Technical University of Denmark

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1 Technical University of Denmark Page 1 of 17 pages Written examination: 29 May 2009 Course name and number: Introduction to Statistics, Aids and facilities allowed: All The questions were answered by no.: There are 30 questions of the multiple chose type included in this exam. To answer the questions you need to fill in the table on the front page (this page), with the number of the answer you think is the correct one. A wrong answer can be corrected by blackening out the wrong answer and adding the correct answer. If there is doubt about the meaning of a correction or more that one number is given as the answer, the question is considered not answered. 5 points are obtained for a correct answer and -1 point is given for a wrong answer. If a question is unanswered or if the answer is 6 (for do not know) 0 points is given. The number of points corresponding to specific marks or needed to pass the examination is ultimately determined during censoring. Exercise I II III IV V VI VII VIII IX X Question Answer Exercise XI XII XII XIII XIV XV XVI XVI XVI XVII Question Answer Exercise XVIII XIX XX XX XX XXI XXII XXIII XXIV XXV Question Answer Remember to write your number on the questionnaire. The questionnaire contains 17 pages. Please check that your questionnaire contains them all. Multiple choice questions: Note that all suggested answers are not necessarily meaningful. In fact, some of them are very wrong. But under all circumstances there is one and only one correct answer to each question.

2 Exercise I For four different types of citrus trees the ratio between leaf area and dry weight of the leaves has been measured. All the studied trees have grown in full sunlight. The results were: Lemon Orange Grape fruit Tangerine Question I.I (1): The ratio between leaf area and dry weight is the same for the four different types of citrus trees can best be analyzed by means of: 1 A test with the binomial distribution 2 A test with the normal distribution 3 A test in the Poisson distribution 4 The sign test 5 The data cannot be satisfactorily analysed based on the given information Exercise II The following table shows reduction in weight of 8 subjects after a period of treatment with a new weight reducing drug. Subject no. Weight reduction (gram) Question II.I (2): Lower quartile (Q 1 ) and median (Q 2 ) weight loss are given as: 1 Q 1 = 2100 and Q 2 = Q 1 = 2345 and Q 2 = Q 1 = 2590 and Q 2 = Q 1 = 2316 and Q 2 = Q 1 = 2345 and Q 2 =

3 Exercise III The following table shows paired values of order size x (number of units) and time consumption y. It is assumed that the observations can be described by the model: y i = α + βx i + ε i, where the deviations ε i have mean 0 and variance σ. 2 ε Observation no. (i) Order size (x) Time consumption (y) Calculations of the data gives: Σ i x i = 550 and Σ i ( x i - x ) 2 = 7040 and a = and b = and s 2 e = Question III.I (3): A 95% confidence interval for the mean at x=50 is: 1 ( ) ± ( ) ± ( ) ± ( ) ± ± Exercise IV In the weekly publication for Danish medical doctors Ugeskrift for Læger February , appeared a paper with the main results of the investigation of: Differences between female and male alcohol abusers and differences in their need for treatment. In the paper the authors conclude among other things that among the alcohol abusers 21.4 percent of the females and 25.2 percent of the males have shown violent behavior. This difference was not regarded as significant in the investigation. Question IV.I (4): Suppose that 1000 males and 1000 females participated in a similar investigation would the difference between 21.4 percent and 25.2 percent become significant at a 5% significance level: 1 Yes, because the test statistic Z is larger than No, because the test statistic Z is larger than Yes, because the test statistic Z is less than No, because the test statistic Z is less than No, because the test statistic Z is less than

