# G242 G G * * MEI STATISTICS Statistics 2 (Z2) ADVANCED SUBSIDIARY GCE. Wednesday 9 June 2010 Afternoon. Duration: 1 hour 30 minutes.

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2 2 1 A birdwatcher has learned to recognise different species of birds by their song. He notices that three particular types of warbler regularly found in his patch prefer to sing in trees. He decides to investigate whether there is any association between the type of warbler and the type of tree from which they are heard singing. 200 warblers, regarded as a random sample, are selected and the numbers of warblers in each category are summarised in the table below. Tree Willow Birch Oak Chiffchaff Warbler Willow Warbler Whitethroat (i) A test is to be carried out to examine whether these data provide any evidence of an association between these classification factors. State clearly the null and alternative hypotheses. The following tables show some of the expected frequencies and contributions to the test statistic. Calculate the remaining expected frequencies and contributions. Carry out the test at the 5% level of significance. [11] Expected frequencies Tree Willow Birch Oak Chiffchaff Warbler Willow Warbler Whitethroat Contributions to the test Tree statistic Willow Birch Oak Chiffchaff Warbler Willow Warbler Whitethroat (ii) For each type of warbler, comment briefly on how its distribution compares with what would be expected if there were no association. [3] (iii) While out for a walk, the birdwatcher hears the song of a whitethroat. Use the given data to estimate the probability that it is singing from a birch tree. [2] OCR 2010 G242 Jun10

3 3 2 A doctor working in a hospital in a poor area of a large city is concerned about the low average birth weight of babies born in the hospital. For babies born in this hospital, the mean birth weight is 2800 grams, which is well below the ideal birth weight. The doctor introduces an extensive prenatal care programme in an attempt to increase the mean birth weight. Following the introduction of the programme, the doctor measures the birth weight of each of a random sample of 12 babies born in the hospital, with results in grams as follows (i) Explain why, in this situation, it would not be appropriate to carry out a hypothesis test for a population mean using the Normal distribution. State the assumption necessary for a test based on the t distribution to be valid. [3] (ii) Use these data to estimate the population mean and the population standard deviation. [3] (iii) Use a t test to examine at the 5% significance level whether this sample provides evidence that the prenatal care programme has been successful in increasing the mean birth weight of babies born in this hospital. State your null and alternative hypotheses clearly. [10] 3 A regional highway authority is concerned about the high numbers of accidents involving cyclists at roundabouts. A random sample of 150 roundabouts is selected, and the number of accidents involving cyclists at each of these roundabouts over a four-week period is recorded. The results are shown in the following frequency table. Number of accidents, x Observed frequency, f (i) The sample standard deviation is 1.734, correct to 3 decimal places. (A) Verify that the sample mean number of accidents is 2.4. [2] (B) Do these statistics give you any reason to doubt the belief that the number of accidents may be modelled using a Poisson distribution? Justify your answer. [2] (ii) The highway authority wishes to carry out a test of the goodness of fit of the Poisson model. The sample mean of 2.4 is used as an estimate of the mean of the underlying population. The following tables show some of the expected frequencies and corresponding contributions to the test statistic. Use the appropriate cumulative probability tables to find the remaining expected frequencies, and calculate the remaining contributions. Carry out the test at the 5% level of significance. [10] Expected frequencies Number of accidents, x Expected frequency Contributions to the test statistic Number of accidents, x Contribution OCR 2010 G242 Jun10 Turn over

5 GCE Statistics (MEI) Advanced Subsidiary GCE G242 Statistics 2 (Z2) Mark Scheme for June 2010 Oxford Cambridge and RSA Examinations

6 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 2010 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone: Facsimile:

7 G242 Mark Scheme June 2010 Q1 (i) H 0 : there is no association between warbler and tree H 1 : there is an association between warbler and tree Expected frequencies Willow Birch Oak Chiffchaff Willow Warbler Whitethroat M1 (ii) Willow Birch Oak Chiffchaff Willow Warbler Whitethroat X 2 = degrees of freedom Critical value for 5% significance level is As > the result is significant There is evidence of an association between the warbler and tree. Chiffchaffs occurred more frequently than expected in Oak trees. Willow Warblers occurred less frequently than expected in Oak trees. Whitethroat occurred more or less as expected. M1 M1 11 E1 E1 E1 3 (iii) P(Birch Whitethroat) = 20/63 M

8 G242 Mark Scheme June 2010 Q2 (i) (ii) This is a small sample The variance is unknown We must assume birth weights are Normally distributed Estimate for population mean = 2965 g 3 Estimate for population standard deviation = = = 316 to 3 sf (iii) H 0 : μ = 2800 & H 1 : μ > 2800 Where μ represents the population mean birth weight of babies born after the introduction of the prenatal care programme t = = (using SD = 316) SD degrees of freedom At 5% level, critical value of t is > so the result is significant. Evidence suggests the mean birth weight has increased. 2 M1 CAO M1 CAO M

9 G242 Mark Scheme June 2010 Q3 (i)a fx f = (= 2.4 A.G.) M1 2 B Variance = = , which seems close to the mean value of 2.4. A Poisson model may be appropriate. E1(compare mean with variance allow arguments either way, with relevant conclusion) 2 (ii) H 0 : The Poisson model is suitable P(X = 1) = & P(X 6) = (both probabilities) Missing expected frequencies are (x = 1), and (x 6) Missing contributions are (x = 2) and (x = 3) X 2 = There are = 5 degrees of freedom. At the 5% significance level the critical value is The result is significant Evidence suggests that the Poisson model is inappropriate. M1 (expected freq) M Q4 H 0 : population median = 210 H 1 : population median Actual differences Associated ranks T - = = 11 T + = = 44 T = 11 From n = 10 tables at the 5% level of significance in a two-tailed Wilcoxon single sample test, the critical value of T is 8 M1 M1 (use of n = 10 in tables) 6 11 > 8 the result is not significant The evidence does not suggest that there is a difference between the median dive duration of adolescent seals and the seal population as a whole. M1 E

