STA301 Past Paper PDF - Spring 2012
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Institute of E-Learning & Modern Studies
2012
VUI
Muhammad Moaaz Siddiq
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This is a STA301 past paper from Spring 2012. The document contains solved subjective questions on various topics in statistics and probability. The paper appears to be from VUI.
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STA301- Statistics and Probability Solved Subjective Feb 16,2013 From Final term Papers MC100401285 [email protected] [email protected]...
STA301- Statistics and Probability Solved Subjective Feb 16,2013 From Final term Papers MC100401285 [email protected] [email protected] PSMD01 STA301- Statistics and Probability Final Term Examination - Spring 2012 1- A sample of size n=3 drawn without replacement experiment from a population N=5 items whose values 0,2,3,6,7 draw possible samples Answer:- N 5 Sample without replacement = 10 n 3 (0,2,3), (0,2,6), (0,2,7), (0,3,6), (0,3,7), (0,6,7), (2,3,6), (2,3,7), (2,6,7), (3,6,7). 2- Calculate sampling error if sample mean is 102 and population mean is 100 Answer:- (Page 14) Sampling error X 102 100 2 3- Define Unbiased estimator Answer:- (Page 258) An estimator is defined to be unbiased if the statistic used as an estimator has its expected value equal to the true value of the population parameter being estimated. E 4- Quartile deviation Answer:- (Page 84) The quartile deviation is defined as half of the difference between the third and first quartiles. Q Q1 Q.D 3 2 1 STA301- Statistics and Probability Final Term Examination - Spring 2012 Q1.decribe about significance level Answer: The significance level is the criterion used for rejecting the null hypothesis. It show you how likely a result is due to chance. The most frequently used values of α, the significance level, are 0.05 and 0.01, i.e. 5 percent and 1 percent. Q2.discuss about center limit theorem Answer:- (Page 243) “If a variable X from a population has mean μ and finite variance σ/2, then the sampling distribution of the sample mean⎯X approaches a normal distribution with mean μ and variance σ2/n as the sample size n approaches infinity.” Q3. What is difference between constant and random variable Answer: A variable whose value cannot be changed once it has been assigned a value while those variable whose values changes when we assign value to it. STA301- Statistics and Probability Final Term Examination - Spring 2012 Q: Define an Un-Biased estimator? 2 Marks Answer: Rep Q: Briefly Explain an experiment design? 2 Marks Answer:- (Page 320) By an experimental design, we mean a plan used to collect the data relevant to the problem under study in such a way as to provide a basis for valid and objective inference about the stated problem. Q: Define an independent and dependent variable in regression? 2 Marks Answer:- (Page 121) In regression Y represents the dependent variable and X represents the independent variable 2 Find the mean and variance for the sampling distribution given below. 5 Marks p̂ No. of Samples Probability f p̂ 0 1 1/20 1/3 9 9/20 2/3 9 9/20 1 1 1/20 20 1 Answer:- (Page 247) p No. of samples f p p. f p p 2. f p 1 0 1 0 0 20 1 9 3 1 9 3 20 20 20 2 9 3 1 9 3 20 10 5 1 1 1 1 1 20 20 20 p. f p 2 p. f p 10 1 2 3 20 1 1 p p. f p 0.5 2 2 2 p p. f p p. f p 2 2 3 1 10 2 3 1 10 4 1 0.05 20 3 STA301- Statistics and Probability Final Term Examination - Spring 2012 What is the difference between an outcome and an event? (2) Answer:- (Page 145) An outcome is a result of a single trial of an experiment while an event is an individual outcome or any number of outcomes. The mean of a Poisson distribution is 5 while its standard deviation is 4. Comment on it (2) Answer:- (Page 223) In Poisson distribution mean and variance are always equal but this statement is not satisfying this property of Poisson distribution. If an automobile is driven on the average no more than 16000 Km per year, then formulate the null and alternative hypothesis. (2) Answer: H 16000 H 1 16000 Q: Discuss three properties of normal distribution? 3 Marks Answer:- (Page 227) 1. Normal distribution is absolutely symmetrical, hence , 3 the third moment about the mean is zero 2. The normal curve is asymptotic to the x-axis as x → ± ∞. 3. For the normal distribution, it can be mathematically proved that 4 3 4 Q: The 90% confidence interval for the population mean is 11 to 20, interpret this result? 3 Marks Answer: - We are 90% sure that our Population mean lie between 11 – 20. Q: Define LSD test? 3 Marks Answer:- (Page 330) According to this procedure, we compute the smallest difference that would be judged significant, and compare the absolute values of all differences of means with it. This smallest difference is called the least significant difference or LSD. And is given by 2 MSE LSD t , v r 2 4 How many parameters are associated with F- distribution and what is the range of the distribution? (3) Answer:- (Page 312) It has two parameters v1 and v2 which are known as the degrees of freedom and it ranging from zero to plus infinity. Which of the following statement represents continuous data and discrete data? (5) i) Number of shopes in a plaza. Discrete data ii) Hourly temperature recorded by whether bureau. Continuous data iii) Inches of rainfall in a city. Continuous data iv) Number of passengers carried by rail every year. Discrete data v) Height measurements of boys studying in a college. Discrete data If the population proportions are gives as: P1 = 0.4, P2 = 0.20 find sigma^2 P-hat 1 - P-hat 2 , where n = 12. Answer:- (Page 256) pq p q 1 1 2 2 p1 p 2 n1 n2 q1 1 p1 1 0.40 0.6 q2 1 p2 1 0.20 0.8 0.4 0.6 0.20 0.80 p1 p 2 10 10 0.024 0.016 0.04 STA301- Statistics and Probability Final Term Examination - Spring 2012 what are steps involved in statistical research Answer: (Page 11) STEPS INVOLVED IN ANY STATISTICAL RESEARCH Topic and significance of the study Objective of your study Methodology for data-collection Source of your data Sampling methodology Instrument for collecting data 5 STA301- Statistics and Probability Final Term Examination - Spring 2012 Q. (Marks 5)Two different dice are thrown. Make the sample space and also find the number of sample points in an even A that the sum is 7. Answer:- 1,1 2,1 3,1 4,1 5,1 6,1 1, 2 2, 2 3, 2 4, 2 5, 2 6, 2 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3 S 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6 A 1, 6 2, 5 3, 4 4, 3 5, 2 6,1 Q. (Marks 2) Define the error sample and also write its formula. Answer:- (Page 13) The difference between the estimate derived from the sample (i.e. the statistic) and the true population value (i.e. the parameter) is technically called the sampling error Sampling error X Q. (Marks 2) Properties of hyper geometric distribution. Answer:- (Page 218) The outcomes of each trial may be classified into one of two categories, success and failure. The probability of success changes on each trial. The successive trials are not independent. The experiment is repeated a fixed number of times 5 number summery Answer:- (Page 97) A five-number summary consists of X0,Q1, Median, Q3, and Xm ; It provides us quite a good idea about the shape of the distribution. 6 STA301- Statistics and Probability Final Term Examination - Spring 2012 1.what s bias?(2) Answer:- (Page 258) E(θˆ )≠ θ, the statistic is said to be a biased estimator 2.advantages and disadvantages of median?(3) Answer:- Click here for detail Advantages (1) It is very simple to understand and easy to calculate. In some cases it is obtained simply by inspection. (2) Median lies at the middle part of the series and hence it is not affected by the extreme values. (3) In grouped frequency distribution it can be graphically located by drawing ogives. (4) It is especially useful in open-ended distributions. Disadvantages (1) In simple series, the item values have to be arranged. If the series contains large number of items, then the process becomes tedious. (2) It is a less representative average because it does not depend on all the items in the series. 3. Mathematical expectation of discrete random variable?(3) Answer:- (Page 179) In probability theory the expected value (or mathematical expectation) of a random variable is the sum of the product of the values within the range of the discrete random variable and their respective probabilities of occurrence. n E X xi f xi i 1 4.any two properties of mathematical expectation?(2) Answer:- (Page 202) The important properties of the expected values of a random variable are as follows: If c is a constant, then E(c) = c. Thus the expected value of a constant is constant itself. This point can be understood easily by considering the following interesting example: Suppose that a very difficult test was given to students by a professor, and that every student obtained 2 marks out of 20! It is obvious that the mean mark is also 2. Since the variable ‘marks’ was a constant, therefore its expected value was equal to itself. If X is a discrete random variable and if a and b are constants, then E(aX + b) = a E(X) + b. 5.what s statistical test?(2) Answer:- (Page 279) A statistic, which provides a basis for testing a null hypothesis, is called a test-statistic. Every test-statistic has a probability distribution (i.e. sampling distribution) which gives the probability that our test-statistic will assume a value greater than or equal to a specified value OR a value less than or equal to a specified value when the null hypothesis is true. 7 6. Decide a small sample and large sample?(2) Answer:- If the sample size ‘n’ is less than or equal to 30 ( 20, and p < 0.05 36. Elaborate the Least Significant Difference (LSD) test. Answer:- Rep 37. Write down the formula of combined or pooled proportion of two samples. Answer:- (Page 290) n1 p1 n2 p2 pc n1 n2 38. If approximate value of class interval is 2.96 and range = 14.8 then find the number of classes. Answer:- (Page 29) range Class int erval number of classes 14.8 2.96 number of classes 14.8 number of classes 5 2.96 8 40. Find the coefficient of variation (C.V) for the following price of a commodity. Price (X): 8, 13, 18, 23, 30 Answer:- X X2 8 64 13 169 18 324 23 529 30 900 X 92 X 2 1986 X X 92 18.4 n 5 X X 2 2 S n n 2 1986 92 S 5 5 S 397.2 338.56 58.64 7.66 S C.V 100 X 7.66 C.V 100 41.63% 18.4 41. Flaws in plywood occur at random with an average of one flaw per 50 square feet. What is the probability that 32 square feet will have no flaws? 9 STA301- Statistics and Probability Final Term Examination – Fall 2011 2- Probability of one electric device failure is 0.01. If a sample is chosen of 400, what is probability that exactly 2 are defective (Marks 5) Answer:- (Page 218) p 0.01 q 1 p q 1 0.01 q 0.99 400 P X 2 0.01 0.99 2 400 2 2 79800*0.0001*0.02 0.1596 4- Two dices are rolled. Find probability that outcome is equal or more than 11. (Marks 5) Answer:- 1,1 2,1 3,1 4,1 5,1 6,1 1, 2 2, 2 3, 2 4, 2 5, 2 6, 2 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3 S 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6 P 5, 6 , (6,5), (6, 6) 3 1 0.83 36 12 5- If X is binomial distribution n=5, p=0.5, q=0.5 then find SD(X) (Marks 3) Answer:- (Page 214) S.D X npq 5 0.5 0.5 1.25 1.12 10 6- n=24, Mean=33, s=15, x=40.4. Compute t statistic (Marks 3) 7- 90% confidence interval of population mean is 11 to 20. Interpret result (Marks 3) Answer:- Rep 8- E(XY)=421, E(X)=42, E(Y)=15. Check independence (Marks 3) Answer:- Cov X , Y E XY E X E Y 421 42 15 209 It is not independent. 9- Difference between statistics and statistic (Marks 3) Answer:- Statistics is the plural of statistic, the science that deals with the collection, classification, analysis, and interpretation of numerical facts or data. While the statistic is a single value or piece of data. 10- Why we call standard deviation the standard error (Marks 2) Answer:- (Page 240) The square root of the variance is the standard deviation, and the standard deviation of a sampling distribution is termed as its standard error. 