4 Exercise V In a Swedish study 16 well trained male runners received a new blood preparation for six weeks. The purpose was to investigate whether the new blood preparation gave an increase in haematocrite (haematocrite is a measure of the oxygen binding capacity of the blood). The 16 runners had their haematocrite measured at the beginning of the study (day 0) and after six weeks (day 42). The following results were: Subject Day 0 Day 42 Subject Day 0 Day Question V.I (5): Suppose the measurements of haematocrite are normally distributed, which of the following methods would be best suited for testing whether the blood preparation had an effect on the haematocrite: 1 The ordinary t-test 2 One way analysis of variance 3 Paired t-test 4 Rank sum test 5 Linear regression analysis Exercise VI It is assumed that there is a linear relation between the concentration of alcohol in blood and the number of driving errors during a given driving test in a closed training ground. Based on results of a large number of tests the correlation coefficient (r) between alcohol concentration and number of driving errors was estimated to be r = Question VI. I (6): Which one of the following statements is correct: 1 When the alcohol concentration increases the number of driving errors increases with r 2 2 When the alcohol concentration increases the number of driving errors increases 3 When the alcohol concentration increases with 1 unit, the number of driving errors increases with 41 percent 4 When the alcohol concentration increases with 0.41 unit, the number of driving errors increases with 1 5 When the alcohol concentration decreases the number of driving errors increases 4

5 Exercise VII In a study of internet purchase of music the following distribution of type of music and age of buyers was observed: Type Age >60 years Age years Age < 30 years Classical Jazz Pop Rock Question VII.I (7): The hypothesis of independence (H 0 ) between type of music and age of buyer is tested at significance level α = The value of the usual test statistic is This means that: 1 H 0 is accepted, because χ 2 (6) = H 0 is rejected, because χ (6) = H 0 is accepted, because χ 2 (6) = H 0 is rejected, becauseχ H 0 is accepted, becauseχ (11) = (11) = Exercise VIII In connection with a study of the strength of concrete 10 experiments were performed with each of two mixtures A and B. Water and cement is added and the mixing is continued until homogeneity is obtained. Perlite and fibers are added while the mixing is ongoing. However, the mixing time should not exceed 4 minutes to avoid crushing of the perlite. Mixture A 50 kg cement 27 kg Chalk (tpe S) 0.2 m 3 Perlite 2.2 kg Alkaline resistant glass fibres Mixture B 50 kg cement 50 kg Plast cement 0.1 m 3 Perlite 0.7 kg Polypropylen fibres Continues on page 6 5

6 Debased on the above 20 measurements of strength (10 with mixture A and 10 with mixture B) the following output from S-plus was obtained: OUTPUT 1 Pooled-Variance Two-Sample t-test data: x: Mixture.A in Exercise VIII, and y: Mixture.B in Exercise VIII t = , df = 18, p-value = alternative hypothesis: difference in means is not equal to 0 95 percent confidence interval: sample estimates: mean of x mean of y OUTPUT 2 Welch Modified Two-Sample t-test data: x: Mixture.A in Exercise VIII, and y: Mixture.B in Exercise VIII t = , df = , p-value = alternative hypothesis: difference in means is not equal to 0 95 percent confidence interval: sample estimates: mean of x mean of y OUTPUT 3 Pooled-Variance Two-Sample t-test data: x: Mixture.A in Exercise VIII, and y: Mixture.B in Exercise VIII t = , df = 18, p-value = alternative hypothesis: difference in means is less than 0 95 percent confidence interval: NA sample estimates: mean of x mean of y OUTPUT 4 Welch Modified Two-Sample t-test data: x: Mixture.A in Exercise VIII, and y: Mixture.B in Exercise VIII t = , df = , p-value = alternative hypothesis: difference in means is less than 0 95 percent confidence interval: NA sample estimates: mean of x mean of y Question VIII.I (8): Decide whether the hypothesis H 0, that the strength with mixture A is equal to the strength with mixture B should be rejected at 5 % significance level relative to H 1 that the strength with mixture A is less than the strength with mixture B: 1 No, because the p-value is Yes, because the t-value is negative 3 No, because the t-value is negative 4 Yes, because the p-value is No, because the p-value is

