Mark Scheme June 2019

Mark Scheme June 2019 Pearson LCCI Certificate in Business Statistics Level 3 (ASE20100)

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General Marking Guidance

• All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last.

• Mark schemes should be applied positively. Candidates must be rewarded

for what they have shown they can do rather than penalised for omissions.

• Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie.

• There is no ceiling on achievement. All marks on the mark scheme should

be used appropriately.

• All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.

• Where some judgement is required, mark schemes will provide the

principles by which marks will be awarded and exemplification may be limited.

• When examiners are in doubt regarding the application of the mark

scheme to a candidate’s response, the team leader must be consulted.

• Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

• For calculation questions full marks can be awarded where correct answer is seen with no workings shown.

• For calculation questions where own figure rule is permitted, marks for accuracy should only be awarded where appropriate method is shown.

Types of mark o M marks: method marks o A marks: dependent accuracy marks o B marks: independent accuracy marks

Abbreviations

o cao – correct answer only o ft – follow through (own figure)

No working If no working is shown then correct answers normally score full marks. If no working is shown then incorrect (even though nearly correct) answers score no marks. With working If there is a wrong answer indicated on the answer line always check the working in the body of the script (and on any diagrams), and award any marks appropriate from the mark scheme. Any case of suspected misread loses A marks on that part, but can gain the M marks. If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makes clear the method that has been used. If there is no answer on the answer line then check the working for an obvious answer. Own figure (or follow-through) marks Own figure marks which involve a single stage calculation can be awarded without working since you can check the answer yourself, but if ambiguous do not award. Own figure marks which involve more than one stage of calculation can only be awarded on sight of the relevant working, even if it appears obvious that there is only one way you could get the answer given. Ignoring subsequent work It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question, e.g. incorrect cancelling of a fraction that would otherwise be correct. It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect, e.g. algebra.

Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer line; mark the correct answer. Probability Probability answers must be given as fractions, percentages or decimals. If a candidate gives a decimal equivalent to a probability, this should be written to at least 2 decimal places (unless tenths).

Incorrect notation should lose the accuracy marks, but be awarded any implied method marks.

If a probability answer is given on the answer line using both incorrect and correct notation, award the marks.

If a probability fraction is given then cancelled incorrectly, ignore the incorrectly cancelled answer.

Linear equations Full marks can be gained if the solution alone is given on the answer line, or otherwise unambiguously indicated in working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified as the solution, the accuracy mark is lost but any method marks can be awarded. Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.

Question Number

Answer (A02) 4 (A03) 1 Mark

1(a)(i) Frequency densities: 1.5, 2.0, 3.0, 2.8, 1.6, 0.4 (or multiple, eg 15, 20, 30, 28, 16, 4)

M1 A1 Horizontal position B1 Vertical position A1ft Linear scales /horizontal labelled axis B1 (A03) (5)

Question Number

Answer (A03) 1 Mark

1(a)(ii) (The number of hours absent from work is) positively skewed

B1ft (1)

Additional guidance Also accept “Skewed to the right”.

Question Number

Answer (A02) 3 Mark

1(a)(iii) Midpoints 5 17.5 37.5 62.5 87.5 125 {(5x15)+(17.5x30)+(37.5x75)+(62.5x70)+(87.5x40)+(125x20}/250 Mean=13787.5/250 = 55.15 hours (allow 55.1 or 55.2)

M1 M1 A1 (3)

Histogram

30

20

10

0160140120100

25

80

5

60

15

0 4020

Adju

sted

freq

uenc

y

Number of hours absent from work

Question Number


1(b) (i) H0: There is no reduction in the mean number of hours absent from work after the Health and Wellbeing programme started H1: There is a reduction in the mean number of hours absent from work after the Health and Wellbeing programme started Differences: 6, -8, -3, 12, 15, 6, -9, -2, 4, 9 Mean difference = 3 Stdev difference = √{606/9} = 8.206 (7.785 if n used) Test statistic = 3/{8.206/√10} = 1.16 Degrees of freedom = 9 Critical value = 1.83 Accept H0

B1 (A03) B1 B1 M1 A1 B1 B1ft (A03) (7)

Question Number

Answer (A04) 1 Mark

1(b) (ii) There is not a significant reduction in the mean number of hours absent from work after the Health and Wellbeing programme started.

