Lösningsförslag till skriftlig tentamen i FINANSIELL STATISTIK, grundnivå, 7,5 hp, torsdagen 15 januari 2009.

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1 Statistiska Institutionen Gebrenegus Ghilagaber (docent) Lösningsförslag till skriftlig tentamen i FINANSIELL STATISTIK, grundnivå, 7,5 hp, torsdagen 5 januari 009. Sannolkhetslära De ne the following two events: A : a randomly selected car has AC (Air Conditioner) B : a randomly selected car has CD.(CD player) We are then given the following pieces of information: P (A) 0:0; P (B) 0:0; P (A \ B) 0:0 Thus, a) P (A [ B) P (A) P (B) P (A \ B) 0:0 0:0 0:0 0:80 b) P (BjA) P (A \ B) P (A) 0:0 0:0 0:50 c) P (AjB) P (A \ B) P (B) 0:0 0:0 0:

2 d) Two events, A and B, are said to be statistically independent if P (A \ B) P (A) P (B) In our case, P (A \ B) 0:0; while P (A)P (B) (0:0)(0:0) 0:: Since 0:0 0:; the condition for statistical independence, P (A \ B) P (A) P (B) is not fu lled. Thus, A and B, are not statistically independent (or, in other words, they are dependent). In other words, having AC and CD are not independent. This, in turn, means that whether a car has AC or not a ects the probability that it will have CD. Diskret sannolikhetsfördelning Let the radnom variable X be the number of correct answers (out of the questions) by the student. Since the student answers at random out of the available alternatives, the probability of answering a question is correctly is : Further, since the questions are independent, this probability is the same for all questions. Thus, X may be viewed as Binomial distributed with parameters n (# trials of the experiement) and p (probability of success in each trial), where by "success" we mean answering a question correctly. X s Bin (; 0:5) : Hence, P (X r) so that a) n r p r ( p) n r r r r ; for r 0; ; ; ; ; 5; P (X ) 5!!! 0:8 0!! ( )!

3 b) P (X 5) P (X 5) P (X ) 5 5 5! 5! ( 5)! 5!! ( )! 0 5! 5! 5!!! 0! : () c) P (X ) P (X ) P (X ) P (X 0) !!!! ( )!! ( )! 0! ( 0)! 0 5! 5! 5! 0!!! 5! 0!! () () :

4 Extra: One way to check if the computations are correct is to compute P (X ) and see if the sum of all (four) probabilities add up to one: and P (X ) 5!!! 0:0959! 9 5 5! ( )! 5 09 X P (X r) P (X ) P (X ) P (X ) P (X 5) r0 0:8057 0:8 0:0959 0:0087 Kontinuerligt sannolikhetsfördelning Let the random variable T represent the lifetime of the components produced by the company. We know (or can assume) that T is normally distributed with mean months and variance 9 months: T s N(; 9): a) We are given that 0 and we given this we look for P (7 < T < ) : 7 P (7 < T < ) P < T < P < Z < P ( < Z < ) P (0 < Z < ) (0:) 0:8 where 0: is obtained from the Table of Standard Normal distribution. b) Now, is unknown (should be computed), but we know that P (T > 5) 0:08: T P (T > 5) 0:08 ) P > 5 0:08 ) P Z > 5 0:08

5 In the Table of Standard Normal distribution, we note that P (0 < Z < ) 0:77 so that P (Z > ) 0:5 0:77 0:08: Thus, the value z and z 5 cover the same area (0.08) to their right-side in the Standard Normal distribution and, hence, should be equal. In other words, 5 Hypotesprövning ) 5 9 Let the random variable X represent the income of an employee in the sector (bransch) under study. We now know that a sample of 900 persons (n 900) gave a sample mean-income of 8000 X 8000 and a standard deviation of 000 (s 000) : The aim is to test the null hypotheis H 0 : against the two-sided alternative H : ; where is the (unknown) mean-income of the entire population of employees in the sector. a) Though the population variance is unknown, we have a very large sample size (n 900). We can,. thus, make use of the Central Limit Theorem (CLT) and use the Standard Normal variable (Z) as a test statistic: Z X 0 ps n p :5 With 5 % signi cance level ( 0:05) and two-sided test, the critical values are Z Z 0:05 Z 0:05 :9: In other words, our Non- Rejection region consists of all points between :9 and :9 while the rejection region consists of all points less than :9 or greater than :9 Since our computed value of the test statistic (Z cal :5) falls within the Rejection region, we reject the null hypothesis H 0 : at 5 % level of signi cance. b) The value of the test statistic computed in (a) above is unchanged (Z cal :5) but the critical values for a two-sided test with 0:0 are now Z Z 0:0 Z 0:005 :575: In other words, our Non- Rejection region consists of all points between :575 and :575 while the rejection region consists of all points less than :575 or greater than :575. 5

