IB exam questions for topic 13

a) 1. $H_0$: the preferred holiday destination is independent of gender.

b) 2. There are $5$ destinations and $2$ genders (men and women): $$\nu =(5-1)(2-1)=4.$$

c) 3. The critical value of ${\chi }^2$ for $\nu =4$ at the $5\%$ significance level (from the ${\chi }^2$ table or the calculator, with area $0.05$ in the right tail) is $${\chi }_{\text{crit}}^2=\mathrm{9.49.}$$

d) 4. We compare: ${\chi }_{\text{calc}}^2=8.73<9.49={\chi }_{\text{crit}}^2$. 5. Therefore $H_0$ is not rejected: there is no evidence, at the $5\%$ level, that the preferred destination depends on gender.

Answer: a) $H_0$: the destination is independent of gender; b) $4$; c) $9.49$; d) as $8.73<9.49$, $H_0$ is not rejected

a) 1. $H_0$: the car colour is independent of the driver's gender.

b) 2. There are $5$ colours and $2$ genders: $$\nu =(5-1)(2-1)=4.$$

c) 3. Critical value for $\nu =4$ and the $5\%$ level (right tail $0.05$): $${\chi }_{\text{crit}}^2=\mathrm{9.49.}$$

d) 4. Compare: ${\chi }_{\text{calc}}^2=8.73<9.49$. 5. $H_0$ is not rejected (strictly speaking, there is not sufficient evidence to reject it): at the $5\%$ level it has not been possible to show that car colour depends on the driver's gender.

Answer: a) $H_0$: the car colour is independent of the driver's gender; b) $4$; c) $9.49$; d) as $8.73<9.49$, $H_0$ is not rejected

a) (i) 1. The ${\chi }^2$ test of independence assumes that there is no relationship between the variables:

$H_0$: the type of coffee drunk is independent of the sex of the interviewee.

(ii) 2. $H_1$: the type of coffee drunk depends on the sex of the interviewee.

b) 3. The contingency table has $2$ rows (men and women) and $5$ columns (types of coffee). The degrees of freedom are $$\nu =(\text{rows}-1)(\text{columns}-1)=(2-1)(5-1)=4.$$

c) 4. With $\nu =4$ and a $5\%$ significance level, the critical value (tables or calculator, inverse ${\chi }^2$ with area $0.05$ to the right) is $${\chi }_{\text{crit}}^2=\mathrm{9.49.}$$

d) 5. We compare the test statistic with the critical value: ${\chi }_{\text{calc}}^2=8.73<9.49={\chi }_{\text{crit}}^2$. 6. Therefore we do not reject $H_0$: there is not enough evidence, at the $5\%$ level, that the type of coffee depends on sex. With these data, the type of coffee a person drinks is independent of their sex.

Answer: a) (i) $H_0$: the type of coffee is independent of sex; (ii) $H_1$: the type of coffee depends on sex; b) $4$; c) $9.49$; d) as $8.73<9.49$, do not reject $H_0$: the type of coffee is independent of sex

a) 1. $H_0$: the type of music chosen is independent of age.

b) 2. There are $3$ age groups and $3$ types of music: $\nu =(3-1)(3-1)=4$.

c) 3. Using the calculator, ${\chi }^2$ test for independence with the observed $3\times 3$ matrix (excluding the row and column of totals): $${\chi }_{\text{calc}}^2=\mathrm{51.6.}$$

d) 4. For $\nu =4$ and the $5\%$ level, the critical value is $9.49$. 5. As $51.6>9.49$ (and $p\approx 1.7\times 10^{-10}<0.05$), we reject $H_0$: there is evidence that the favourite type of music depends on age.

Answer: a) $H_0$: the type of music is independent of age; b) $4$; c) $51.6$; d) as $51.6>9.49$, reject $H_0$

a) 1. The ${\chi }^2$ test of independence tests whether two categorical variables are related. Here, the examiners not differing in their level of awards means that the result does not depend on the examiner.

$H_0$: the examination result (credit, pass or fail) is independent of the examiner.

b) 2. The contingency table has $3$ rows (examiners) and $3$ columns (results): $$\text{d.f.}=(3-1)(3-1)=4.$$

c) 3. The expected frequency of a cell is $$\frac{\text{row total}\times \text{column total}}{\text{grand total}}=\frac{45\cdot 30}{135}=10.$$

d) 4. We compare the p-value with the significance level: $0.0327<0.05$. 5. As $p<\alpha$, we reject $H_0$ (it is not accepted): there is evidence, at $5\%$, that the result depends on the examiner, i.e. the examiners differ in their level of awards.