B1ft (1)

Question Number

Answer (A03) 1 Mark

1(b)(iii) The data were naturally paired, because the before and after values were taken (from the same subjects).

B1 (1)

Question Number

Answer (A01) 1 Mark

1(c) (i) Type II error is the error where we accept H0 (reject H1) when H1 is true (H0 is false).

B1 (1)

Question Number

Answer (A03) 1 Mark

1(c) (ii) As we accepted H0, then we may have made a type II error.

B1ft (1)

Total for Question 1 = 20 marks

Question Number


2(a)(i) H0: There is no association between contact method and attendance at the sales event H1: There is an association between contact method and attendance at the sales event Degrees of freedom (r-1)(c -1) = 2 Critical value chi-squared (χ2) = 5.99 As 8.94 > 5.99 Reject H0 or Accept H1 at the 5% level of significance.

B1(A03) B1 B1ft(A03) (3)

Question Number

Answer (A04) 1 Mark

2(a)(ii) There is a significant association between contact method and attendance at the sales event.

B1ft (1)

Question Number

Answer (A02) 2 Mark

2(a)(iii) Proportions attending the sales event: Letter: 35% Email: 25% Text: 20% (As proportions not asked for directly, approximations may be given) Hence using a letter seems to give the highest proportion attending

B1 B1ft (2)

Question Number

Answer (A02) 2 Mark

2(b)(i) p = 131/500 = 0.262

M1 A1 (2)

Question Number

Answer (A02) 3 Mark

2(b)(ii) Standard error = √{0.262x0.738/500} = 0.01966 95% confidence interval is 0.262 ± 1.96 x 0.01966 = (0.223, 0.301) Allow (0.224, 0.300) or any value between

M1 M1 A1 (3)

Question Number

Answer (A02) 3 Mark

2(c) 2.6x √{0.262x0.738/n} < 0.03 n > 2.62 {0.262 x 0.738}/0.032 n > 1452 (allow answers in the range 1420 to 1460)

M1 M1 A1 (3)

Question Number


2(d) (i) H0: The mean car value = 6000 H1: The mean car value ≠ 6000 Degrees of freedom = 14 Critical values = ± 2.15

t

= (6350 – 6000)/(2000/√15) = 0.678 (allow 0.655 if n-1 used) Accept H0

B1 (A03) B1 M1 A1 B1ft A03) (5)

Question Number

Answer (A04) 1 Mark

2(d) (ii) There is insufficient evidence to reject the Sale Manager’s claim (that the average value of a used car is $6000 and so it might be correct).

B1ft (1)


Question Number

Answer (A03) 2 Mark

3(a)

Any two from: There may be no sample frame available High cost Long time (due to geographical spread) Not representative oe

B1 B1 (2)

Question Number

Answer (A03) 2 Mark

3(b)(i) Sample 40 from each age group at convenience OR not randomly

B1 B1 (2)

Question Number

Answer (A01) 2 Mark

3(b)(ii) Advantage: One from More convenient, quicker, cheaper No sample frame needed More representative Disadvantage: One from Lack of randomness means the results cannot be used in statistical inference (so sampling error cannot be measured) Possible biased sample because not selected at random

B1 B1 (2)

Question Number

Answer (A03) 2 Mark

3(c)

(Divide the nation into areas and take a sample of areas) Choose a sample of charity shops from each area, then sample customers from within the selected charity shops

B1 B1 (2)

Question Number

Answer (A03) 1 Mark

3(d)(i) There is decreasing trend oe

B1 (1)

Question Number

Answer (A03) 2 Mark

3(d)(ii) Any two from e.g.