11- if E(Y)=0.5 then find E(10.5+2Y)... (Marks 2) Answer:- E 10.5 2Y 10.5 2 E Y 10.5 2*0.5 11.5 12- What is the meaning of 'b' in Y=a+bX (Marks 2) Answer:- (Page 121) b represents the slope of the line 13- Explain level of significance (Marks 2) Answer:- rep 11 STA301- Statistics and Probability Final Term Examination – Fall 2011 Q# 28: can all deciles be expressed as percentile? Explain (2) Answer: Yes all deciles cab be expressed as percentiles As the D1 is 10 percent of data and P10 is 10% of data. D2 is 20% of data whereas in percentile it is P20 and so on … Q# 29: what is meant by sampling distribution? (2) Answer: Page 237 The probability distribution of any statistic (such as the mean, the standard deviation, the proportion of successes in a sample, etc.) is known as its sampling distribution. Q# 30: The department claims that the exceeds Rs. 2500 at the 0.05 level, then formulate null alternative hypothesis?(2) Answer: Page 283 H 0 : 2500 H1 : 2500 Q# 31: How we decide that the drawn sample is “Small Sample ” or a “Large Sample”?(2) Answer: rep Q# 32: If E(X)=0.7 then find E(2X)? (2) Answer: E 2 X 2E X 2*0.7 1.4 Q# 33: State the Baye‟s theorem? (3) Answer: Page 166 If events A1, A2… Ak form a PARTITION of a sample space S (that is, the events Ai are mutually exclusive and exhaustive (i.e. their union is S)), and if B is any other event of S such that it can occur ONLY IF ONE OF THE Ai OCCURS, then for any i, P Ai P B / Ai P Ai / B k P Ai P B / Ai i 1 Q# 35: “The 95% confidence interval for population mean is 1.3 to 4.7”. Interrupt this result (3) Answer: We are 95% sure that the population mean lies between 1.3 to 4.7 12 Q# 37: If E(XY)=045 E(X)=0.50 and E(Y)=0.90 then X and Y are independent?(3) Answer: Cov X , Y E XY E X E Y 0.45 0.50 0.90 0.45 0.45 0 Therefore it is independent STA301- Statistics and Probability Final Term Examination – Fall 2011 2 Marks 2) If we draw a card from an ordinary deck of 52 cards. Can king and diamond be mutually exclusive events? Give reason to support your answer. Answer:- (Page 146) No, because If the two events can occur at the same time, they are not mutually exclusive, Therefore, kings and diamonds are not mutually exclusive. 3) Suppose that we toss a fair coin three times. What is its sample space? Answer:- {(H,H,H),(H,T,H),(H,H,T),(T,H,H),(T,T,H),(T,H,T),(H,T,T),(T,T,T)} 4) Why we take B2 =3 as the criteria for measuring the kurtosis of any distribution? Answer:- (Page 228) For the normal distribution, β2 comes out to be 3, this is why this value has been taken as a criterion for measuring the kurtosis of any distribution. 5) Define level of significance? Answer:- rep 6) How we decide that the drawn sample is “Small sample” or a “large sample”? Answer:- Rep 13 3 Marks 7) If mean of a distribution is 0.925 and standard deviation is 0.132. Find out the limits by applying chebychev‟s inequality for k=2. How much fraction of data will lie between the two limits? Answer:- (Page 95) X kS , X kS 0.925 2 0.132 , 0.925 2 0.132 0.925 0.264, 0.925 0.264 0.661,1.189 95% of the measurements will fall within these limits. 8) If E(XY)=7.5, E(X)=2.4 and E(Y)=4.3, calculate covariance of X and Y. Answer:- Cov X , Y E XY E X E Y 7.5 2.4 4.3 7.5 10.32 2.82 11) Write down the two properties of sampling distribution of proportion p‟, when the sampling is performed with replacement? Answer:- (Page 241) Property 1: X Property 2: In case of sampling with replacement: X n 12) What is natural pairing in observations? Give example. Answer:- (Page 302) Natural pairing occurs whenever measurement is taken on the same unit or individual at two different times. For example, suppose ten young recruits are given a strenuous physical training programme by the Army. Their weights are recorded before they begin and after they complete the training. The two observations obtained for each recruit i.e. the before-and-after measurement constitute natural pairing. The above is natural pairing. 