S5.3 Test for a Normal mean

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (b) most students stated their hypotheses correctly in terms of with very few using a one-tailed test. To score the method mark in this type of question the students simply need to select and write down the correct model (distribution). A number seemed to be listing numbers they needed for their calculator, but we expect to see the appropriate mathematical notation used. Fortunately, many obtained a correct probability and would therefore receive the method mark by implication. Some chose to standardise and find a z value (1.9088… which should be compared with 1.96) or a critical region

Common mistakes

• two-tailed: H1: μ ≠ 175.4, compare 0.028 with 0.025 (or z = 1.909 with 1.96)
• sample mean ~ N(175.4, 6.8²/52) under H0, not mean 177.2
• conclusion in context: insufficient evidence of a difference

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(𝑆̄ ⩾ 177.248. . . ) but given that a p-value was required in part (c) this was not the most efficient route. A common error in stating the model is to use an incorrect value for the 2 6.8 mean. On this occasion we condoned for the M1 and first A1 but ~ N 177.2, S 52 penalised the final accuracy mark. In a hypothesis test the probability (or critical region) is always calculated assuming H0 is true and so the mean should be 175.4 and the probability calculated is ( ) S Some students stopped after stating that the result P 177.2 . was not significant or not rejecting H0 but a good number gave a correct conclusion in context. Part (c) was not answered well with only a few scoring this mark. The common error was to use 0.028 or 0.025 but others used values like 175.4

Common mistakes

• two-tailed p-value = 2 × 0.0281 = 0.056
• common errors 0.028 or 0.025

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

rather than P(160 < F < 175). Part (d) was a familiar hypothesis test and many students scored something here. The hypotheses were usually stated in terms of mu and most of the time 166.5 rather than 164.6 was used. Selecting the correct model caused a number of problems: some used 164.6 as the mean and others used 7.4 (or occasionally 6.1) as the standard deviation, simply stating the correct model will score the method mark here but a number of students didn’t show us the model they were using and we had to award this mark by implication from correct working. Most students successfully used the probability approach (as we would recommend) with a few attempting a critical region or using a z value and comparing with 1.6449 from the tables. The final mark required a correct interpretation of their calculation in context and some students simply stopped after stating that the result was significant and failed to score this final mark.

Common mistakes

• H₀: μ = 166.5, H₁: μ < 166.5
• X̄ ~ N(166.5, 7.4²/50), not mean 164.6 or s.d. 7.4 or 6.1
• P(X̄ < 164.6) = 0.0347 < 0.05
• conclusion in context about Mia's belief

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

It was disappointing that over 20% of candidates scored no marks here as there were some straightforward marks available to those who had studied this part of the students could use their calculator efficiently to score this mark. Part (b) was a fairly standard hypothesis test and a good number made progress here. Some still did not use 10 T = mu for the hypotheses and a null hypothesis of was all too common. Most attempts used a normal distribution with a correct variance but a significant number of students used a mean of 11.5 rather than 10. This is not a correct method for a hypothesis test (the probabilities should be calculated on the assumption that H0 is true) but, on this occasion, for those who still managed to reach a correct conclusion a special case enabled them to secure one mark. A variety of different approaches were used in part (b), some finding the critical value for X and others calculating the test statistic of 1.677… but some of these candidates compared this value to the significance level rather than the critical value of 1.6669 from the tables. The first part of (c) was another

Common mistakes

• hypotheses in terms of μ: H₀: μ = 10, H₁: μ > 10
• sample mean ~ N(10, 4²/20) – use μ = 10, not 11.5
• P = 0.0468 < 5% (or z = 1.677 > 1.6449)
• conclusion in context about the complaint

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question as whole proved to be one of the more challenging questions on the paper with many candidates unable to make any meaningful attempt at any part of the question. Part (c) was a standard hypothesis test for a sample mean from a Normal distribution that should have been relatively straight forward but in many cases candidates struggled. It was common for candidates to use the wrong parameter when defining the null and alternative hypotheses; they need to be fully aware that they must use the standard parameters when setting up hypotheses, in this case 𝜇. For those who did make progress failing to divide the variance by n was a common error and candidates often gave large probabilities that were nowhere near 0.05 which should be a warning sign that they likely have a mistake in their working. Some used 0.16 or 𝜇= 24.94. Those using a critical regions approach were rare and these candidates often had issues when rounding the critical region to 2 decimal places which meant it was equal to the test statistic. The majority of candidates made an attempt to interpret their results and give a contextual conclusion. Those candidates that opted to give their contextual conclusion in terms of ‘Hannah’s belief’ were usually more successful than those who tried to write a statement about the mean amount of liquid being less than 25ml. Some candidates wrote ‘the mean is less than 25’ which does not mention the context and was therefore not acceptable.

Common mistakes

• H0: μ = 25, H1: μ < 25 (μ not other symbols)
• D̄ ~ N(25, 0.16²/20) – divide the variance by n
• p = 0.047 < 0.05 (or z = −1.677 < −1.6449)
• conclusion in context: evidence to support Hannah's belief

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Whilst parts (a) and (d) were often answered well, parts (b) and (c) proved much more challenging and only the strongest 10% made significant progress here. Part (a) was answered well and most two new batteries as well. Part (d) was often answered quite well with many stating the hypotheses correctly and identifying the correct model. Most calculated the appropriate probability correctly and many gave their solution in context too. SECTION B: MECHANICS Introduction Overall the quality of the scripts seemed very mixed with some clear and entirely correct solutions but a substantial number were well below standard, particularly for Question 9 and Question 10. There may have been some evidence of time being a limiting factor as some Question 10 answers seemed rushed or unfinished, although it is difficult to be sure whether time or ability was the main issue here. Question 7 was the best answered with 54% of the students scoring full marks, closely followed by question 6 where 55% of students scored full marks. Question 9 was by far the most challenging where the modal mark was zero and just under half the students scored 6, or fewer, out of 13. In calculations the numerical value of g which should be used is 9.8 unless otherwise stated. Final answers should then be given to 2 (or 3) significant figures – more accurate answers will be penalised, including fractions but exact multiples of g are usually accepted. If there is a printed answer to show then students need to ensure that they show sufficient detail in their working to warrant being awarded all of the marks available. In all cases, as stated on the front of the question paper, students should show sufficient working to make their methods clear to the examiner and correct answers without working may not score all, or indeed, any of the marks available. If a student runs out of space in which to give his/her answer than he/she is advised to use a supplementary sheet.

Common mistakes

• H0: μ = 18, H1: μ > 18 (in terms of μ)
• sample mean ~ N(18, 4²/20)
• P(mean > 19.2) = 0.0899 > 0.05 (or CR mean > 19.5, z = 1.34 < 1.6449)
• not significant, insufficient evidence for Alice's belief, in context