7 Exercise IX The distribution function of the discrete stochastic variable X is given in the following table: x F(x) Question IX.I (9): The mean value µ of X is: 1 µ = µ = µ = µ = µ = Exercise X In a large laboratory 80 randomly selected mice were analyzed for concentration of the stress hormones adrenaline, cortisol, indorfin and oxytocin in blood. It was found that 20 of the mice had an elevated concentration of the stress hormones. Question X.I (10): Indicate the 99 % confidence interval for the fraction of mice with elevated concentration of stress hormones: 1 I 0.99 = 0.25 ± (1 0.25) / 80 2 I 0.99 = 0.25 ± (1-0.25)/20 3 I 0.99 = 0.25 ± (1 0.25) / 80 4 I 0.99 = 0.25 ± (1 0.25) / 20 5 I 0.99 = 0.25 ± (1-0.25)/80 7

8 Exercise XI The sale of mobile phones is not only determined by technical specifications but rather by non-technical features of the phones. A manufacturer has on an experimental basis developed three phones (model I, II and III) with identical technical specifications but with different size and design of their push-buttons. During a trial period of one month all three types of phones are sold in four different geographical regions. The profit of the sales is shown in the following table. Region 1 Region 2 Region 3 Region 4 Model I 23,300 29,600 18,200 19,000 Model II 17,000 20,700 19,000 26,300 Model III 27,500 29,000 17,500 26,500 Question XI.I (11): Which of the following methods is best suited for analysis of possible differences in profit from sale of the three phone models? 1 Test of independence of cross-classified categorical data 2 One-way analysis of variance 3 Paired t-test 4 Randomized block experiment 5 Sign test Exercise XII A company controls continuously the quality of a product by means of a test procedure. A sample of 20 units is drawn, and the production is stopped if more than one defective unit is found. Question XII.I (12): Suppose that the probability of producing a defective unit is 15 %. Indicate the probability p that the test procedure mentioned above will result in a production stop: 1 p = p = p = p = p = Do not know Continues on page 9 8

9 Question XII.II (13): The probability of committing a type I error by using a test procedure is: 1 The probability of rejecting H 0 when H 0 is false 2 The probability of accepting H 1 when H 1 is false 3 The probability of rejecting H 0 when H 0 is true 4 The probability of rejecting H 1 when H 1 is true 5 The probability of accepting H 1 when H 0 is false Exercise XIII A company makes a certain product by mass production. In order to analyse how the percent defective units varies within the production sizes of 4,000 to 6,000 units, the number of defective units during a period was counted and the percent defective units was calculated. The results were as follows: Defective % Production size Defective % Production size Question XIII.I (14): If the objective is to analyse whether production size influences the percentage defective units this should be carried out by means of one of the following methods: 1 A usual t-test 2 A rank sum test 3 A paired t-test 4 Two-way analysis of variance 5 Linear regression analysis 9

10 Exercise XIV A production process makes elements with a nominal length of 90 mm. A random sample of 50 elements is drawn from the process, and the usual two-sided 95 % confidence interval for the mean length µ is calculated. It is found that the interval includes the value 90 mm. Question XIV.I (15): Consider the usual test for the hypothesis H 0 : µ = 90 mm against the alternative H 1 : µ 90 mm. Which one of the following statements is correct? 1 It is not possible to draw conclusions about H 0 on the basis of the confidence interval 2 The data is evidence for rejection of H 0 at 5% significance level 3 The hypothesis H 0 is rejected at 5 % significance level 4 The hypothesis H 0 cannot be rejected at 5% significance level 5 None of the above statements is correct. Exercise XV It may be of importance for certain types of toys that the packing is felt hard. A producer of toys wants to study whether there is a difference in the hardness between two types of packing A and B. Eight randomly selected customers are asked to evaluated the two types of packing and rate their hardness on a scale from 1 to 10. The data which cannot be assumed to follow a normal distribution are listed in the following table: Customer Packing A Packing B Question XV.I (16): Which one of the following methods would be appropriate for analysis whether the two packing are rated equal? 1 The sign-test 2 One-way analysis of variance 3 The t-test 4 The rank sum test 5 The F-test 10