6 Since our computed value of the test statistic (Z cal :5) falls within the Non-Rejection region, we don t reject the null hypothesis H 0 : at % level of signi cance. c) No, one does not draw the same conclusions from (a) and (b). This is because of the simple reason that the tests in (a) and (b) are based on two di erent signi cance levels. We have a smaller signi cance-level, %, in (b). The following comment may also added (with some advantage) of the ambitious student, but it is not an absolute requirement to get full-points in the sub-question: Going back to the original de nition of sigini cance-level, we know that represents the probability of type-i error (rejecting a true hypothesis). Thus. minimizing this risk (of commiting type-i error) from 5 % to % amounts to becoming more conservative (reject less) and keep the original hypothesis. That is what happened here. 5 Regression a) nx t Y i n nx t n t i Y t : :9

7 Thus, b b and nx t t i t Y i Y nx t t i t ( 5:5) (0 :9) ( 5:5) (8 :9) ::: (0 5:5) ( :9) ( 5:5) ( 5:5) ( 5:5) ::: (0 5:5) 5:50 8:50 :5 b 0 b 0 Y b t :9 ( :5 5:5) :9 8: :7 so that the estimated regression equation is by t :7 :5 t b) The estimated values of Y t may be obtained for all values of t from the equation Y b t :7 :5 t: Thus, by :7 :5 0:7 by :7 :5 9:5 by :7 :5 7:70 : : : by 0 :7 :5 0 7:055 7

8 Månad (t) Antal bilar (Y t ) Estimated ( Y b t ) 0:7 9:5 7:70 7:055 The unexplained sum of squares is then given by SSE X Y t b Yt (0 0:7) ::: ( 7:055) while the total sum of squares is given by SST X Y t Y t (0 :9) ::: ( :9) 07 so that the unexplained portion of the variability in Y t is given by c) R SSE SST 07 0:077 7:7 % by t :7 :5t ) b Y :7 :5 cars in month Tidsserie a) S t Y t ( ) S t 0: Y t ( 0:) S t Thus, S 0: Y ( 0:) S 0 0: 0 ( 0:) 0 0 S 0: Y ( 0:) S 0: 8 ( 0:) 0 9:00 S 0: Y ( 0:) S 0: 8 ( 0:) 9: 9:80 S 0: Y ( 0:) S 0: ( 0:) 9:8 : S 5 0: Y 5 ( 0:) S 0: 7 ( 0:) : :99 8

9 S 0: Y ( 0:) S 5 0: ( 0:) :99 5:959 S 7 0: Y 7 ( 0:) S 0: ( 0:) 5:959 5:57 S 8 0: Y 8 ( 0:) S 7 0: 9 ( 0:) 5:57 :5 S 9 0: Y 9 ( 0:) S 8 0: 8 ( 0:) :5 :00 S 0 0: Y 0 ( 0:) S 9 0: ( 0:) :00 :0 b) Månad (t) Y t S t (Sing) S t (Doub) One can see that Double smoothing gives estimates closer to the true values. One can compute MAD or MSD to compare the deviations. But, one can also argue that there was a strong trend component as we saw in Problem 5 above (with R 9%) and we need to take due account of this trend in smoothing. Single Exponential Smoothing not include a term for Trend while Double exponential smoothign does. Thus, Double Exponential Smoothing would t the data better. 9