Answer: a) $H_0$: the result is independent of the examiner; b) $4$; c) $10$; d) as $0.0327<0.05$, reject $H_0$

a)

The ${\chi }^2$ test for independence assumes that there is no relationship between the two variables:

$H_0$: gender is independent of the chosen professional field.

Answer: $H_0$: gender is independent of the chosen professional field

b)

1. Table totals: men $55+30+25=110$; engineering $55+35=90$; grand total $220$.
2. The expected frequency of a cell is $\frac{\text{row total}\cdot \text{column total}}{\text{grand total}}$:

$$E=\frac{110\cdot 90}{220}=45.$$

Answer: $45$ male engineers

c)

1. It is a table with $2$ rows and $3$ columns, so the degrees of freedom are $(2-1)(3-1)=2$.
2. Using the GDC, carry out a ${\chi }^2$ test for independence with the observed matrix $\left(\begin{array}{ccc}55& 30& 25\\ 35& 45& 30\end{array}\right)$: ${\chi }^2\approx 7.899$ and $p\approx 0.0193$.

Answer: $p\approx 0.0193$

d)

1. $H_0$ is rejected only if $p<\alpha$. Here $\alpha =0.01$ and $p=0.0193>0.01$.
2. Therefore, at a $1\%$ significance level there is not sufficient evidence to reject $H_0$: Abhinav is wrong to reject it (we cannot conclude that gender influences the professional field).

Answer: a) $H_0$: gender is independent of the chosen professional field; b) $45$; c) $p\approx 0.0193$; d) as $p=0.0193>0.01$, $H_0$ should not be rejected, so Abhinav is wrong

a)

The Grade 10 row sums to $74$: $31+x+13=74$, so $x=74-44=30$.

Check with the juice column: $34+30+35+36=135$.

Answer: $x=30$

b)

The table has $4$ rows (grades) and $3$ columns (drinks):

$$\text{d.f.}=(4-1)(3-1)=6.$$

Answer: $6$ degrees of freedom

c)

Using the GDC, carry out a ${\chi }^2$ test for independence with the observed $4\times 3$ matrix (without the totals):

$$\left(\begin{array}{ccc}25& 34& 15\\ 31& 30& 13\\ 18& 35& 17\\ 9& 36& 26\end{array}\right)$$

This gives ${\chi }^2\approx 18.96\approx 19.0$ (with $p\approx 0.00422$).

Answer: ${\chi }^2\approx 19.0$

d)

1. We compare the statistic with the critical value: $19.0>12.6$ (equivalently, $p=0.00422<0.05$).
2. Therefore $H_0$ is rejected: at a $5\%$ significance level there is evidence that favourite drink is not independent of grade.

Answer: a) $x=30$; b) $6$; c) ${\chi }^2\approx 19.0$; d) as $19.0>12.6$, $H_0$ is rejected: favourite drink depends on grade

a)

$H_0$: preferred destination is independent of the traveller's age.

Answer: $H_0$: preferred destination is independent of age

b)

Age has $4$ categories and destination has $5$, so the contingency table is $4\times 5$:

$$\text{d.f.}=(4-1)(5-1)=12.$$

Answer: $12$ degrees of freedom

c)

$H_0$ is rejected when the statistic exceeds the critical value (or, equivalently, when it falls in the rejection region):

$${\chi }_{\text{calc}}^2>\mathrm{21.026.}$$

Answer: $H_0$ is rejected for ${\chi }^2>21.026$

d)

1. Travellers aged $61$ or over: $285$. Travellers who prefer Tokyo: $420$. Total: $1200$.
2. Expected frequency:

$$E=\frac{285\cdot 420}{1200}=99.75\approx \mathrm{99.8.}$$

Answer: a) $H_0$: preferred destination is independent of age; b) $12$; c) $H_0$ is rejected when ${\chi }^2>21.026$; d) $99.8$ travellers

a)

1. Women: $93$. Documentaries: $54$. Total: $180$.
2. Expected frequency:

$$E=\frac{93\cdot 54}{180}=\mathrm{27.9.}$$

Answer: $27.9$ women

b)

A $2\times 4$ table (without the totals): using the GDC, carry out a ${\chi }^2$ test for independence with the matrix $\left(\begin{array}{cccc}20& 24& 32& 11\\ 18& 30& 20& 25\end{array}\right)$. This gives ${\chi }^2\approx 8.80$, with $3$ degrees of freedom, and a $p$-value of $p\approx 0.0321$.

Answer: $p\approx 0.0321$

c)

1. $H_0$: the preferred type of TV programme is independent of gender.
2. As $p=0.0321<0.05$, $H_0$ is rejected: at the $5\%$ significance level there is evidence that the preferred type of TV programme depends on gender.

