. p-value: 0.002229

1. Scale the expected ratios to the total count N, so expected and observed add up the same: E = N × (ratio ÷ sum of ratios)

   $N$$=467$

2. The expected count (E) for each category

   $\text{262.6875, 87.5625, 87.5625, 29.1875}$

3. For every category, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(225-262.6875\right)}^2}{262.6875}+\frac{{\left(100-87.5625\right)}^2}{87.5625}+\frac{{\left(100-87.5625\right)}^2}{87.5625}+\frac{{\left(42-29.1875\right)}^2}{29.1875}$$=14.564597$

4. Count the degrees of freedom (df): the number of categories − 1, since the last one is fixed by the total

   $\mathrm{df}$$=4-1$$=3$

5. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 3 df)

   $p$$=\text{P(χ² >}14.564597\text{) =}$$=0.0022291973$

6. p is at or below α, so the counts differ significantly from the expected pattern

   $\text{Significant: reject H₀ (p ≤ 0.04)}$

4. p-value: 9.466972e-10

1. Scale the expected ratios to the total count N, so expected and observed add up the same: E = N × (ratio ÷ sum of ratios)

   $N$$=475$

2. The expected count (E) for each category

   $\text{267.1875, 89.0625, 89.0625, 29.6875}$

3. For every category, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(200-267.1875\right)}^2}{267.1875}+\frac{{\left(130-89.0625\right)}^2}{89.0625}+\frac{{\left(100-89.0625\right)}^2}{89.0625}+\frac{{\left(45-29.6875\right)}^2}{29.6875}$$=44.953216$

4. Count the degrees of freedom (df): the number of categories − 1, since the last one is fixed by the total

   $\mathrm{df}$$=4-1$$=3$

5. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 3 df)

   $p$$=\text{P(χ² >}44.953216\text{) =}$$=0.0000000009$

6. p is at or below α, so the counts differ significantly from the expected pattern

   $\text{Significant: reject H₀ (p ≤ 0.04)}$

5. p-value: 6.941503e-11

1. Add up each row and each column, since the expected counts are built from these totals. Row totals 78, 50; column totals 64, 64; grand total N

   $N$$=128$

2. If the two things weren't related, each cell would expect its row total times its column total, divided by the grand total N

   $\text{E = (row total × column total) ÷ N}$

3. For example, the expected count (E) for the top-left cell

   $\mathrm{E₁₁}$$=\frac{78\times 64}{128}$$=39$

4. For every cell, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(21-39\right)}^2}{39}+\frac{{\left(57-39\right)}^2}{39}+\frac{{\left(43-25\right)}^2}{25}+\frac{{\left(7-25\right)}^2}{25}$$=42.535385$

5. Count the degrees of freedom (df), how many cells are free to vary once the totals are fixed: (rows − 1) × (columns − 1)

   $\mathrm{df}$$=\left(2-1\right)\times \left(2-1\right)$$=1$

6. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 1 df)

   $p$$=\text{P(χ² >}42.535385\text{) =}$$=6.941502928e-11$

7. p is at or below α, so the two variables appear to be linked (not independent)

   $\text{Significant: reject H₀ (p ≤ 0.06)}$

8. Measure how strong the link is, from 0 to 1, with Cramér's V = √(χ² ÷ (N × (min(rows, columns) − 1)))

   $V$$=\sqrt{\frac{42.535385}{\mathrm{N(k\; -\; 1)}}}$$=0.576461$

6. p-value: 0

1. With 4 equally likely categories, each one expects the total count N ÷ 4

   $N$$=503$

2. The expected count (E) for each category

   $\text{125.75, 125.75, 125.75, 125.75}$

3. For every category, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(300-125.75\right)}^2}{125.75}+\frac{{\left(100-125.75\right)}^2}{125.75}+\frac{{\left(75-125.75\right)}^2}{125.75}+\frac{{\left(28-125.75\right)}^2}{125.75}$$=343.194831$

4. Count the degrees of freedom (df): the number of categories − 1, since the last one is fixed by the total

   $\mathrm{df}$$=4-1$$=3$

5. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 3 df)

   $p$$=\text{P(χ² >}343.194831\text{) =}$$=0$

6. p is at or below α, so the counts differ significantly from the expected pattern

   $\text{Significant: reject H₀ (p ≤ 0.03)}$

7. p-value: 0

1. Scale the expected ratios to the total count N, so expected and observed add up the same: E = N × (ratio ÷ sum of ratios)

   $N$$=534$

2. The expected count (E) for each category

   $\text{314.1176, 62.8235, 94.2353, 62.8235}$

3. For every category, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(280-314.117647\right)}^2}{314.117647}+\frac{{\left(130-62.823529\right)}^2}{62.823529}+\frac{{\left(86-94.235294\right)}^2}{94.235294}+\frac{{\left(38-62.823529\right)}^2}{62.823529}$$=86.064919$

