Answer key. Statistics March Section A

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1 Answer key Statistics March 2018 Section A One mark 1. Cohort is a group of individuals who are born at the same time and who experience the same mortality conditions Base year quantities 4. Arrangement of statistical data in chronological order or in accordance with time is known as a time series. 5. When λ is large, Poisson distribution can be considered to be normal distribution 6. (0, ) 7. Sample mean 8. The set of all the admissible values of the parameters is called parameter space 9. H 0 : µ = Statistical Quality Control is the method of controlling the quality of the products using statistical techniques. 11. Feasible region is the area which satisfies all the constraints simultaneously and also the nonnegativity restriction. 12. Transportation problem is said to be balanced if total availability is equal to total requirement Two Marks Section B 13. Census enumeration method and registration method 14. Limitations of index numbers a. Since index numbers are based on the sample data, they are only approximate indicators. b. There is a likelihood of error being introduced at each stage of construction of the index number 15. V 01 = Trend values semi average Year Profit ('000) Trend values Interpolation It is a technique of estimating the value of dependent variable for the known value of independent variable which is within the given range of given series. Extrapolation - It is a technique of estimating the value of dependent variable for the known value of independent variable which is outside the given range of given series

2 18. Bernoulli Distribution X 0 1 Total p(x) Features a. For t distribution mean = median = mode = 0 b. It is leptokurtic 20. If a single value is proposed as an estimate of the unknown parameter then it is point estimation If an interval is proposed as an estimate of the unknown parameter then it is interval estimation Controlling the quality of the product during the manufacturing process itself is called process control. Controlling the quality of the finished products or manufactured products is called product control. 23. Point Z=5x+4y (12, 10) 100 (14, 4) 86 Value of x = 12, y= 10 and maximum value of Z = marks 24. m+n-1=5 and number of solutions = 4. Hence given solution is degenerate. Section C 25. Age group (in years) Population Number of live births Male Female and above !.!# $%&' (%)*+,%-.'-) /!*-$!0$-*%! %*+'.'-) Total population in the child bearing age group is, GFR= 01(')!# $%&' (%)*+, 20)%3 -.'-) /!*-$ 01(')!# 4!1'!# 5+%$2 ('-)%3-3'%*+'.'-) 1000= 26. unweighted geometric mean Items Price (2014) Price (2016) P log P A B C D E =

3 P 01 = antilog ( 678 9:;<=>?@ A.B Consumer price index number is the index number of the cost met by a specified class of consumers in buying a basket of goods and services. Uses of consumer price index numbers are a. It is commonly used in fixation of salary, dearness allowance or grant of bonus to the employees. b. They are used by the government for the formulation of price policy, wage policy and general economic policies. c. They are used for comparing changes in the cost of living of different classes of people. d. Cost of living index is used widely in wage negotiations and wage contracts 28. Trend values by 3 yearly moving average year Sales ( 000) 3 yearly moving sum Trend values values known 5 y 0 = 0 y 5-5y 4 +10y 3-10y 2 +5y 1 -y 0 = 0 y 2 = 67.8 y 6-5y 5 +10y 4-10y 3 +5y 2 -y 1 = 0 y 6 = Mean λ=2.5 Let X denotes number of accidents in a year. Hence X~P(λ=2.5) Pmf is given by, EFG9 'Hλ Fλ9 I 'H.L F.9 I,G0,1,2,. J! J! (i) P(1 accident) = 'H.L F ! Number of accidents = = O411 P[more than 2 accidents] = 1-P[X 2] = 1-p(0)- p(1)-p(2) 1P QR. F2.59 0! = Number of accidents = = O Pmf is given by,efg9= a C b x Cn x a b Cn +, G0,1,.min F:,;9 P QR. F2.59 1! P QR. F2.59 2!

4 P(X=2) = C 9 x C4 x 15 C 4, G0,1,.4 WXY ;:ZF:ZP;9 F:Z9 F:ZP Given P 0 = 0.02, n = 400, x=12. p=x/n = 12/400 = 0.03 H 0 : P = 0.02 H 1 : P< 0.02 Test statistic is given by, Z^_6 `Ra ~NF0,19 under H b c d 0.03P At 1% level of significance, the critical value is -k = We reject H 0 if Z cal < -k. Otherwise we accept H 0. On comparison, we accept H 0. Hence manufacturer s claim is not justifiable. 33. Given n=16, xm=53, µ 0 =56, FGPGn9 =150 s = FJRJn9 H 0 : µ = 56 H 1 : µ 56 Test statistic is given by, = t^_6 xmpµ s/ np1 ~t FtR9 under H 53P56. P R Given for n-1 =15 degrees of freedom, the critical value (k) is ±2.13 We reject H 0 if t^_6 > k or t^_6 vw. Otherwise we accept H 0. On comparison, we reject H 0. Sample cannot be regarded as taken from the population having mean Given n=50, k=10 and En 2'#'5*%&', 0.03 x Central line -CL = ;En Lower control limit LCL= ;EnP3y;Enzm500.03P P (Since control limits cannot be negative) Upper control limit UCL= ;En3y;Enzm Row minimums are circled and column maximums are boxed Player B Player A { B 1 B 2 B 3 B 4 A 1 P P} P A 2 ~ } A 3 P P

