King Saud University College of Science Statistics and Operations Research Department Stat 319 Theory of Statistics (2) Exercises References: 1. 2. 3. 4. Introduction to Mathematical Statistics, Sixth Edition, by R. Hogg, J. McKean, and A. Craig, Prentice Hall. Introduction to Theory of Statistics, A. Mood, F. Graybill and B. Boes, McGrow-Hill. Introduction to Mathematical Statistics, Fourth Edition, by Hogg and Craig, Macmillan Publishing Co., Inc. Statistical Inference, second Edition, by G. Casella and R. Berger, Duxbury. By Dr. Samah Alghamdi 1 Stat 319 Exercises: Theory of Statistics 2 Dr. Samah Alghamdi Confidence Intervals 1. Let πΜ be the mean of a random sample from the exponential distribution, πΈπ₯π(π). (a) Show that πΜ is an unbiased point estimator of π. (b) Using the MGF technique determine the distribution of πΜ . (c) Use (b) to show that π = 2ππΜ ⁄π has a π 2 distribution with 2n degrees of freedom. (d) Based on Part (c), find a 95% confidence interval for π if π = 10. Hint: Find c and d such that π (π < 2ππΜ π < π) = 0.95 and solve the inequalities for π. 2. Let π1 , … , ππ be a random sample from the Π(2, π) distribution, where π is unknown. Let π = ∑ππ=1 ππ . (a) Find the distribution of Y and determine c so that cY is an unbiased estimator of π. (b) If π = 5, show that π (9.59 < 2π π < 34.2) = 0.95. (c) Using Part (b), show that if y is the value of Y once the sample is drawn, then the 2π¦ 2π¦ interval (34.2 , 9.59) is a 95% confidence interval for π. 3. 4. 5. 6. (d) Suppose the sample results in the values, 44.8079 1.5215 12.1929 12.5734 43.2305 Based on these data, obtain the point estimate of π as determined in Part (a) and the computed 95% confidence interval in Part (c). What does the confidence interval mean? Let the observed value of the mean πΜ of a random sample of size 20 from a distribution that π(π, 80) be 81.2. Find a 95% confidence interval for π. Let πΜ be the mean of a random sample of size n from a distribution that is π(π, 9). Find n such that π(πΜ − 1 < π < πΜ + 1) = 0.90, approximately. Let a random sample of size 17 from the normal distribution π(π, π 2 ) yield π₯Μ = 4.7 and π 2 = 5.76. Determine a 90% confidence interval for π. Let πΜ denote the mean of a random sample of size n from a distribution that has mean π and variance π 2 = 10. Find n so that the probability is approximately 0.954 that the random 1 1 interval (πΜ − < π < πΜ + ) includes π. 2 2 7. Let Y be π΅ππ(300, π). If the observed value of Y is π¦ = 75, find an approximate 90% confidence interval for p. 8. Let π₯Μ be the observed mean of a random sample of size n from a distribution having mean π π π and known variance π 2 . Find n so that π₯Μ − 4 to π₯Μ + 4 is an approximate 95% confidence interval for π. 9. Let π1 , … , ππ be a random sample from π(π, π 2 ), where both parameters π and π 2 are unknown. A confidence interval for π 2 can be found as follows: 2 We know that (π − 1)π 2 ⁄π 2 is a random variable with a π(π−1) distribution. Thus, we can find constants a and b so that π((π − 1)π 2 ⁄π 2 < π) = 0.975 and π(π < (π − 1)π 2 ⁄π 2 < π) = 0.95. (a) Show that this second probability statement can be written as π((π − 1)π 2 ⁄π < π 2 < (π − 1)π 2 ⁄π) = 0.95. 2 Stat 319 Exercises: Theory of Statistics 2 Dr. Samah Alghamdi (b) If π = 9 and π 2 = 7.93, find a 95% confidence interval for π 2 . 10. Let π1 , … , ππ be a random sample from gamma distributions with known parameter πΌ = 3 and unknown π½ > 0. Discuss the construction of a confidence interval for π½. π Hint: What is the distribution of 2 ∑ππ=1 π½π ? Follow the procedure outlined in Exercise 9. 11. When 100 tacks were thrown on a table, 60 of them landed point up. Obtain a 95% confidence interval for the probability that a tack of this type will land point up. Assume independence. 12. Let two independent random samples, each of size 10, from two normal distributions π(π1 , π 2 ) and π(π2 , π 2 ) yield π₯Μ = 4.8, π 12 = 8.64, π¦Μ = 5.6, π 22 = 7.88. Find a 95% confidence interval for π1 − π2 . 13. Let two independent random variables, π1 and π2 , with binomial distribution that have parameters π1 = π2 = 100, π1 and π2 , respectively, be observed to be equal to π¦1 = 50 and π¦2 = 40. Determine an approximate 90% confidence interval for π1 − π2 . 14. Let πΜ and πΜ be the means of two independent random samples, each of size n, from the respective distributions π(π1 , π 2 ) and π(π2 , π 2 ), where the common variance is known. π π Find n such that π (πΜ − πΜ − < π1 − π2 < πΜ − πΜ + ) = 0.90. 5 5 15. If 8.6, 7.9, 8.3, 6.4, 8.4, 9.8, 7.2, 7.8, 7.5 are the observed values of a random sample of size 9 from a distribution that is π(8, π 2 ), construct a 90% confidence interval for π 2 . 16. Let π1 , … , ππ and π1 , … , ππ be two independent random samples from the respective normal distributions π(π1 , π12 ) and π(π2 , π22 ), where the four parameters are unknown. To construct a confidence interval for the ratio, π12 π22 , of the variances, from the quotient of the two independent chi-square variables, each divided by its degrees of freedom, namely πΉ = π22 ⁄π22 π12 ⁄π12 , where π12 and π22 are the respective sample variances. (a) What kind of distribution does F have? (b) From the appropriate table, a and b can be found so that π(πΉ < π) = 0.975 and π(π < πΉ < π) = 0.95. π2 π2 π2 2 2 2 (c) Rewrite the second probability statement as π (π π12 < π12 < π π12 ) = 0.95. The observed values, π 12 and π 22 , can be inserted in these inequalities to provide a 95% π2 confidence interval for π12 . 2 17. Let two independent random samples of sizes π = 16, π = 10, taken from two independent normal distributions π(π1 , π12 ) and π(π2 , π22 ), respectively, yield π₯Μ = 3.6, π2 π 12 = 4.14, π¦Μ = 13.6, π 22 = 7.26. Fin a 90% confidence interval for π22 when π1 and π2 are 1 unknown. 18. Find a 90 percent confidence interval for the mean of a normal distribution with π = 3 given the sample (3.3, -0.3, -0.6, -0.9). What would be the confidence interval if π were unknown? 3 19. The breaking strengths in pounds of five specimens of manila rope of diameter 16 inch were found to be 660, 460, 540, 580 and 550. 3 Stat 319 Exercises: Theory of Statistics 2 Dr. Samah Alghamdi (a) Estimate the mean breaking strength by a 95% confidence interval assuming normality. (b) Estimate π 2 by a 90% confidence interval; also π. 20. A sample was drawn from each of five populations assumed to be normal with the same variance. The values of (π − 1)π 2 = ∑(ππ − πΜ ) and n, the sample size, were π 2 : 40 30 20 42 50 π: 6 4 3 7 8 Find 98% confidence interval for the common variance. 21. To test two promising new lines of hybrid corn under normal farming conditions, a seed company selected eight farms at random in Iowa and planted both lines in experimental plots on each farm. The yields (converted to bushels per acre) for the eight locations were Line A: 86 87 56 93 84 93 75 79 Line B: 80 79 58 91 77 82 74 66 Assuming that the two yields are jointly normally distributed, estimate the difference between the mean yields by a 95% confidence interval. 