A student is chosen at random from the group. Find the probability that the student

(i)     studied Mathematics HL;

(ii)    was examined in French;

(iii)   studied Mathematics HL and was examined in French;

(iv)   did not study Mathematics SL and was not examined in English;

(v)    studied Mathematical Studies SL given that the student was examined in Spanish.

Pam believes that the Mathematics course a student chooses is independent of the language in which the student is examined.

Using your answers to parts (a) (i), (ii) and (iii) above, state whether there is any evidence for Pam’s belief. Give a reason for your answer.

Pam decides to test her belief using a Chi-squared test at the \(5\% \) level of significance.

(i)     State the null hypothesis for this test.

(ii)    Show that the expected number of Mathematical Studies SL students who took the examination in Spanish is \(41.3\), correct to 3 significant figures.

Write down

(i)     the Chi-squared calculated value;

(ii)    the number of degrees of freedom;

(iii)   the Chi-squared critical value.

State, giving a reason, whether there is sufficient evidence at the \(5\% \) level of significance that Pam’s belief is correct.

(i)     \(\frac{{100}}{{400}}{\text{ }}\left( {\frac{1}{4}{\text{, }}0.25{\text{, }}25\% } \right)\)     (A1)

(ii)    \(\frac{{90}}{{400}}{\text{ }}\left( {\frac{9}{{40}}{\text{, }}0.225{\text{, }}22.5\% } \right)\)     (A1)

(iii)   \(\frac{{20}}{{400}}{\text{ }}\left( {\frac{1}{{20}}{\text{, }}0.05{\text{, }}5\% } \right)\)     (A1)(A1)

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

(iv)   \(\frac{{120}}{{400}}{\text{ }}\left( {\frac{3}{{10}}{\text{, }}0.3{\text{, }}30\% } \right)\)     (A1)(A1)

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

(v)    \(\frac{{30}}{{110}}{\text{ }}\left( {\frac{3}{{11}}{\text{, }}0.273{\text{, }}27.3\% } \right)\) (\(0.272727 \ldots \))     (A1)(A1)

Note: Award (A1) for numerator, (A1) for denominator. Accept \(0.27\), do not accept \(0.272\), do not accept \(0.3\).

[8 marks]

\(\frac{1}{{20}} \ne \frac{1}{4} \times \frac{9}{{40}}\)     (R1)(ft)

Note: The fractions must be used as part of the reason. Follow through from (a)(i), (a)(ii) and (a)(iii).

Pam is not correct.     (A1)(ft)

Notes: Do not award (R0)(A1). Accept the events are not independent (dependent).

[2 marks]

(i)     The mathematics course and language of examination are independent.     (A1)

Notes: Accept “There is no association between Mathematics course and language”. Do not accept “not related”, “not correlated”, “not influenced”.

(ii)    \(\frac{{110}}{{400}} \times \frac{{150}}{{400}} \times 400{\text{ }}\left( { = \frac{{110 \times 150}}{{400}}} \right)\)     (M1)

 \( = 41.25\)     (A1)

 \( = 41.3\)     (AG)

Note: \(41.25\) and \(41.3\) must be seen to award final (A1).

[3 marks]

(i)     \(7.67\) (\(7.67003 \ldots \))     (G2)

Note: Accept \(7.7\), do not accept \(8\) or \(7.6\). Award (G1) if formula with all nine terms seen but their answer is not one of those above.

(ii)    \(4\)     (G1)

(iii)   \(9.488\)     (A1)(ft)

Notes: Accept \(9.49\) or \(9.5\), do not accept \(9.4\) or \(9\). Follow through from their degrees of freedom.

[4 marks]

\(7.67 < 9.488\)     (R1)

OR

\(p = 0.104 \ldots , p > 0.05\)     (R1)

Accept (Do not reject) \({H_0}\) (Pam’s belief is correct)     (A1)(ft)

Notes: Follow through from part (d). Do not award (R0)(A1).

[2 marks]

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and p-values as the basis for accepting the null hypothesis.

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and \(p\)-values as the basis for accepting the null hypothesis.

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and \(p\)-values as the basis for accepting the null hypothesis.

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and \(p\)-values as the basis for accepting the null hypothesis.

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and \(p\)-values as the basis for accepting the null hypothesis.