11 Exercise XVI The results from a completely randomized experiment regarding decreases of blood sugar concentration at different concentrations of insulin are listed in the following output from S-plus. Previous experiments have shown that the insulin hormone which is produced by the pancreas gland decreases the blood sugar concentration. *** Linear Model *** Call: lm(formula = Bloodsugardecrease ~ Concentration, data = Spm17, na.action = na.exclude) Residuals: Min 1Q Median 3Q Max Coefficients: Value Std. Error t value Pr(> t ) (Intercept) Concentration Residual standard error: 40.8 on 12 degrees of freedom Multiple R-Squared: F-statistic: on 1 and 12 degrees of freedom, the p-value is Analysis of Variance Table Response: Bloodsugardecrease Terms added sequentially (first to last) Df Sum of Sq Mean Sq F Value Pr(F) Concentration VALUE MISSING Residuals Question XVI.I (17): Which test statistic (value) from the S-plus output above is relevant for testing whether there is a relation between the decrease in blood sugar and the insulin concentration? 1 Value= Value= Value= Value= Value= Continues on page 12 11

12 Question XVI.II (18): The fraction of the variation in the blood sugar decrease which can be explained by the variation in insulin concentration is according to the S-plus output: percent percent percent percent percent Question XVI.III (19): In the output Analysis of Variance Table the F-test statistic is missing. The missing F-value can be calculated as: 1 Test statistic F = Test statistic F = Test statistic F = Test statistic F = Test statistic F = Exercise XVII Cars arrive independently at a terminal according to a Poisson process with an intensity of 3 cars/minute. Question XVII.I (20): A car has just arrived at the terminal. What is the probability p that exactly 3 cars will arrive during the next 2 minutes? 1 p = p = p = p = p =

13 Exercise XVIII In connection with production of electrical circuits for high voltage installations it is important to determine the strongest way to fasten condensators. Three types of glues A, B and C are each tried 9 times, and the results were as follows (the higher the number the stronger adhesion): A A A A A A A A A B B B B B B B B B C C C C C C C C C Question XVIII.I (21): Suppose a statistical test at 5 % significance level is performed to assess whether there is differences between the adhesive strength of the three types of glue (A, B, and C), which of the following F-values should the appropriate test statistic be compared to? 1 F 0.05 (24, 3) 2 F 0.05 (25, 2) 3 F 0.05 (3, 24) 4 F 0.05 (2, 24) 5 F 0.05 (2, 25) Exercise XIX A cutting machine in a large slaughter-house is used for cutting sausages of suitable length. It is assumed that the weight X of a randomly selected sausage is normally distributed with mean µ = 200 gram and variance σ 2 = 10 2 gram 2. (X ~ N (200, 10 2 )). Question XIX.I (22): A sample of 10 sausages is randomly drawn for assessment of the average weight X. It is known that the average weight X is distributed as: 1 X ~ N (2000, 10 3 ) 2 X ~ N (2000, 10) 3 X ~ N (2000, 10 2 ) 4 X ~ N (200, 10 2 ) 5 X ~ N (200, 10) 13

14 Exercise XX Two different types of water filters are compared by measuring the average reduction of the impurities expressed as parts per million (ppm). Filter 1 was tested with 14 samples of water and filter 2 was tested with 11 water samples (25 samples in all). The following data were obtained: Number of observations Mean estimate Estimated variance s 2 Filter 1 = x Filter2 = x Question XX.I (23): A 95% confidence interval for the variance of the measurements with filter 1: 1 I 0.95 (σ 2 x ) = [ I 0.95 (σ 2 x ) = [ > σ 2 1 > σ 2 1 x > ] x > ] I 0.95 (σ 2 x ) = [ > σ 2 1 x > ] I 0.95 (σ 2 x ) = [ I 0.95 (σ 2 x ) = [ > σ 2 1 > σ 2 1 x > ] x > ] Question XX.II (24): The usual test statistic for testing whether the variance of filter 1 is significantly different from the variance of filter 2 becomes: 14 1 F = F = F = F = F = Continues on page 15 14