• Sales increase from quarter 3 to quarter 4 oe • highest sales in quarter 4 • lowest sales in quarter 1

B1 B1 (2)

Question Number

Answer (A02) 3 Mark

3(e) Q1 (4320 + 2x4960 + 2x3560 + 2x4120 + 3800)/8 = $4175 Q2 (4960 + 2x3560 + 2x4120 + 2x3800 + 4400)/8 = $4040

M1 A1 A1 (3)

Question Number

Answer (A02) 4 Mark

3(f)(i) Trend increase = (4040 – 4700)/7 = -94.3 Estimated trend = 4040 – 3x94.3 = 3757 Estimated sales revenue = 3757 x 0.86 = $3231

M1 M1 M1 A1 (4)

Question Number

Answer (A03) 2 Mark

3(f)(ii) The forecast is in the future and so it is possible that there is a change in economic conditions…to influence the actual value… However, it is only just in the future and you might expect the forecast to be fairly reliable.

B1 B1 (2)


Question Number


4 (a) (i) H0: There is no significant correlation between the quality score and number made H1: There is a significant correlation between the quality score and number made Degrees of freedom (n-2) = 10 critical t = 2.23 Calculated t = r√(n-2) √(1-r² ) = (0.68)(√10) √(1-0.682) t = 2.93 Reject H0

B1(A03) B1 M1 A1 B1ft(A03) (5)

Question Number

Answer (A04) 1 Mark

4 (a)(ii) There is evidence that the product moment correlation coefficient is different from zero…( or there is evidence of significant correlation between quality score and number made)

B1ft (1)

Question Number

Answer (A03) 2 Mark

4 (b) The correlation coefficient is positive which indicates that when the number made increases then the quality increases (so the Production Manager’s belief is clearly not true)

B1 B1 (2)

Question Number

Answer (A02) 4 Mark

4 (c) Slope = 30/10 = 3 Intercept = 50 – 6x3 = 32 The least squares regression equation is y = 32 + 3x

B1 M1 A1 B1ft (4 )

Question Number

Answer (A02 ) 5 (A03) 2 Mark

4 (d)(i) H0: There is no difference in the mean completion time between the two age groups H1: There is a difference in the mean completion time between the two age groups Standard error = √{5.32/35 + 6.22/40} = 1.328 Test statistic = (79.6 – 75.2)/1.328 = 3.31 Critical value = 2.60 (or 2.58, 2.575) Reject H0

B1(A03) M1 A1 M1 A1 B1 B1ft(A03) (7)

Question Number

Answer (A04) 1 Mark

4(d) (ii) There is a significant difference in the mean completion time between the two age groups

B1ft (1)


Question Number

Answer (A02) 3 Mark

5(a)(i)

σµ−×

=Z

P(X>24.7) = P{Z>(24.7-25)/0.4} = P{Z>-0.75} = 0.773

M1 A1 A1 (3)

Question Number

Answer (A02) 2 Mark

5(a)(ii) P{Z<1.25} = 0.8944

M1 A1 (2)

Question Number

Answer (A02) 4 Mark

5(b) Batch mean = 16 x 25 = 400 Batch standard deviation = 0.4 x √16 = 1.6 P{Z>(396-400)/1.6} = P{Z>-2.5} = 0.994

B1 B1 M1 A1 (4)

Question Number

Answer (A02) 3 Mark

5(c) Mean = 25.2 g 1.645s = 0.5 s = 0.5/1.645 = 0.304 g (±0.001)

B1 M1 A1 (3)

Question Number


5(d)(i) H0: The proportion of faulty containers remains at 0.08 H1: The proportion of faulty containers does not remain at 0.08 Critical value = 1.96 Test statistic = (12/80 – 0.08)/√{0.08x0.92/80} = 2.31 Reject H0

B1(A03) B1 M1 A1 B1ft (A03) (5)

Question Number

Answer (A04) 1 Mark

5(d)(ii) The proportion of faulty items manufactured by the new machine is significantly different to that of the current machine.

B1ft (1)

Question Number

Answer (A03) 2 Mark

5(e) The critical value is now 2.33 So the conclusion has changed and we conclude that the new machine’s proportion of faulty items is not higher than that of the current machine.

B1 B1ft (2)


TOTAL FOR EXAMINATION = 100 marks

Mark Scheme June 2019

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