4. Count the degrees of freedom (df): the number of categories − 1, since the last one is fixed by the total

   $\mathrm{df}$$=4-1$$=3$

5. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 3 df)

   $p$$=\text{P(χ² >}86.064919\text{) =}$$=0$

6. p is at or below α, so the counts differ significantly from the expected pattern

   $\text{Significant: reject H₀ (p ≤ 0.04)}$

8. p-value: 0.000057

1. Scale the expected ratios to the total count N, so expected and observed add up the same: E = N × (ratio ÷ sum of ratios)

   $N$$=556$

2. The expected count (E) for each category

   $\text{278, 92.6667, 139, 46.3333}$

3. For every category, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(315-278\right)}^2}{278}+\frac{{\left(108-92.666667\right)}^2}{92.666667}+\frac{{\left(101-139\right)}^2}{139}+\frac{{\left(32-46.333333\right)}^2}{46.333333}$$=22.284173$

4. Count the degrees of freedom (df): the number of categories − 1, since the last one is fixed by the total

   $\mathrm{df}$$=4-1$$=3$

5. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 3 df)

   $p$$=\text{P(χ² >}22.284173\text{) =}$$=0.0000569261$

6. p is at or below α, so the counts differ significantly from the expected pattern

   $\text{Significant: reject H₀ (p ≤ 0.05)}$

9. p-value: 0.000441

1. Add up each row and each column, since the expected counts are built from these totals. Row totals 28, 38; column totals 46, 20; grand total N

   $N$$=66$

2. If the two things weren't related, each cell would expect its row total times its column total, divided by the grand total N

   $\text{E = (row total × column total) ÷ N}$

3. For example, the expected count (E) for the top-left cell

   $\mathrm{E₁₁}$$=\frac{28\times 46}{66}$$=19.515152$

4. For every cell, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(26-19.515152\right)}^2}{19.515152}+\frac{{\left(2-8.484848\right)}^2}{8.484848}+\frac{{\left(20-26.484848\right)}^2}{26.484848}+\frac{{\left(18-11.515152\right)}^2}{11.515152}$$=12.350997$

5. Count the degrees of freedom (df), how many cells are free to vary once the totals are fixed: (rows − 1) × (columns − 1)

   $\mathrm{df}$$=\left(2-1\right)\times \left(2-1\right)$$=1$

6. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 1 df)

   $p$$=\text{P(χ² >}12.350997\text{) =}$$=0.000440751$

7. p is at or below α, so the two variables appear to be linked (not independent)

   $\text{Significant: reject H₀ (p ≤ 0.06)}$

8. Measure how strong the link is, from 0 to 1, with Cramér's V = √(χ² ÷ (N × (min(rows, columns) − 1)))

   $V$$=\sqrt{\frac{12.350997}{\mathrm{N(k\; -\; 1)}}}$$=0.432593$

10. p-value: 3.461109e-11

1. Add up each row and each column, since the expected counts are built from these totals. Row totals 57, 56; column totals 34, 79; grand total N

   $N$$=113$

2. If the two things weren't related, each cell would expect its row total times its column total, divided by the grand total N

   $\text{E = (row total × column total) ÷ N}$

3. For example, the expected count (E) for the top-left cell

   $\mathrm{E₁₁}$$=\frac{57\times 34}{113}$$=17.150442$

4. For every cell, square the gap between observed (O) and expected (E), divide by E, and add them up

   $\mathrm{\chi ²}$$=\frac{{\left(1-17.150442\right)}^2}{17.150442}+\frac{{\left(56-39.849558\right)}^2}{39.849558}+\frac{{\left(33-16.849558\right)}^2}{16.849558}+\frac{{\left(23-39.150442\right)}^2}{39.150442}$$=43.897045$

5. Count the degrees of freedom (df), how many cells are free to vary once the totals are fixed: (rows − 1) × (columns − 1)

   $\mathrm{df}$$=\left(2-1\right)\times \left(2-1\right)$$=1$

6. Find the p-value: the chance of a χ² at least this big if the hypothesis were true (chi-square curve with 1 df)

   $p$$=\text{P(χ² >}43.897045\text{) =}$$=3.461109177e-11$

7. p is at or below α, so the two variables appear to be linked (not independent)

   $\text{Significant: reject H₀ (p ≤ 0.04)}$

8. Measure how strong the link is, from 0 to 1, with Cramér's V = √(χ² ÷ (N × (min(rows, columns) − 1)))

   $V$$=\sqrt{\frac{43.897045}{\mathrm{N(k\; -\; 1)}}}$$=0.623273$