ENM SIMULATION PROBABILITY AND STATISTICS REVIEW 1

Question 1 2 The probability that the Red River will flood in any given year has been estimated from 200 years of historical data to be one in four. This means: a)The Red River will flood every four years b)In the next 100 years, the Red River will flood exactly 25 times c)In the last 100 years, the Red River flooded exactly 25 times d)In the next 100 years, the Red River will flood about 25 times e)In the next 100 years, it is very likely that the Red River will flood exactly 25 times

Question 2 3 A random variable X has a probability distribution as follows: Then the probability that X is less than 2, i.e., P(X<2),is equal to: a)0.90 b)0.25 c)0.65 d)0.15 e)1.00 X 0123 P(X=x)=P X (x) 2k3k13k2k

Question 3 4 Cans of soft drinks cost $0.30 in a certain vending machine. What is the expected value and variance of daily revenue ( Y ) from the machine if X, the number of cans sold per day has E(X)=125 and Var(X)=50 ? a)E(Y)=37.5, Var(Y)=50 b)E(Y)=37.5, Var(Y)=4.5 c)E(Y)=37.5, Var(Y)=15 d)E(Y)=125, Var(Y)=4.5 e)E(Y)=125, Var(Y)=15

Question 4 - i 5 The average length of stay in a hospital is useful for planning purposes. Suppose that the following is the distribution of length of stay in a hospital after a certain operation: What is the probability that the length of stay is 6? a)0.15 b)0.17 c)0.20 d)0.25 e)0.05 Days Probabilities ?

Question 4 - ii 6 The average length of stay in a hospital is useful for planning purposes. Suppose that the following is the distribution of length of stay in a hospital after a certain operation: The average length of stay is: a)0.15 b)0.17 c)3.3 d)4.0 e)4.2 Days23456 Probabilities

Question 5 7 Many professional schools require applicants to take a standardized test. Suppose that 1000 students take the test, and you find that your mark of 63 (out of 100) was the 73 rd percentile. This means: a)At least 73% of the students got 63 or better b)At least 270 students got 73 or better c)At least 270 students got 63 or better d)At least 27% of the students got 73 or worse e)At least 730 students got 73 or better.

Question 6 8 To determine the reliability of experts used in interpreting the results of polygraph examinations in criminal investigations, 280 cases were studied. The results were: If hypotheses were H: Suspect is innocent versus A: Suspect is guilty, then we could estimate the probability of making a type II error as: a)15/280 b)9/280 c)15/140 d)9/140 e)15/146 True status InnocentGuilty Examiner’s decision Innocent13115 Guilty9125 Type-I: Reject Ho when Ho is true Type-II: Cannot reject Ho when Ho is not true

Question 7 9 In a statistical test of hypothesis, what happens to the rejection region when , the probability of type-I error or the level of significance, is reduced? a)The answer depends on the value of  (probability of type-II error) b)The rejection region is reduced in size c)The rejection region is increased in size d)The rejection region is not changed e)The answer depends on the form of the alternative hypothesis Type-I: Reject Ho when Ho is true Type-II: Cannot reject Ho when Ho is not true

Question 8 10 Which of the following is not correct? a)The probability of type-I error is controlled by the selection of the significance level, . b)The probability of type-II error is controlled only by the sample size c)The power of a test depends upon the sample size and the distance between the null and alternative hypotheses d)The p-value measure the probability that the null hypothesis is true when Ho is rejected by the current data. e)The rejection region is controlled by the a level and the alternate hypothesis. Type-I: Reject Ho when Ho is true Type-II: Cannot reject Ho when Ho is not true

Question 9 11 In a statistical test for the equality of a mean, such as Ho:  =10, if  =0.05: a)We will make an incorrect inference 95% of the time, b)We will say that there is a real difference 5% of the time when there is no difference c)We will say that there is no real difference 5% of the time when there is no difference d)95% of the time the null hypothesis will be correct e)5% of the time we will make a correct inference Type-I: Reject Ho when Ho is true Type-II: Cannot reject Ho when Ho is not true

