| Date | November 2014 | Marks available | 2 | Reference code | 14N.1.sl.TZ0.10 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Calculate | Question number | 10 | Adapted from | N/A |
Question
Minta surveyed students from her school about their preferred morning snack from a choice of an apple, a fruit salad or a smoothie.
She surveyed 350 students, of whom 210 are female.
She performed a \({\chi ^2}\) test at the 5% significance level to determine whether there is a relationship between the choice of morning snack and gender.
State Minta’s null hypothesis.
State the number of degrees of freedom.
150 students showed a preference for a smoothie.
Calculate the expected number of female students who chose a smoothie.
Minta found that the calculated value of the \({\chi ^2}\) test was 3.576. The critical value at the 5% significance level is \(5.99\).
State Minta’s conclusion. Give a reason for your answer.
Markscheme
\({{\text{H}}_0}\): Choice of morning snack is independent of (not dependent on) gender. (A1) (C1)
Note: Accept there is “no association” between snack chosen and gender.
Do not accept “not related” or “not correlated” or “influenced”.
\(2\) (A1) (C1)
\(\frac{{210 \times 150}}{{350}}\) (M1)
Note: Award (M1) for correct substitution in the correct formula.
\( = 90\) (A1) (C2)
Null hypothesis is accepted (not rejected). (A1)
OR
Choice of morning snack is independent of gender (A1)
\(3.576 < 5.99\;\;\;{\mathbf{OR}}\;\;\;\chi _{{\text{calc}}}^2 < \chi _{{\text{crit}}}^2\) (R1) (C2)
Note: Do not award (A1)(R0).