22. Let πΜ denote the mean of a random sample of size 25 from a gamma distribution with πΌ = 4 and π½ > 0. Use the Central Limit Theorem to find an approximate 0.954 confidence interval for π, the mean of the gamma distribution. Hint: Use the random variable (πΜ −4π½) 23. √4π½ ⁄25 Let π12 5πΜ = 2π½ − 10. and π22 denote, respectively, the variances of random samples, of sizes n and m, from two independent distributions that are π(π1 , π 2 ) and π(π2 , π 2 ). (a) Derive (1 − πΌ)100% confidence interval for the common unknown variance π 2 . (b) Find 99% confidence interval for π 2 if π = 12, π 12 = 4.74, π = 15, π 22 = 5.66. 4 Stat 319 Exercises: Theory of Statistics 2 Dr. Samah Alghamdi Type I Error, Type II Error and Power Function 1. Let π have a binomial distribution with the number of trials π = 10 and with π either 1/4 1 or 1/2. The simple hypothesis π»0 : π = 2 is rejected, and the alternative simple hypothesis 1 π»1 : π = 4 is accepted, if the observed value of π1 , a random sample of size 1, is less than or equal to 3. Find the significance level and the power of the test. 2. Let π1 , π2 … . , π25 be a random sample from π(π, 16). If the test: Reject π»0 : π = 1 and accept π»1 : π = 3 when π₯Μ ≥ 2. Find the Type I error and the power function of the test. 3. Let π1 , π2 be a random sample of size π = 2 from the distribution having pdf 1 formπ(π₯; π) = π π −π₯⁄π , π₯ > 0, π > 0. We reject π»0 : π = 2 π£π π»1 : π = 1 if the observed values of π1 , π2 , say π₯1 , π₯2 are such that π(π₯1 , 2)π(π₯2 , 2) 1 ≤ π(π₯1 , 1)π(π₯2 , 1) 2 Here β¦ = {π: π = 1,2}. Find the significance level of the test. 4. Let π have a Poisson distribution with mean π. Consider the simple hypothesis π»0 : π = 1 2 1 1 and the alternative composite hypothesis π»1 : π < 2. Thus πΊ = {π: 0 < π ≤ 2}. Let π1 , … . , π12 denote a random sample of size 12 from this distribution. We reject π»0 if and only if observed value of π = π1 + β― . +π12 ≤ 2. If πΎ(π) is the power function of the test, 1 1 1 1 1 find the power πΎ (2) , πΎ (3) , πΎ (4) , πΎ (6) and πΎ (12). What is the significance level of the 5. 6. 7. 8. test? Let π1 , π2 … . , π8 be a random sample of size π = 8 from a Poisson distribution with mean π. Reject the simple null hypothesis π»0 : π = 0.5 and accept π»1 : π > 0.5 if the observed sum ∑8π=1 π₯π ≥ 8. (a) Compute the significance level α of the test. (b) Find the power function πΎ(π) of the test as a sum of Poisson probabilities. (c) Using the Poisson Table, determine πΎ(0.75), πΎ(1) and πΎ(1.25). Let X be a Bernoulli random variable with probability of success p. suppose we want to test at size πΌ, π»0 : π = 0.7 π£π π»1 : π = 0.5. Reject π»0 if ∑20 π=1 π₯π ≤ π. Find c if the size of test is equal to 0.048. Then, find the power function of the test. Let π1 , … , ππ is normally distributed with mean π and variance 100. Reject π»0 : π = 75 π£π π»1 : π = 77 if ∑ππ=1 π₯π > ππ. Determine n and c so that the power function πΎ(π) of the test has the values πΎ(75) = 0.159 and πΎ(77) = 0.841. Let π have a pdf of the form π(π₯, π) = ππ₯ π−1 , 0 < π₯ < 1, zero elsewhere, where π ∈ {π: π = 1, 2}. To test the simple hypothesis π»0 : π = 1 against the alternative simple hypothesis π»1 : π = 2, use a random sample π1 , π2 of size π = 2 and define the critical 3 region to be πΆ = {(π₯1 , π₯2 ): 4 ≤ π₯1 π₯2 }. Find the power function of the test. 5 Stat 319 Exercises: Theory of Statistics 2 Dr. Samah Alghamdi 9. Let π1 < π2 < π3 < π4 be the order statistics of a random sample of size π = 4 from a distribution with pdf π(π₯; π) = 1⁄π , 0 < π₯ < π, zero elsewhere, where 0 < π. The hypothesis π»0 : π = 1 is