Question Which of the following statements is correct? a)An extremely small p-value indicates that the actual data differs markedly from that expected if the null hypothesis were true b)The p-value measures the probability that hypothesis is true c)The p-value measures the probability of making a Type-II error. d)A large p-value indicates that the data is consistent with the alternative hypothesis e)The larger the p-value, the stronger the evidence against the null hypothesis Type-I: Reject Ho when Ho is true Type-II: Cannot reject Ho when Ho is not true

Question In a test of Ho:  =100, versus Ha:  ≠100, a sample of size 10 produces a sample mean of 103 and a p-value of Thus at the significance level  =0.05: a)There is sufficient evidence to conclude that  ≠100 b)There is sufficient evidence to conclude that  =100 c)There is insufficient evidence to conclude that  =100 d)There is insufficient evidence to conclude that  ≠100 e)There is sufficient evidence to conclude that  =103 Type-I: Reject Ho when Ho is true Type-II: Cannot reject Ho when Ho is not true

14 Formulas Expectation and variance formulas 1. E[cX] = cE[X] 2. E[c 1 X 1 + c 2 X 2 + … + c n X n ] = c 1 E[X 1 ] + c 2 E[X 2 ] + … + c n E[X n ] 3. Var(X)  0 4. Var(cX) = c 2 Var(X) 5. Var(c 1 X 1 + c 2 X 2 + … + c n X n ) = c 1 2 Var[X 1 ] + c 2 2 Var[X 2 ] + … + c n 2 Var[X n ] if X i ’s are independent (or uncorrelated) Covariance and correlation formulas 1. Cov(X i, X j ) = C ij = E[(X i -  i )(X j -  j )] = E[X i X j ] -  i  j 2. C ij = C ji, C ii =  i 2 = Var(X i ) 3.

15 Joint Distributions Joint Distributions of Discrete Random Variables Joint Distributions of Continuous Random Variables

16 Example

17 Example

18 Important Families of Distributions Normal Distribution We can obtain almost all other important distributions in (parametric) statistics by the following transformations. Standard Normal Distribution

19 Standard Normal PDF

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23 Cumulative distribution function (CDF):

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30 Chi-Square Distribution Z is a random variable with standard nrmal distribution

31 Chi-Square PDF

32 F Distribution

33 Student’s t Distribution

34 Student’s t PDF

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36 Estimation of Means, Variances and Correlations Suppose that X 1, X 2, …, X n are IID random variables (observations) with finite population mean  and variance  2. Sample Mean is an unbiased estimator of the population mean 

37 Sample Variance is an unbiased estimator of the population variance  2 The Central Limit Theorem

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39 Confidence Intervals and Hypothesis Tests for the Mean Suppose that X 1, X 2, …, X n are IID random variables (observations) with finite population mean  and variance  2 We want to find a confidence interval [l(n,  ), u(n,  )] so that P{ l(n,  )    u(n,  )} = 1 -  The length of the confidence interval is longer for the t distribution since

40 Skewness Actual coverage of the confidence interval depends on the sample size as well as the shape of the distribution in which skewness (a measure of symmetry) plays an important role.

41 Example 4.26 Suppose that the 10 observations 1.20, 1.50, 1.68, 1.89, 0.95, 1.49, 1.58, 1.55, 0.50, and 1.09 are from a normal distribution. Then, a 90% confidence interval is found as follows

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43 Hypothesis Testing In hypothesis testing, we need to choose among two competing hypothesis Status quoClaim

44 Testing Hypothesis on the Mean The decision rule with the maximum power (minimum Type II error probability) for which Type I error probability is at most  is given by the following t test The critical region of rejection is given by a confidence interval

45 Example 4.27 For the data of Example 4.26, suppose that we want to test H 0 :  = 1 at level  = 0.90.

46 The Strong Law of Large Numbers Theorem 4.2: Suppose that X 1, X 2, … are IID random variables with finite mean , then Example 4.29:

47 Replacing a Random Variable by Its Mean In general, it is not a good practice to replace random quantities by their means in a simulation study. Example 4.30: Suppose that mean interarrival time is 1 minute and the mean service time is 0.99 minute in an M/M/1 queue. If simulation is done with the means, then all delays are 0 and the queue is always empty. But, in the actual M/M/1 model  = /  = 0.99 and the average delay is computed as W q =  /  (1-  ) = 0.99(0.99)/0.01 = minutes.