Answer: a) $27.9$ women; b) $p\approx 0.0321$; c) as $p<0.05$, $H_0$ is rejected (the preferred programme is not independent of gender)

a)

1. Infected patients: $100$. Nightingale ward: $330$. Total: $1000$.
2. Expected frequency:

$$E=\frac{100\cdot 330}{1000}=33.$$

Answer: $33$ patients

b)

Using the GDC, carry out a ${\chi }^2$ test for independence with the matrix $\left(\begin{array}{cccc}44& 27& 13& 16\\ 281& 303& 182& 134\end{array}\right)$ ($2\times 4$, $3$ degrees of freedom):

$${\chi }^2\approx 8.215\approx \mathrm{8.21.}$$

(The associated $p$-value is $\approx 0.0418$.)

Answer: ${\chi }^2\approx 8.21$

c)

1. Compare with the critical value: $8.215>7.815$ (equivalently, $p\approx 0.0418<0.05$).
2. $H_0$ is rejected: at the $5\%$ significance level there is evidence that contracting an infection depends on the ward in which the patient is admitted.

Answer: a) $33$; b) ${\chi }^2\approx 8.21$; c) as $8.21>7.815$, $H_0$ is rejected

a)

$H_0$: the preferred type of Latin dance is independent of age (whether the viewer is 20 years old or under, or over 20).

Answer: $H_0$: preferred dance is independent of age

b)

1. Over $20$ years old: $20+17+18+15=70$. Rumba: $12+18=30$. Total: $150$.
2. Expected frequency:

$$E=\frac{70\cdot 30}{150}=14.$$

Answer: $14$ viewers

c)

Using the GDC, carry out a ${\chi }^2$ test for independence with the matrix $\left(\begin{array}{cccc}35& 23& 12& 10\\ 20& 17& 18& 15\end{array}\right)$ ($3$ degrees of freedom): ${\chi }^2\approx 6.55$ and $p\approx 0.0876$.

Answer: $p\approx 0.0876$

d)

1. As $p=0.0876>0.05$, $H_0$ is not rejected.
2. There is not sufficient evidence, at the $5\%$ level, to reject independence: the producer's claim is justified (the data are consistent with the preferred dance not depending on age).

Answer: a) $H_0$: preferred dance is independent of age; b) $14$; c) $p\approx 0.0876$; d) as $p>0.05$, $H_0$ is not rejected, so the claim is justified

a)

Boys: $15+15+10=40$. Girls: $10+10+30=50$. Total: $40+50=90$.

Answer: $90$ students

b)

$H_0$: gender and preferred drink are independent.

Answer: $H_0$: gender and preferred drink are independent

c)

Using the GDC, carry out a ${\chi }^2$ test for independence with the matrix $\left(\begin{array}{ccc}15& 15& 10\\ 10& 10& 30\end{array}\right)$ ($2$ degrees of freedom):

$${\chi }^2=11.025\approx 11.0(p\approx 0.00404).$$

Answer: ${\chi }^2\approx 11.0$

d)

1. Compare with the critical value: $11.025>5.99$.
2. $H_0$ is rejected: at the $5\%$ significance level there is evidence that preferred drink depends on gender.

Answer: a) $90$; b) $H_0$: gender and preferred drink are independent; c) ${\chi }^2\approx 11.0$; d) as $11.0>5.99$, $H_0$ is rejected

a)

$H_0$: the country chosen for university studies is independent of gender.

Answer: $H_0$: the country chosen is independent of gender

b)

The table has $2$ rows and $3$ columns: $\text{d.f.}=(2-1)(3-1)=2$.

Answer: $2$ degrees of freedom

c)

Using the GDC, carry out a ${\chi }^2$ test for independence with the matrix $\left(\begin{array}{ccc}55& 26& 40\\ 25& 31& 41\end{array}\right)$:

(i) ${\chi }^2\approx 9.17$.

(ii) $p\approx 0.0102$.

Answer: (i) ${\chi }^2\approx 9.17$; (ii) $p\approx 0.0102$

d)

1. Compare with the critical value: $9.17<9.210$ (equivalently, $p=0.0102>0.01$).
2. $H_0$ is not rejected (it is accepted): at the $1\%$ level there is not sufficient evidence that the country chosen depends on gender.

