In the past, the mean time taken by Freda for a particular daily journey was 39.2 minutes. Following the introduction of a one-way system, Freda wishes to test whether the mean time for the journey has decreased. She notes the times, t minutes, for 40 randomly chosen journeys and summarises the results as follows. n = 40 Σt = 1504 Σt² = 57760
- (a)Calculate unbiased estimates of the population mean and variance of the new journey time.[3]
- (b)Test, at the 5% significance level, whether the population mean time has decreased.[5]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.
- [B1] est (µ) = 1504/40 = 37.6 or 188/5
- [M1] est (σ²) = [1/(40-1)] * [57760 - (1504^2)/40]; Correct substitution in any correct formula [57760 - 1504^2/40]
- [A1] = 31.(0) (3 sf); Accept 2016/65 or 31 1/65
- [B1] H₀: Pop mean (or µ) = 39.2; H₁: Pop mean (or µ) < 39.2; Both. Not just 'mean'
- [M1] ('37.6' - 39.2) / sqrt('31.0154' / 40); Allow use of biased variance (30.2), must have √40
- [A1] = -1.817; SC FT use of biased = -1.840 for A1
- [M1] '1.817' > 1.645 OE; Valid comparison 'their 1.817' with 1.645 or valid area comparison 0.0346 < 0.05 OE
- [A1FT] There is evidence that mean time has decreased; FT their 1.817; in context, not definite, no contradictions; SC For 2 tail test: H₁: µ ≠ 39.2 and comp 1.96, max B0M1A1M1A0 (no FT for final mark)
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →