| Date | November 2017 | Marks available | 2 | Reference code | 17N.3sp.hl.TZ0.2 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Justify and Interpret | Question number | 2 | Adapted from | N/A |
Question
Anne is a farmer who grows and sells pumpkins. Interested in the weights of pumpkins produced, she records the weights of eight pumpkins and obtains the following results in kilograms.
\[{\text{7.7}}\quad {\text{7.5}}\quad {\text{8.4}}\quad {\text{8.8}}\quad {\text{7.3}}\quad {\text{9.0}}\quad {\text{7.8}}\quad {\text{7.6}}\]
Assume that these weights form a random sample from a \(N(\mu ,{\text{ }}{\sigma ^2})\) distribution.
Anne claims that the mean pumpkin weight is 7.5 kilograms. In order to test this claim, she sets up the null hypothesis \({{\text{H}}_0}:\mu = 7.5\).
Determine unbiased estimates for \(\mu \) and \({\sigma ^2}\).
Use a two-tailed test to determine the \(p\)-value for the above results.
Interpret your \(p\)-value at the 5% level of significance, justifying your conclusion.
Markscheme
UE of \(\mu \) is \(8.01{\text{ }}( = 8.0125)\) A1
UE of \({\sigma ^2}\) is 0.404 (M1)A1
Note: Accept answers that round correctly to 2 sf.
Note: Condone incorrect notation, ie, \(\mu \) instead of UE of \(\mu \) and \({\sigma ^2}\) instead of UE of \({\sigma ^2}\).
Note: M0 for squaring \(0.594 \ldots \) giving 0.354, M1A0 for failing to square \(0.635 \ldots \)
[3 marks]
attempting to use the \(t\)-test (M1)
\(p\)-value is 0.0566 A2
Note: Accept any answer that rounds correctly to 2 sf.
[3 marks]
\(0.0566 > 0.05\) R1
we accept the null hypothesis (mean pumpkin weight is 7.5 kg) A1
Note: Apply follow through on the candidate’s \(p\)-value.
Note: Do not award A1 if R1 is not awarded.
[2 marks]