Answer: a) $H_0$: the country chosen is independent of gender; b) $2$; c) (i) ${\chi }^2\approx 9.17$, (ii) $p\approx 0.0102$; d) as $9.17<9.210$ (or $p>0.01$), $H_0$ is accepted

a) 1. If $H_0$ is true (the way of travelling to school is independent of gender), the expected frequency of each cell is $$E=\frac{\text{row total}\cdot \text{column total}}{\text{grand total}}.$$ 2. Girls who travel by public transport: row “Girls” ($48$) and column “Public transport” ($30$): $$E=\frac{48\cdot 30}{100}=\mathrm{14.4.}$$

b) 3. We calculate all the expected frequencies with the same method:

4. We apply ${\chi }^2=\sum \frac{(O-E{)}^2}{E}$ to the eight cells (or, directly, the GDC ${\chi }^2$ test for independence with the matrix of observed frequencies): $${\chi }^2=\frac{(18-12.96{)}^2}{12.96}+\frac{(13-14.4{)}^2}{14.4}+\cdots +\frac{(16-9.88{)}^2}{9.88}\approx \mathrm{12.955.}$$ That is, ${\chi }^2\approx 13.0$.

c) 5. Decision rule: $H_0$ is rejected if ${\chi }_{\text{calc}}^2>{\chi }_{\text{crit}}^2$. 6. Here $12.955>7.815$, so $H_0$ is rejected (it is not accepted). There is evidence, at the $5\%$ level, that the way of travelling to school depends on the student's gender.

Answer: a) $14.4$; b) ${\chi }^2\approx 13.0$; c) as $12.955>7.815$, $H_0$ is rejected: the way of travelling to school depends on gender

a) 1. The null hypothesis states that there is no relationship between the variables:

$H_0$: the choice of morning snack is independent of the student's gender.

b) 2. There are $3$ snacks (apple, fruit salad, milkshake) and $2$ genders. The degrees of freedom are $$\nu =(\text{rows}-1)(\text{columns}-1)=(3-1)(2-1)=2.$$

c) 3. Total students: $350$; women: $210$; milkshakes: $150$. Using the expected frequency formula under $H_0$: $$E=\frac{210\cdot 150}{350}=90.$$ So $90$ women are expected to choose a milkshake.

d) 4. We compare the statistic with the critical value: ${\chi }_{\text{calc}}^2=3.576<5.99$. 5. As it does not exceed the critical value, $H_0$ is not rejected. At the $5\%$ significance level there is not sufficient evidence of a relationship: the choice of morning snack is independent of the students' gender.

Answer: a) $H_0$: the morning snack is independent of gender; b) $2$; c) $90$; d) as $3.576<5.99$, $H_0$ is not rejected: the snack choice does not depend on gender

a) 1. $H_0$: the crop yield is independent of the fertiliser brand.

b) 2. Table with $3$ rows (yields) and $3$ columns (brands): $$\nu =(3-1)(3-1)=4.$$

c) 3. For $\nu =4$ and a $1\%$ significance level (right tail $0.01$) the critical value is $${\chi }_{\text{crit}}^2=\mathrm{13.277.}$$

d) 4. Medium-yield crops (total $50$) with fertiliser C (total $40$), out of $120$ crops: $$E=\frac{50\cdot 40}{120}=16.\overline{6}\approx 17.$$

e) 5. Using the GDC, carry out a ${\chi }^2$ test for independence with the matrix of observed frequencies ($3\times 3$):

(i) ${\chi }_{\text{calc}}^2\approx 3.86$ (exact: $3.8613\dots$).

(ii) $p\approx 0.425$.

f) 6. ${\chi }_{\text{calc}}^2=3.86<13.277$ and, equivalently, $p=0.425>0.01$. 7. $H_0$ is not rejected: at the $1\%$ level there is no evidence that the yield depends on the fertiliser brand.

Answer: a) $H_0$: the yield is independent of the fertiliser; b) $4$; c) $13.277$; d) $16.67\approx 17$; e) (i) $3.86$; (ii) $0.425$; f) $H_0$ is not rejected ($3.86<13.277$; $p>0.01$)

a) 1. $H_0$: the preferred swimming style is independent of the spectator's gender.

b) 2. There are $4$ styles and $2$ genders: $$\nu =(4-1)(2-1)=3.$$

c) 3. Enter the table of observed frequencies ($4\times 2$) as a matrix in the GDC and carry out the ${\chi }^2$ test for independence. It can also be done by hand: the totals are $60$ men and $120$ women, and the totals per style are $35,50,45,50$. The expected frequencies are, for example, $E=\frac{35\cdot 60}{180}=11.67$ for men and freestyle. Summing $\frac{(O-E{)}^2}{E}$ gives $${\chi }_{\text{calc}}^2\approx 16.4(16.43\dots ).$$

d) 4. Compare with the critical value: $16.4>7.815$. 5. $H_0$ is rejected: at the $5\%$ level there is evidence that the preferred swimming style depends on gender.

