Jorge applies a \({\chi ^2}\) test at the \(5\% \) level to investigate whether wearing a seat belt is associated with the time a driver has had their licence.

(i)     Write down the null hypothesis, \({{\text{H}}_0}\).

(ii)    Write down the number of degrees of freedom.

(iii)   Show that the expected number of drivers that wear a seat belt and have had their driving licence for more than \(15\) years is \(22\), correct to the nearest whole number.

(iv)   Write down the \({\chi ^2}\) test statistic for this data.

(v)    Does Jorge accept \({{\text{H}}_0}\) ? Give a reason for your answer.

Consider the \(200\) drivers surveyed. One driver is chosen at random. Calculate the probability that

(i)     this driver wears a seat belt;

(ii)    the driver does not wear a seat belt, given that the driver has held a licence for more than \(15\) years.

Two drivers are chosen at random. Calculate the probability that

(i)     both wear a seat belt.

(ii)    at least one wears a seat belt.

(i)    \({{\text{H}}_0} = \) wearing of a seat belt and the time a driver has held a licence are independent.     (A1)

Note: For independent accept 'not associated' but do not accept 'not related' or 'not correlated'

(ii)    \(2\)     (A1)

(iii)   \(\frac{{98 \times 45}}{{200}} = 22.05 = 22\) (correct to the nearest whole number)     (M1)(A1)(AG)

Note: (M1) for correct formula and (A1) for correct substitution. Unrounded answer must be seen for the (A1) to be awarded.

(iv)   \({\chi ^2} = 8.12\)     (G2)

Note: For unrounded answer award (G1)(G0)(AP). If formula used award (M1) for correct substituted formula with correct substitution (6 terms) (A1) for correct answer.

(v)   “Does not accept \({{\text{H}}_0}\)”     (A1)(ft)

\(p{\text{-}}value < 0.05\)     (R1)(ft)

Note: Allow “Reject \({{\text{H}}_0}\)” or equivalent. Follow through from their \({\chi ^2}\) statistic. Award (R1)(ft) for comparing the appropriate values. The (A1)(ft) can be awarded only if the conclusion is valid according to the comparison given. If no reason given or if reason is wrong the two marks are lost.

[8 marks]

(i)     \(\frac{{98}}{{200}}( = 0.49{\text{, }}49\% )\)     (A1)(A1)(G2)

Note: (A1) for numerator, (A1) for denominator.

(ii)    \(\frac{{15}}{{45}}( = 0.333{\text{, }}33.3\% )\)     (A1)(A1)(G2)

Note: (A1) for numerator, (A1) for denominator.

[4 marks]

(i)     \(\frac{{98}}{{200}} \times \frac{{97}}{{199}} = 0.239{\text{ }}(23.9\% )\)     (A1)(M1)(A1)(G3)

Note: (A1) for correct probabilities seen, (M1) for multiplying two probabilities, (A1) for correct answer.

(ii)     \(1 - \frac{{102}}{{200}} \times \frac{{101}}{{199}} = 0.741{\text{ }}(74.1\% )\)     (M1)(M1)(A1)(ft)(G2)

Note: (M1) for showing the product, (M1) for using the probability of the complement, (A1) for correct answer. Follow through for consistent use of with replacement.

OR

\(\frac{{98}}{{200}} \times \frac{{97}}{{199}} + \frac{{98}}{{200}} \times \frac{{102}}{{199}} + \frac{{102}}{{200}} \times \frac{{98}}{{199}} = 0.741{\text{ }}(74.1\% )\)     (M1)(M1)(A1)(ft)(G2)

Note: (M1) for adding three products of fractions (or equivalent), (M1) for using the correct fractions, (A1) for correct answer. Follow through for consistent use of with replacement.

[6 marks]

The first part of the question was relatively well done. The null hypothesis and the degrees of freedom were well answered by the majority of the students. In the show that question some students used the GDC to find the expected values table and highlighted the correct value \(22.05\). This procedure gained no mark; the expected value formula was expected to be used here. Also those who did use the formula were expected to show the unrounded value \(22.05\) to gain full marks in this part question. Many lost the answer mark for not doing so. GDC was used by most of the students to find the chi-squared test though some students attempted to find this value by hand which made them waste time. Correct values were compared when deciding whether to accept or not the null hypothesis and follow through marks were awarded from their degrees of freedom and chi-squared test when incorrect.

The second part was not as successful as the first one. Simple probability was well answered. Not all the students changed the denominator to \(45\) for the second probability showing their weaknesses in conditional probability. It would have been useful for the students to use a tree diagram to help them solve the last part of this question but very few did so. Some of those students that reached the last part of the question forgot to add one of the three terms. Very few used the probability of the complement.

The first part of the question was relatively well done. The null hypothesis and the degrees of freedom were well answered by the majority of the students. In the show that question some students used the GDC to find the expected values table and highlighted the correct value \(22.05\). This procedure gained no mark; the expected value formula was expected to be used here. Also those who did use the formula were expected to show the unrounded value \(22.05\) to gain full marks in this part question. Many lost the answer mark for not doing so. GDC was used by most of the students to find the chi-squared test though some students attempted to find this value by hand which made them waste time. Correct values were compared when deciding whether to accept or not the null hypothesis and follow through marks were awarded from their degrees of freedom and chi-squared test when incorrect.

The second part was not as successful as the first one. Simple probability was well answered. Not all the students changed the denominator to \(45\) for the second probability showing their weaknesses in conditional probability. It would have been useful for the students to use a tree diagram to help them solve the last part of this question but very few did so. Some of those students that reached the last part of the question forgot to add one of the three terms. Very few used the probability of the complement.

The first part of the question was relatively well done. The null hypothesis and the degrees of freedom were well answered by the majority of the students. In the show that question some students used the GDC to find the expected values table and highlighted the correct value \(22.05\). This procedure gained no mark; the expected value formula was expected to be used here. Also those who did use the formula were expected to show the unrounded value \(22.05\) to gain full marks in this part question. Many lost the answer mark for not doing so. GDC was used by most of the students to find the chi-squared test though some students attempted to find this value by hand which made them waste time. Correct values were compared when deciding whether to accept or not the null hypothesis and follow through marks were awarded from their degrees of freedom and chi-squared test when incorrect.

The second part was not as successful as the first one. Simple probability was well answered. Not all the students changed the denominator to \(45\) for the second probability showing their weaknesses in conditional probability. It would have been useful for the students to use a tree diagram to help them solve the last part of this question but very few did so. Some of those students that reached the last part of the question forgot to add one of the three terms. Very few used the probability of the complement.