| Date | May 2018 | Marks available | 4 | Reference code | 18M.1.hl.TZ0.14 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Calculate | Question number | 14 | Adapted from | N/A |
Question
At an early stage in analysing the marks scored by candidates in an examination paper, the examining board takes a random sample of 250 candidates and finds that the marks, \(x\) , of these candidates give \(\sum {x = 10985} \) and \(\sum {{x^2} = 598736} \).
Calculate a 90% confidence interval for the population mean mark μ for this paper.
The null hypothesis μ = 46.5 is tested against the alternative hypothesis μ < 46.5 at the λ% significance level. Determine the set of values of λ for which the null hypothesis is rejected in favour of the alternative hypothesis.
Markscheme
\(\bar x = 43.94\) (A1)
unbiased variance estimate = 466.0847 (A1)
Note: Accept sample variance = 464.2204.
⇒ 90% confidence interval is (41.7,46.2) A1A1
[4 marks]
Z-value is −1.87489 or −1.87866 (A1)
probability is 0.0304 or 0.0301 (A1)
⇒ λ ≥ 3.01 (M1)A1
[4 marks]