Answer: a) $H_0$: the preferred style is independent of gender; b) $3$; c) ${\chi }_{\text{calc}}^2\approx 16.4$; d) as $16.4>7.815$, $H_0$ is rejected

a) 1. $H_0$: the choice of profession (engineering, medicine or law) is independent of the sex of the person.

b) 2. There are $3$ professions and $2$ sexes: $$\nu =(3-1)(2-1)=2.$$

c) 3. Total $400$; women $180$; engineers (both sexes) $80$. Expected frequency of female engineers under $H_0$: $$E=\frac{180\cdot 80}{400}=36.$$

d) 4. We compare the p-value with the significance level: $p=0.0634>0.05$. 5. Do not reject $H_0$: at the $5\%$ level there is not enough evidence that the chosen profession depends on sex.

Answer: a) $H_0$: the chosen profession is independent of sex; b) $2$; c) $36$; d) do not reject $H_0$ ($p=0.0634>0.05$)

a) 1. The table has $3$ rows (ages) and $4$ columns (drinks): $$\nu =(3-1)(4-1)=6.$$

b) 2. $H_0$: the preferred breakfast drink is independent of age.

c) 3. With $\nu =6$ and a $5\%$ significance level, the critical value is $${\chi }_{\text{crit}}^2=\mathrm{12.6.}$$

d) 4. The test statistic is ${\chi }_{\text{calc}}^2=146$, much greater than the critical value $12.6$. 5. Equivalently, the p-value $6.62\times 10^{-29}$ is less than $\alpha =0.05$. 6. Reject $H_0$: at the $5\%$ level there is evidence that the breakfast drink chosen depends on age.

Answer: a) $6$; b) $H_0$: the breakfast drink is independent of age; c) $12.6$; d) reject $H_0$ (because $146>12.6$, or because $p<0.05$)

a) 1. $H_0$: the height of the plants is independent of the use of fertiliser.

$H_1$: the height of the plants depends on the use of fertiliser (the fertiliser has an effect).

b) 2. Expected frequency $=\frac{\text{row total}\times \text{column total}}{\text{grand total}}$. 3. Plants over $75$ cm: $115+80=195$. Fertilised plants: $115+45+20=180$. Total: $195+110+55=360$. $$E=\frac{195\cdot 180}{360}=\mathrm{97.5.}$$

c) 4. Using the calculator, ${\chi }^2$ test of independence with the observed $3\times 2$ matrix: $${\chi }_{\text{calc}}^2=14.0$$ (unrounded value $14.009\dots$).

d) 5. $\nu =(3-1)(2-1)=2$.

e) 6. For $\nu =2$ and a $1\%$ significance level: ${\chi }_{\text{crit}}^2=9.21$.

f) 7. As ${\chi }_{\text{calc}}^2=14.0>9.21={\chi }_{\text{crit}}^2$, we reject $H_0$ (equivalently, $p=0.000908<0.01$). 8. There is evidence, at the $1\%$ level, that height depends on the fertiliser: the manufacturer's claim is justified.

Answer: a) $H_0$: height is independent of fertiliser; $H_1$: height depends on fertiliser; b) $97.5$; c) $14.0$; d) $2$; e) $9.21$; f) reject $H_0$ ($14.0>9.21$): the claim is justified

a) 1. Total males: $14+6+8=28$.

b) 2. Total students who prefer blue: $14+31=45$; grand total: $100$. $$E=\frac{28\cdot 45}{100}=\mathrm{12.6.}$$

c) (i) 3. $H_0$: preferred car colour is independent of gender.

(ii) 4. The table has $2$ rows and $3$ columns: $\nu =(2-1)(3-1)=2$.

(iii) 5. For $\nu =2$ and a $5\%$ level: ${\chi }_{\text{crit}}^2=5.99$.

(iv) 6. As ${\chi }_{\text{calc}}^2=1.367<5.99$ (the p-value is $0.505>0.05$), we do not reject $H_0$. 7. Accept $H_0$: the data do not show, at $5\%$, a difference between males and females in preferred car colour; the apparent difference may be due to chance.

Answer: a) $28$; b) $12.6$; c) (i) $H_0$: preferred colour is independent of gender; (ii) $2$; (iii) $5.99$; (iv) as $1.367<5.99$, do not reject $H_0$

a) 1. $H_0$: the size of the dog is independent of the time of day.

b) 2. Using the calculator, ${\chi }^2$ test of independence with the $3\times 3$ matrix of the data: $${\chi }_{\text{calc}}^2=4.33$$ (unrounded value $4.328\dots$; $p=0.363$).

c) 3. The table has $3$ rows and $3$ columns: $$\nu =(3-1)(3-1)=2\cdot 2=4.$$

d) 4. As $4.33<9.488$, the statistic does not exceed the critical value, so we do not reject $H_0$. 5. Conclusion: at the $5\%$ level there is no evidence of a relationship between dog size and time of day; they are independent.