14 5 Marks 13) Interpret the concept of five numbers summary and also explain the purpose of five number summaries. Answer:- (Page 97) A five-number summary consists of X0,Q1, Median, Q3, and Xm ; It provides us quite a good idea about the shape of the distribution If the data were perfectly symmetrical, the following would be true: 1. The distance from Q1 to the median would be equal to the distance from the median to Q3 2. The distance from X0 to Q1 would be equal to the distance from Q3 to Xm. 3. The median, the mid-quartile range, and the midrange would all be equal. All these measures would also be equal to the arithmetic mean of the data On the other hand, for non-symmetrical distributions, the following would be true: 1. In right-skewed distributions the distance from Q3 to Xm greatly exceeds the distance from X0 to Q1 2. in right-skewed distributions, median < mid-quartile range < midrange: THE POSITIVELY SKEWED CURVE 14) What is the probability that a poker hand of 5 cards contain exactly 2 aces (hyper geometric distribution)? Answer:- 4 48 P X 2 2 3 52 5 6 17296 103776 0.0399 2598960 2598960 16)From the table given below find the value of chi square 15 STA301- Statistics and Probability Final Term Examination – Fall 2011 the mean of a population is 6 a sample of size '20' is taken with replacement from this population.which one of the following is the mean of sampling distribution of mean. 1) 2 2) 8 3) 6 4) 3 Answer:- 6 Q no 2 Write down the significance of experimental design? marks :2 Answer:- Rep Q No: 03 Explain the formula of hyper geometric probability distribution. marks :2 Answer:- (Page 219) k N k n n x P X x N n Where N = number of units in the population, n = number of units in the sample, and k = number of successes in the population. The hyper geometric probability distribution has three parameters N, n and k. Q no 5 Suppose that we toss a fair coin three times. What is the sample space? marks :2 Answer:- Rep Compute the following formula which is used to find the sample size. Answer:- For finding sample space of coin the formula is: 2n =23 = 8 If x is a passion random variable with parameter value „4.5‟, find p(x=1)? Marks: 03 Answer:- (Page 222) e4.5 4.5 1 P X 1 1! 0.0499 16 Q no: 09 When the value of chi-square is equal to zero? Marks : 03 Answer:- (Page 338) If the observed frequencies are exactly equal to the expected ones, then chi-square will be exactly equal to zero. Q no: 10 If x= 255, n=634, Po=0.60 then find the z-test statistics for probability. Marks: 03 Answer:- (Page 289) 1 X np0 Z 2 np0 1 p0 np0 634*0.60 380.4 1 here X np0 therefore we use X 2 1 X np0 Z 2 np0 1 p0 1 225 380.4 Z 2 380.4 1 0.60 154.9 Z 12.34 Z 12.55 STA301- Statistics and Probability Final Term Examination – Fall 2010 Describe the formula for hypergeometric distribution? 2 marks Answer:- Rep What is graphical representation? 2 marks Answer:- The plot of the points in the plane which constitute the graph of a given real function or a pictorial diagram depicting interdependence of variables. 17 What is the t formula for paired distribution? 3marks. Answer:- (Page 302) d d t sd / n What is the probability that at least one head occurs if a coin tossed 6 times successively? 5 marks Answer:- 6 1 6 1 1 1 P( X 1) . . 1 2 2 6 1 1 5 1 . . 