15 Question XX.III (25) It is now assumed that the variances of filter 1 and filter 2 are not significantly different. The best estimate for the common (pooled) variance s p 2 becomes: 1 s p 2 = s p 2 = s p 2 = s p 2 = s p 2 = Exercise XXI A manufacturer produces ropes for yachts which should have a breaking strength of more than one ton. In order to check this, 20 ropes were loaded until they broke. From these data the sample mean and variance is calculated as x = ton and s 2 = ton 2. The observations can reasonably be assumed to be independent and normally distributed. The manufacturer wants to test whether the above requirement for the breaking strength is fulfilled. Question XXI.I (26): How can the correct hypotheses H 0 and H 1 be stated in this case? 1 H 0 : µ = and H 1 : µ < H 0 : µ = and H 1 : µ > H 0 : µ > and H 1 : µ = H 0 : µ < and H 1 : µ = H 0 : µ = and H 1 : µ > Exercise XXII Consider a stochastic variable X which has a binomial distribution with n = 500 and p = 0.05, i.e. X ~ B(500, 0.05). Question XXII.I (27): Which one of the following distributions is useful as an approximation to B(500, 0.05): 1 N(µ = n / p, σ 2 = n /( 1 p) ) 2 N(µ = n p, σ 2 = n p( 1 p) ) 3 N(µ = n p, σ 2 = n p( 1 p) ) 4 P(λ = n p( 1 p) ) 5 P(λ = n /( 1 p) ) 15

16 Exercise XXIII The height of 772 men has been measured at an examination of army recruits. None was smaller than 165 cm none was higher than 2 meter. The heights of the men were recorded in intervals as shown in the following table. Height interval in cm Observed number of men It was of interest to study whether the height could be assumed to be normally distributed. Therefore the sample mean and variance were estimated based on the midpoints of the height intervals, and the test statistic Q was calculated. All e i > 5. Question XXIII.I (28): Which of the following distributions can Q be assumed to follow, and what is the corresponding critical value when α = 0.05? 1 Q has approximately a χ 2 -distribution with 6 degrees of freedom and χ (6) = Q has approximately a χ 2 -distribution with 5 degrees of freedom and χ (5) = Q has approximately a χ 2 -distribution with 4 degrees of freedom and χ (4) = Q has approximately a χ 2 -distribution with 4 degrees of freedom and χ (4) = Q has approximately a χ 2 -distribution with 6 degrees of freedom and χ (6) = Exercise XXIV The adverse effects per patient (X) during treatment with a new type of drug have been studied. For 1000 patients the individual number of adverse effects was registered, and the distribution of the adverse effects per patient is given in the following table. x x = 0 x = 1 x = 2 x = 3 x = 4 x = 5 x 6 f(x) It is seen that the density for x = 1 is missing. Calculate and insert the missing value and answer the following question. Question XXIV.I (29): Which one of the following values of p estimates the probability that a randomly selected patient has a maximum of two adverse effects? 1 p = p = p = p = p =

17 Exercise XXV The Nordic Cancer Union s analyses for the period showed that less than 5 percent of the Danish male patients with lung cancer were alive 5 years after they received the diagnosis lung cancer. A new method of treatment for lung cancer patients has been tested during a number of years. Out of 100 randomly selected patients treated with the new method, seven were alive 5 years after the time of diagnosis. Question XXV.I (30): Indicate the usual statistic for testing whether the survival rate for lung cancer patients has increased: 1 Z = X n p0 S n 2 t = 3 Z = D 0 S n D 4 χ 2 = i= X n p 0 n p0 ( 1 p0 ) ( o i e ) e i i 5 Z = n 0 p 0 p 0 p 1 n p 1 1 End of the exercise. Have a nice summer! 17