Answer: a) $H_0$: dog size is independent of time of day; b) $4.33$; c) $(3-1)(3-1)=4$; d) as $4.33<9.488$, do not reject $H_0$

a) 1. Ignoring age, we add the rows for each gender:

Men: $22+36=58$; $65+54=119$; $90+67=157$; $35+17=52$.

Women: $22+64=86$; $59+39=98$; $82+38=120$; $15+46=61$.

Check: total men $386$, total women $365$, grand total $751$.

b) 2. $H_0$: the most-watched type of programme is independent of gender.

$H_1$: the most-watched type of programme depends on gender.

c) 3. Women: $365$. "Comedy" column: $119+98=217$. $$E=\frac{365\cdot 217}{751}=105.5\dots \approx 105.$$

d) 4. Using the calculator, ${\chi }^2$ test of independence with the observed $2\times 4$ matrix: $${\chi }_{\text{calc}}^2=12.6$$ (unrounded value $12.558\dots$).

e) (i) 5. $\nu =(2-1)(4-1)=3$.

(ii) 6. For $\nu =3$ and the $5\%$ level: ${\chi }_{\text{crit}}^2=7.81$.

(iii) 7. As $12.6>7.81$ (and $p=0.00570<0.05$), we reject $H_0$: there is evidence that the most-watched type of programme depends on gender.

Answer: a) $2\times 4$ table with the totals by gender; b) $H_0$: the type of programme is independent of gender; $H_1$: it depends on gender; c) $105$; d) $12.6$; e) (i) $3$; (ii) $7.81$; (iii) as $12.6>7.81$, reject $H_0$

a) 1. $H_0$: the customer's age (adult or child) is independent of the dish chosen.

b) 2. The table has $2$ rows and $3$ columns: $$\text{d.f.}=(2-1)(3-1)=2.$$

c) 3. Expected number of children who choose seafood $=\frac{\text{total children}\times \text{total seafood}}{\text{total}}$: $$\frac{69\cdot 67}{150}=30.82\dots \approx 31.$$

d) 4. Using the graphic display calculator, ${\chi }^2$ test of independence with the matrix of observed frequencies ($2\times 3$):

(i) ${\chi }^2=2.66$ (unrounded value $2.6575$).

(ii) $p=0.265$.

e) 5. The statistic is less than the critical value: $2.66<4.605$ (equivalently, $p=0.265>0.10$). 6. We do not reject $H_0$: there is no evidence, at the $10\%$ level, that the dish chosen depends on the customer's age.

f) 7. (i) There are $81$ adults among $150$ customers: $$P(\text{adult})=\frac{81}{150}=\mathrm{0.54.}$$

8. (ii) By the inclusion–exclusion principle, we subtract the adults who chose seafood ($32$): $$P(\text{adult or seafood})=\frac{81+67-32}{150}=\frac{116}{150}=\mathrm{0.773.}$$

9. (iii) Given that the customer is a child, the sample space is the $69$ children; of these, $20+14=34$ chose pasta or fish: $$P(\text{pasta or fish}\mid \text{child})=\frac{34}{69}=\mathrm{0.493.}$$

Answer: a) $H_0$: age is independent of the dish chosen; b) $2$; c) $30.82\approx 31$; d) (i) $2.66$; (ii) $0.265$; e) do not reject $H_0$ ($2.66<4.605$); f) (i) $0.54$; (ii) $0.773$; (iii) $0.493$

a) 1. $H_0$: the language studied by the student is independent of their sex.

b) 2. The table has $2$ rows (sex) and $3$ columns (language): $$\text{d.f.}=(2-1)(3-1)=2.$$

c) 3. (i) Expected frequency of female students who study Chinese: $$\frac{43\cdot 42}{110}=16.418\dots \approx \mathrm{16.4.}$$ (The calculator shows it in the matrix of expected frequencies when the ${\chi }^2$ test is carried out.)