1 2 2 3.03 Find first 2 moments for function with respect to mean? 5 marks Answer:- (Page 192) 16)From the table given below find the value of chi square Obvervation Frequency Oij Expected Frequency Eij 120 100 130 150 80 100 170 150 Answer:- Rep STA301- Statistics and Probability Final Term Examination – Fall 2010 Q no 5 Suppose that we toss a fair coin three times. what is the sample space? marks :2 Answer: Rep 18 Compute the following formula which is used to find the sample size. Z n /2 e Page 276 FINALTERM EXAMINATION Spring 2010 STA301- Statistics and Probability (Session - 4) Question No: 31 ( Marks: 2 ) What is the mean and variance of Poisson distribution? Answer:- (Page 233) If the random variable X has a Poisson distribution with parameter μ, then its mean and variance are given by E(X) = μ and Var(X) = μ. Question No: 32 ( Marks: 2 ) Explain the Chi-square test of goodness of fit. Answer:- (Page 332) The chi-square test of goodness-of-fit is a test of hypothesis concerned with the comparison of observed frequencies of a sample, and the corresponding expected frequencies based on a theoretical distribution. Question No: 33 ( Marks: 2 ) Define level of significance? Answer:- Rep Question No: 34 ( Marks: 3 ) Find the value of F(table value), when n1 7 , n 2 10 and α= 0.05 Answer:- (Page 316) Question No: 35 ( Marks: 3 ) Define Null and Alternative hypothesis. Answer:- Rep 19 Question No: 36 ( Marks: 3 ) A random sample of 100 is taken from a population with mean 30 and standard deviation 5. The probability distribution of the parent population is unknown, find the mean and standard error of the sampling distribution of X. Question No: 37 ( Marks: 5 ) What is the probability that a poker hand of 5 cards contain exactly 2 aces (hypergeometric distribution)? Answer:- Rep Question No: 38 ( Marks: 5 ) A random sample of size n is drawn from normal population with mean 5 and variance . 2 If n=25, s=10 and t=2, what is the values of x ? Question No: 39 ( Marks: 5 ) Describe the main steps of General Procedure for Testing Hypothesis. Answer:- (Page 281) Step-1: Formulation of the Null and Alternative Hypotheses: Step-2: Decision Regarding the Level of Significance Step-3: Test Statistic (that statistic that will enable us to test our hypothesis): Step-4: Calculations: Step-5: Critical Region (that portion of the X-axis which compels us to reject the null hypothesis): Step-6: Conclusion: FINALTERM EXAMINATION Spring 2010 STA301- Statistics and Probability (Session - 4) Question No: 31 ( Marks: 2 ) How many parameters are involved in hyper geometric distribution? Answer:- (Page 291) The hyper geometric probability distribution has three parameters N, n and k. 20 Question No: 32 ( Marks: 2 ) If an automobile is driven on the average no more than 16000 Km per year, then formulate the null and alternative hypothesis. Answer:- Rep Question No: 33 ( Marks: 2 ) Write down the test statistic when chi- square goodness of fit test is performed. Answer:- (Page 334) oi ei 2 χ 2 i ei Question No: 34 ( Marks: 3 ) Find the value of F(table value), when n1 7 , n 2 10 and α= 0.05 Answer:- (Page 316) F0.05(n1-1, n2-1) = (6,9) =3.37 F0.05(n2-1, n1-1) = (9,6) V2=6 , v1=8 => 4.15, V2=6, v1=12 => 4.00 Mean = (4.15+4)/2 = 4.075 Hence value of F0.05(9,6) = 4.075 Question No: 35 ( Marks: 3 ) p0 If X = 327, n = 634, 0.50 then find the z-test statistic for proportion. Answer:- (Page 289) 1 X np0 Z 2 np0 1 p0 np0 634 * 0.50 317 1 here X np0 therefore we use X 2 1 X np0 Z 2 np0 1 p0 1 327 317 Z 2 317 1 0.50 9.5 Z 12.58 Z 0.75 21 Question No: 36 ( Marks: 3 ) If population proportions are given as: P1 0.30, P2 0.20. p2ˆ pˆ 1 2 Find ,where n = 10 Answer:- (Page 256) p1q1 p2 q2 p1 p2 n1 n2 q1 1 p1 1 0.30 0.7 q2 1 p2 1 0.20 0.8 p1 0.30 0.7 0.20 0.80 p2 10 10 0.021 0.016 0.037 Question No: 37 ( Marks: 5 ) A candidate for mayor in a large city believes that he appeals to at least 10 per cent more of the educated voters than the uneducated voters. He hires the services of a poll-taking organization, and they find that 62 of 100 educated voters interviewed support the candidate, and 69 of 150 uneducated voters support him at the 0.05 significance level. Answer:- (Page 292) Step-1: The null and alternative hypothesis is H0 : p1 – p2 > 0.10, and H1 : p1 – p2 < 0.10, where p1 = proportion of educated voters, and p2 = proportion of uneducated voters. Step-2: Level of Significance: α = 0.05. 