10 G242 Mark Scheme June 2010 Q5 (i) P(X < 500) = P(Z < ) = P(Z <-1.550) Φ(1.550) = = (awrt 0.061) (ii) From tables Φ -1 ( 0.99 ) = μ = µ = = 503 M1 standardising M1 correct tail 3 for seen M1 for equation in µ and negative z-value 3 (iii) 9.05 ± 1.96 (9.03, 9.07) centred on 9.05 for 1.96 M1 structure 5 (iv) As the lower limit of the interval in part (iii) is more than 9 gallons, this does not suggest that the mean volume is below 9 gallons for this month. Allow sensible alternatives E1 E

11 OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge C 2EU OCR Customer Contact Centre Qualifications (General) Telephone: Facsimile: For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; 1 Hills Road, Cambridge, C 2EU Registered Company Number: OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 2010

12 Report on the Unit taken in June 2010 Chief Examiner s Report The Principal Examiners' reports that follow discuss the candidates' performances on the individual modules. There is one matter that should be discussed in a general way as it applies to all the statistics modules. This is in respect of arithmetical accuracy in intermediate working and in quotation of final answers. Most candidates are sensible in their arithmetical work, but there is some unease as to exactly what level of accuracy the examiners are expecting. There is no general answer to this! The standard rubric for all the papers sums the situation up by including "final answers should be given to a degree of accuracy appropriate to the context". Three significant figures may often be the norm for this, but this always needs to be considered in the context of the problem in hand. For example, in quoting from Normal tables, some evidence of interpolation is generally expected and so quotation to four decimal places will often be appropriate. But even this does not always apply quotations of the standard critical points for significance tests such as 1.96, 1.645, (maybe even 2.58 but not 2.57) will commonly suffice. Talking now in general terms, the examiners always exercise sensible discretion in cases of small variations in the degree of accuracy to which an answer is given. For example, if 3 significant figures are expected (either because of an explicit instruction or because the general context of a problem demands it) but only 2 are given, a candidate is likely to lose an Accuracy mark; but if 4 significant figures are given, there would normally be no penalty. Likewise, answers which are slightly deviant from what is expected in a very minor manner are not penalised (for example, a Normal probability given, after an attempt at interpolation, as whereas was expected). However, there are increasing numbers of cases where candidates give answers which are grossly over- or under-specified, such as insistence that the value of a test statistic is (say) merely because that is the value that happens to come off the candidate's calculator. Such gross over-specification indicates a lack of appreciation of the nature of statistical work and, with effect from the January 2011 examinations, will be penalised by withholding of associated Accuracy marks. Candidates must however always be aware of the dangers of premature rounding if there are several steps in a calculation. If, say, a final answer is desired that is correct to 3 decimal places, this can in no way be guaranteed if only 3 decimal places are used in intermediate steps; indeed, it may not be safe to carry out the intermediate work even to 4 decimal places. The issue of over-specification may arise for the final answer but not for intermediate stages of the working. It is worth repeating that most candidates act sensibly in all these respects, but it is hoped that this note may help those who are perhaps a little less confident in how to proceed. 1

13 Report on the Unit taken in June 2010 G242 Statistics 2 General comments This year saw another small entry, similar in size to last year. The majority of this year s candidates were very well prepared and many high marks were produced. Overall, the candidates demonstrated very good understanding of the statistical methods required and communicated their responses using appropriate statistical terms and in sufficient detail. The parts of questions requiring candidates to interpret information, explain or comment were not as well answered as the parts involving calculation. Some candidates lost marks through incorrect use of their calculator; there were several cases where a correct method was seen but the final answer did not match what was written. Problems identifying the correct number of degrees of freedom were again common. Comments on individual questions 1) (Chi-squared test for Association) (i) (ii) (iii) Some candidates mixed up the hypotheses, leading to contradictory conclusions and loss of marks. Some candidates did not include context in either their hypotheses or in their concluding remarks. A few slips with degrees of freedom were seen and incorrect critical values were fairly common. It is expected that candidates should state the number of degrees of freedom used some did not and were penalised. This part was poorly understood. Few candidates showed an understanding of the link between the size of the contribution to the test statistic and the level of association. For willow warblers and chiffchaffs, candidates were expected to identify the cells containing relatively large contributions and comment whether this provided evidence that the warblers were seen more frequently or less frequently than expected in the corresponding tree. For whitethroats, the candidates were expected to comment that the small contributions indicated that they occurred in numbers that would be expected if there were no association between warbler and type of tree. This too was poorly answered. Generally, candidates interpreted the question incorrectly, not realising the importance of the condition that the bird heard was a whitethroat. A small number reversed the question, finding the probability that the bird was a whitethroat given that it was singing from a birch tree. 2) (Hypothesis test using the t distribution) (i) (ii) This required an understanding of the differences between the situations leading to hypothesis tests based on the Normal distribution and the t distribution. In general, this was not well answered. Several candidates did not comment on the assumption necessary for a t test to be valid. Confusion between population and sample was evident. This required candidates to provide estimates for population mean and population standard deviation. This led to full marks in most cases. 6

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