4. (ii) Using the graphic display calculator, ${\chi }^2$ test of independence with the observed $2\times 3$ matrix: ${\chi }^2=8.69$ (unrounded value $8.685$).

d) 5. Compare with the critical value: $8.69>5.99$. 6. Reject $H_0$: there is evidence, at the $5\%$ level, that the language chosen depends on the student's sex.

e) 7. (i) Students not studying Spanish: $110-22=88$. $$P(\text{not Spanish})=\frac{88}{110}=\mathrm{0.8.}$$

8. (ii) The second student is a different person, so the first is not replaced: there are $109$ students left, $87$ of whom do not study Spanish: $$P(\text{neither studies Spanish})=\frac{88}{110}\cdot \frac{87}{109}=\mathrm{0.639.}$$

9. (iii) "At least one female" is the complement of "both are male" (there are $67$ males): $$P=1-\frac{67}{110}\cdot \frac{66}{109}=1-0.369=\mathrm{0.631.}$$

Answer: a) $H_0$: the language is independent of sex; b) $2$; c) (i) $16.4$; (ii) $8.69$; d) reject $H_0$ ($8.69>5.99$); e) (i) $0.8$; (ii) $0.639$; (iii) $0.631$

a) 1. (i) There are $18$ Calculus answers out of $90$: $$P=\frac{18}{90}=\mathrm{0.2.}$$

2. (ii) Satisfactory Calculus answers: $10$: $$P=\frac{10}{90}=\mathrm{0.111.}$$

3. (iii) Conditioned on Calculus, the sample space is its $18$ answers: $$P=\frac{10}{18}=\mathrm{0.556.}$$

b) 4. In Logic there are $20$ answers, $6$ of them unsatisfactory. Two are chosen without replacement: $$P=\frac{6}{20}\cdot \frac{5}{19}=\frac{30}{380}=\mathrm{0.0789.}$$

c) 5. $H_0$: the quality of the answer (satisfactory or unsatisfactory) is independent of the question (unit) it corresponds to.

d) 6. Expected number of satisfactory Calculus answers: $$\frac{60\cdot 18}{90}=12.$$

e) 7. The table has $2$ rows and $4$ columns: $$\text{d.f.}=(2-1)(4-1)=3.$$

f) 8. Using the graphic display calculator, ${\chi }^2$ test of independence with the observed $2\times 4$ matrix: ${\chi }^2=1.46$ (unrounded value $1.4636$).

g) 9. As $1.46<7.815$, we do not reject $H_0$: there is no evidence, at the $5\%$ level, that the quality of the answer depends on the question.

Answer: a) (i) $0.2$; (ii) $0.111$; (iii) $0.556$; b) $0.0789$; c) $H_0$: quality is independent of the question; d) $12$; e) $3$; f) $1.46$; g) do not reject $H_0$ ($1.46<7.815$)

a) 1. $H_0$: choice of sport is independent of gender.

b) 2. Expected value for female football players: $$\frac{85\cdot 48}{120}=34.$$

c) 3. Table $2\times 3$: $$\text{d.f.}=(2-1)(3-1)=2.$$

d) 4. With $2$ degrees of freedom and a $5\%$ level, the critical value is ${\chi }_{\text{crit}}^2=5.99$.

e) 5. Using the GDC, ${\chi }^2$ test for independence with the observed $2\times 3$ matrix: ${\chi }_{\text{calc}}^2=2.42$ (unrounded, $2.421$).

f) 6. Since $2.42<5.99$, we do not reject $H_0$: there is no evidence, at the 5% level, that choice of sport depends on gender.

g) 7. (i) $P(\text{male})=\frac{35}{120}=0.292$.

8. (ii) $P(\text{tennis})=\frac{25}{120}=0.208$.

h) 9. (i) Two different students, without replacement ($48$ play football): $$P=\frac{48}{120}\cdot \frac{47}{119}=\mathrm{0.158.}$$

10. (ii) Students who do not play basketball: 120-47=73: $$P=\frac{73}{120}\cdot \frac{72}{119}=\mathrm{0.368.}$$

Answer: a) $H_0$: choice of sport is independent of gender; b) $34$; c) $2$; d) $5.99$; e) $2.42$; f) do not reject $H_0$ ($2.42<5.99$); g) (i) $0.292$; (ii) $0.208$; h) (i) $0.158$; (ii) $0.368$

a) 1. (i) Those who do not prefer Zucos: 400-120=280 people: $$P=\frac{280}{400}=\mathrm{0.7.}$$

2. (ii) Conditional on Montevideo (sample space: 210 people), 57 prefer Chocos: $$P=\frac{57}{210}=\mathrm{0.271.}$$

b) 3. 180 people prefer Fruti. Two different people are chosen (without replacement): $$P=\frac{180}{400}\cdot \frac{179}{399}=\mathrm{0.202.}$$

c) 4. H0: the preferred cereal brand is independent of the city.

d) 5. Table 2×3: d.f.=(2-1)(3-1)=2.

e) 6. Expected number of people from Montevideo who prefer Zucos: $$\frac{210\cdot 120}{400}=63.$$

f) 7. Using the GDC, chi-squared test for independence with the observed 2×3 matrix: ${\chi }^2=39.3$ (unrounded, $39.31$).

g) 8. At the 5% level and with 2 degrees of freedom, the critical value is 5.99. Since 39.3>5.99 (the p-value is practically 0), we reject H0. 9. The organisation would not accept the null hypothesis: there is evidence that the preferred brand is associated with the city.

Answer: a) (i) $0.7$; (ii) $0.271$; b) $0.202$; c) $H_0$: the preferred brand is independent of the city; d) $2$; e) $63$; f) $39.3$; g) reject $H_0$ ($39.3>5.99$)

a)

(i) 1. The ${\chi }^2$ test for independence tests whether two categorical variables are associated. The null hypothesis states that they are not: $H_0$: seatbelt use is independent of the length of time the driver has held their driving licence.

(ii) 2. The table has 3 rows and 2 columns, so the degrees of freedom are $$\nu =(3-1)(2-1)=2.$$

(iii) 3. Expected frequency = (row total · column total) / grand total. The row "more than 15 years" sums to 30+15=45; the column "wears a seatbelt" sums to 38+30+30=98: $$E=\frac{45\cdot 98}{200}=22.05\approx 22.$$

(iv) 4. Using the GDC, ${\chi }^2$ test for independence with the observed 3×2 matrix (without continuity correction): $${\chi }_{calc}^2=8.12(\text{p-value }p=0.0172).$$

(v) 5. We compare the p-value with the significance level: 0.0172<0.05. 6. Therefore Jorge rejects H0: there is evidence, at the 5% level, that seatbelt use is associated with the length of time the driver has held their licence.

b)

(i) 7. The number who wear a seatbelt is 38+30+30=98 out of 200 drivers: $$P=\frac{98}{200}=\mathrm{0.49.}$$

(ii) 8. Conditional probability: we restrict attention to the 45 drivers who have held their licence for more than 15 years, of whom 15 do not wear a seatbelt: $$P(\text{no seatbelt}\mid >15\text{ years})=\frac{15}{45}=\frac{1}{3}\approx \mathrm{0.333.}$$

c) Two drivers are chosen without replacement, so the second probability changes.

(i) 9. $$P(\text{both})=\frac{98}{200}\cdot \frac{97}{199}\approx \mathrm{0.239.}$$

(ii) 10. We use the complementary event ("none wears a seatbelt"), with 102 drivers who do not wear one: $$P(\text{at least one})=1-\frac{102}{200}\cdot \frac{101}{199}\approx 1-0.259=\mathrm{0.741.}$$

Answer: a) (i) seatbelt use is independent of the length of time the driver has held their licence; (ii) $2$; (iii) $22.05\approx 22$; (iv) ${\chi }_{calc}^2=8.12$; (v) no, we reject $H_0$ (p=0.0172<0.05); b) (i) $0.49$; (ii) $0.333$; c) (i) $0.239$; (ii) $0.741$

a) 1. If the distribution were uniform, each of the 5 colours would have 20% of the 600 sweets: 0.2·600=120.

b) 2. In a goodness of fit test, $\nu =\text{(number of categories)}-1=5-1=4$.

c) 3. $H_0$: the data fit a uniform distribution (the five colours are equally likely). $H_1$: the data do not fit a uniform distribution.

d) 4. Test statistic: $${\chi }_{calc}^2=\sum \frac{(O-E{)}^2}{E}=\frac{(104-120{)}^2+(132-120{)}^2+(98-120{)}^2+(129-120{)}^2+(137-120{)}^2}{120}=\frac{1254}{120}=\mathrm{10.45.}$$ 5. Using the GDC (goodness of fit test, $\nu =4$) we obtain $p=0.0335$.

e) 6. Since p=0.0335<0.05 (equivalently, ${\chi }_{calc}^2=10.45>9.488$), we reject H0. 7. Conclusion: there is evidence, at the 5% level, that the colours are not distributed uniformly across the bags.

Answer: a) $120$ for each colour; b) $4$; c) $H_0$: uniform distribution, $H_1$: not uniform; d) ${\chi }_{calc}^2=10.45$, $p=0.0335$; e) reject $H_0$

a) 1. With 12 equally likely months, the expected frequency for each month is $60/12=5$.

b) 2. $\nu =12-1=11$.

c) 3. $H_0$: the data fit a uniform distribution (birthdays are spread equally across the months). $H_1$: the data do not fit a uniform distribution.

d) 4. Test statistic: $${\chi }_{calc}^2=\sum \frac{(O-E{)}^2}{E}=\frac{4+0+1+1+0+1+1+4+9+1+4+4}{5}=\frac{30}{5}=6.$$ 5. Using the GDC (goodness of fit test, $\nu =11$): $p=0.873$.