22 Step-5: Critical Region: As this is a one-tailed test, therefore the critical region is given by Z < -z0.05 = -1.645 Step-6: Conclusion: Since the calculated value z = 0.95 does not fall in the critical region, so we accept the null hypothesis H0 : p1 – p2 > 0.10.The data seems to support the candidate’s view. Until now, we have discussed in considerable detail interval estimation and hypothesis-testing based on the standard normal distribution and the Z-statistic. Next, we begin the discussion of interval estimation hypothesis-testing based on the t-distribution. 23 Question No: 38 ( Marks: 5 ) If we have RCBD with MSE=3.19, no.of.treatments = 4, no.of.blocks = 5; then find the value of LSD (least significant difference) for treatments by using α=0.05 and error degrees of freedom is 12. Answer:- Rep Question No: 39 ( Marks: 5 ) Find the mean and variance for the sampling distribution given below. p̂ No. of Samples Probability f p̂ 0 1 1/20 1/3 9 9/20 2/3 9 9/20 1 1 1/20 20 1 Answer:- Rep FINALTERM EXAMINATION Spring 2010 STA301- Statistics and Probability (Session - 3) Question No: 31 ( Marks: 2 ) Define normal distribution. Answer:- (Page 226) A continuous random variable is said to be normally distributed with mean μ and standard deviation σ if its probability density function is given by 1 x 2 1 f x e 2 2 Question No: 32 ( Marks: 2 ) Fill up the missing values in the formula. /2 2 ˆˆ pq n 24 Answer:- (Page 277) z p q 2 n /2 e2 Question No: 33 ( Marks: 2 ) Explain the Chi-square test of goodness of fit. Answer:- Rep Question No: 34 ( Marks: 3 ) What is natural pairing in observations? Also give an example. Answer:- Rep Question No: 35 ( Marks: 3 ) If population proportions are given as: P1 0.30, P2 0.20. p2ˆ pˆ 1 2 Find ,where n = 10 Answer:- Rep Question No: 36 ( Marks: 3 ) Explain the method of Maximum Likelihood in Point Estimation? Answer:- Page 263 The method of maximum likelihood is regarded as the MOST important method of estimation, and is the most widely used method. This method was introduced in 1922 by Sir Ronald A. Fisher (1890-1962).The mathematical technique of finding Maximum Likelihood Estimators is a bit advanced, and involves the concept of the Likelihood Function. Question No: 37 ( Marks: 5 ) The following data was obtained for a randomized block design involving five treatments and three blocks SST=430, SSTR=310, SSB=85, Setup the ANOVA table. Answer:- (Page 332) Source d.f SS MS Between 4 430 107.5 treatments Between 2 85 42.5 Blocks Error 8 310 38.75 Total 14 825 188.75 25 Question No: 38 ( Marks: 5 ) In a Standard Normal Distribution, find: i. Mean and Standard Deviation ii. Lower Quartile iii. Upper quartile iv. Inter-quartile range v. Mean Deviation Answer:- In normal distribution i. Mean and Standard Deviation Mean is 0 and S.D is 1 ii. Lower Quartile: Q1 0.6745 =0-0.6745(1) Q1 = -0.6745 iii. Upper Quartile: Q3 0.6745 =0.6745 iv. Inter-quartile range Q3 - Q1 /2 = 0.6745 - (-0.6745)/2 = 0.6745 v. Mean Deviation = 0.7979 = 0.7979(1) = 0.7979 Question No: 39 ( Marks: 5 ) n 13, x 34, 2 70, 0.10 If 31 Test the hypothesis 26 Answer:- H 0 30 H1 30 Level of significance =0.10 /2=0.05 Critical region=z>1.645 X z n 34 31 Z= 2.32 Z=1.29 Conclusion: Since the value Z=1.29 is less than the value of critical region so we accepted H 0 at 0.10 Level of significance. FINALTERM EXAMINATION Fall 2009 STA301- Statistics and Probability (Session - 4) Question No: 21 ( Marks: 1 ) Write down the formula for binomial distribution. Answer:- (Page 212) n P X x p x q n x x Question No: 22 ( Marks: 2 ) Write down the formula for testing the equality of two population proportions. Answer:- (Page 290) Z 1 2 p p 0 p q 1 1 c c n1 n2 27 Question No: 23 ( Marks: 3 ) Define moment ratios. In which unit they are expressed? Answer:- (Page 119) m 2 m4 b1 3 3 and b2 m2 m2 2 They are independent of origin and units of measurement, i.e. they are pure numbers. 28