5 Player B 1 B 2 B 3 B 4 A Player A A A Saddle point exists at (1, 3) Strategy of player A is A 1 Strategy of Player B is B 3 Value of the game is zero Game is fair 36. Give purchase price is Rs and resale value is S n is Rs Year (P-S n ) C i ƒ % F;9 F PW 9% % ; % Since annual average cost is minimum for the 5 th year, the machine should be replaced at the end of 5 th year. 10 mark questions 37. a) Section D Age group ASDR ASDR Standard PA PB (in year) (TOWN A) = A (TOWN B) = B population &above STDR A 14.7 and WˆX Town A is healthier b) GRR Age group (in yrs) Female population Female births WSFR

6 GRR = i Σ WSFR = = a) Commodities p 0 q 0 p 1 q 1 p 0 q 0 p 0 q 1 p 1 q 0 p 1 q 1 A B C D Fisher s price index number Š Œ100 Œ100= Any index number satisfies TRT if P 01 x P 10 = 1 We have Š Œ :; Š Œ Consider Š Š YŽ E z E z YŽ E z E z Y 562 E z E z E z E z Y Hence fisher s price index number satisfies TRT An index number satisfies FRT if PPPPF19 We have Š Œ :; Š Œ Consider LHS of (1) P x Q YŽ E z E z YŽ z E z E Y 562 E z E z z E z E Y Consider RHS, LHS=RHS. Hence Fisher s index number satisfies FRT 39. The parabolic trend fitted to the given data is Y=a+bx+cx 2 The normal equations are, Σy = n a + b Σx +cσx 2 Σxy=aΣx + b Σx 2 + cσx 3 Σx 2 y=aσx 2 + bσx 3 + cσx 4 Year y x x 2 x 3 x 4 xy x 2 y

7 Total Substituting the values in normal equation, 115 = 5a + b(0) + c(10) 5a + 10c = (1) -25 = a(0) + b(10) + c(0) b=-25/10= = a (10) + b (0) + c(34) 10a + 34c = (2) On solving (1) and (2) a=22 and c = 0.5 Hence the parabolic trend fitted to the given data is, Y= x x For binomial distribution, Parameter p = Jn, n=5 X f (O i ) fx E i F % P % Total , Gn #J = 600/200 = 3 and p=3/5 = 0.6 n x n x Pmf is given by, p( x)= Cx p q, x= 0, 1,....n Expected frequencies E i s are given by N.p(x), H 0 : Binomial distribution is a good fit H 1 : Binomial distribution is not a good fit Test statistic is given by, 5-$ F šr š 9 š ~ R5 œ; Q ž 5-$ Here n is number of expected frequencies calculated which are greater than 5. Hence at 5% level of significance critical value (right tail) for (5-1-1[p is estimated])=3 degrees of freedom is k 2 = We reject H 0 if 5-$ Ÿw otherwise we accept H 0. On comparison we accept H 0. Binomial distribution is a good fit 5 mark questions Section E 41. P(X<56) = P(z< R 9= P(Z<1.5) = = P(52<X<55) = P(0.5<Z<1.25) = = Given n 1 =64, xm = 63, s 2 1 =64 n 2 =48, xm = 60, σ 2 2 =144 H 0 : µ 1 = µ 2 H 1 : µ 1 > µ 2 [right tailed test] Test statistic is given by, Z^_6 m R m Y d d ~NF0,19 under H 1.5 At 5% level of significance, the critical value k=1.65. We reject H 0 if Z cal > k. Otherwise we accept H 0. On comparison, we accept H 0 Mean weight of boys is not greater than mean weight of girls. %

8 43. H 0 : vaccination and attack of TB are independent. H 1 : vaccination and attack of TB are not independent. Attacked Not attacked Total Inoculated Not Inoculated Total Test statistic is given by, 5-$ F-2R(59 F- (9F5 29F- 59F( 29 ~ œ; Q ž 60F126P Hence at 1% level of significance critical value (right tail) for 1 degree of freedom is k 2 = We reject H 0 if 5-$ Ÿw otherwise we accept H 0. On comparison we reject H 0. vaccination and attack of TB are not independent 44. Given R = 3000 unit per year. P = Rs.7 per unit, I = 20% of capital cost per year, C 3 = Rs.150 Hence C 1 = PI =7 0.2 = Rs.1.4 per item per year Since shortage cost is not mentioned, we use I model. Economic order quantity The optimum number of orders Y œ;=< < Q: 3000 ; 1 1 < ? Q EQ Q: