Draw a Venn Diagram, using sets labelled \(A\) , \(S\) and \(P\) , that shows this information.

There were 48 customers of Pete’s Eats that day. Calculate the number of people who were customers of all three cafés.

There were 50 customers of Sarah’s Snackbar that day. Calculate the total number of people who were customers of Alan’s Diner.

Write down the number of customers of Alan’s Diner that were also customers of Pete’s Eats.

Find \(n[(S \cup P) \cap A']\).

One of the survey forms was chosen at random, find the probability that the form showed “Dissatisfied”;

One of the survey forms was chosen at random, find the probability that the form showed “Satisfied” and was completed at Sarah’s Snackbar;

One of the survey forms was chosen at random, find the probability that the form showed “Dissatisfied”, given that it was completed at Alan’s Diner.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

Write down the null hypothesis, \({{\text{H}}_0}\) , for the \({\chi ^2}\) test.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

Write down the number of degrees of freedom for the test.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

Using your graphic display calculator, find \({\chi ^2}_{calc}\) .

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

State, giving a reason, the conclusion to the test.

(A1) for rectangle and three labelled intersecting circles
(A1) for \(15\), \(27\) and \(17\)
(A1) for \(10\) and \(8\)     (A3)

[3 marks]

\(48 - (8 +10 +17)\) or equivalent     (M1)

\( = 13\)     (A1)(ft)(G2)

[2 marks]

\(50 - (27 +10 +13)\)     (M1)

Note: Award (M1) for working seen.

\( = 0\)     (A1)
number of elements in A \(= 36\)     (A1)(ft)(G3)

Note: Follow through from (b).

[3 marks]

\(21\)     (A1)(ft)

Note: Follow through from (b) even if no working seen.

[1 mark]

\(54\)     (M1)(A1)(ft)(G2)

Note: Award (M1) for \(17\), \(10\), \(27\) seen. Follow through from (a).

[2 marks]

\(\frac{{40}}{{120}}{\text{ }}\left( {\frac{1}{3}{\text{, }}0.333{\text{, }}33.3\% } \right)\)     (A1)(A1)(G2)

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

[2 marks]

\(\frac{{34}}{{120}}{\text{ }}\left( {\frac{{17}}{{60}}{\text{, }}0.283{\text{, }}28.3\% } \right)\)     (A1)(A1)(G2)

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

[2 marks]

\(\frac{8}{{28}}{\text{ }}\left( {\frac{2}{7}{\text{, }}0.286{\text{, }}28.6\% } \right)\)     (A1)(A1)(G2)

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

[2 marks]

customer satisfaction is independent of café     (A1)

Note: Accept “customer satisfaction is not associated with the café”.

[1 mark]

\(2\)     (A1)

[1 mark]

\(0.754\)     (G2)

Note: Award (G1)(G1)(AP) for \(0.75\) or for correct answer incorrectly rounded to 3 s.f. or more, (G0) for \(0.7\).

[2 marks]

since \({\chi ^2}_{calc} < {\chi ^2}_{crit}5.991 accept (or Do not reject) H₀     (R1)(A1)(ft)

Note: Follow through from their value in (e).

OR

Accept (or Do not reject) H₀ as \(p\)-value \((0.686) > 0.05\)     (R1)(A1)(ft)

Notes: Do not award (A1)(R0). Award the (R1) for comparison of appropriate values.

[2 marks]

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

A surprising number seemed unfamiliar with set notation in (e) and thus did not attempt this part.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

The chi-squared test was well done by the great majority, however, it was clear that a number of centres do not teach this subject, since there were a number of scripts which either were left blank or showed no understanding in the responses seen.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

The chi-squared test was well done by the great majority, however, it was clear that a number of centres do not teach this subject, since there were a number of scripts which either were left blank or showed no understanding in the responses seen.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

The chi-squared test was well done by the great majority, however, it was clear that a number of centres do not teach this subject, since there were a number of scripts which either were left blank or showed no understanding in the responses seen.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

The chi-squared test was well done by the great majority, however, it was clear that a number of centres do not teach this subject, since there were a number of scripts which either were left blank or showed no understanding in the responses seen.