rejected and π»1 : π > 1 is accepted if the observed π4 ≥ π. (a) Find the constant π so that the significance level is πΌ = 0.05. (b) Determine the power function of the test. 10. Let π be a single observation from the density π(π₯, π) = (2ππ₯ + 1 − π), 0 < π₯ < 1, where −1 ≤ π ≤ 1. To test π»0 : π = 0 vs π»1 : π > 0, the following procedure was used: Reject π»0 1 if X exceeds 2. Find the power and the size of the test 11. Let π1 , … . , ππ denote a random sample from π(π₯; π) = (1⁄π), 0 < π₯ < π, and let π1 , … . , ππ be the corresponding ordered sample. To test π»0 : π = π0 versus π»1 : π ≠ π0 , the π following test was used: Accept π»0 if π0 ( √πΌ ) ≤ ππ ≤ π0 ; otherwise reject. Find the power function of this test. 12. Let π1 , … . , ππ be a random sample of size π from π(π₯; π) = π 2 π₯π −ππ₯ π₯ > 0. In testing π»0 : π = 1 versus π»1 : π > 1 for π = 1 (a sample of size 1) the following test was used: Reject π»0 if and only if π1 ≤ 1. Find the power function and size of the test. 6 Stat 319 Exercises: Theory of Statistics 2 Dr. Samah Alghamdi Testing Hypotheses Case 1: Simple Hypotheses (Neyman-Pearson Lemma) −π₯ 1. Let the random variable π have the pdf π(π₯; π) = (1⁄π)π ⁄π , 0 < π₯ < ∞ , zero elsewhere. Consider the simple hypothesis π»0 : π = 2 and the alternative hypothesis π»1 : π = 4. Let π1 , π2 denote a random sample of size 2 from this distribution. Show that the best of π»0 against π»1 may be carried out by use of the statistic π1 + π2. 2. Let π1 , π2 , … . . π10 be a random sample of size 10 from a normal distribution π(0, π 2 ). Find a best critical region of size πΌ = 0.05 for testing π»0 : π 2 = 1 against π»1 : π 2 = 2. 3. If π1 , π2 , … . . ππ is a random sample from a distribution having pdf of the formπ(π₯; π) = ππ₯ π−1 , 0 < π₯ < 1, zero elsewhere, show that a best critical region for testing π»0 : π = 1 against π»1 : π = 2 is πΆ = {(π₯1 , π₯2 , … . . π₯π ): π ≤ ∏ππ=1 π₯π }. 4. Let π1 , π2 , … . . π10 be a random sample from a distribution that is π(π1 , π2 ). Find a best test of the simple hypothesis π»0 : π1 = 0, π2 = 1 against the alternative simple hypothesis π»1 : π1 = 1, π2 = 4. 5. Let π1 , π2 , … . . ππ denote a random sample from a normal distribution π(π, 100). Show that πΆ = {(π₯1 , π₯2 , … . . π₯π ): π ≤ π₯Μ } is a best critical region for testing π»0 : π = 75 against π»1 : π = 78 . Find π and π so that π[(π1 , π2 , … . , ππ ) ∈ πΆ; π»0 ] = π(πΜ ≥ π; π»0 ) = 0.05 and π[(π1 , π2 , … . , ππ ) ∈ πΆ; π»1 ] = π(πΜ ≥ π; π»1 ) = 0.90, approximately. 6. If π1 , π2 , … . . ππ is a random sample from a beta distribution with parameters πΌ = π½ = π > 0, find a best critical region for testing π»0 : π = 1 against π»1 : π = 2. 7. Let π1 , π2 , … . . ππ be iid with pmf π(π₯; π) = π π₯ (1 − π)1−π₯ , π₯ = 0, 1, zero elsewhere. 1 Show that πΆ = {(π₯1 , π₯2 , … . . π₯π ): ∑π1 π₯π ≤ π} is a best critical region for testing π»0 : π = 2 1 against π»1 : π = 3. Use the Central Limit Theorem to find π and π so that approximately π(∑π1 ππ ≤ π; π»0 ) = 0.10 and π(∑π1 ππ ≤ π; π»1 ) = 0.80. 8. Let π1 , π2 , … . . π10 denote a random sample of size 10 from a Poisson distribution with mean π. Show that the critical region πΆ defined by ∑10 1 π₯π ≥ π is a best critical region for testing π»0 : π = 0.1 against π»1 : π = 0.5. Determine, for this test, c and the power at π = 0.5 when the significance level πΌ = 0.08. 9. Let X be a random variable has the pdf π½ π(π₯) = π½+1 , π₯ ≥ 1, π½>0 π₯ Find the best critical region of size πΌ for testing π»0 : π½ = π½0 versus π»1 : π½ = π½1 , π½1 > π½0. 7 Stat 319 Exercises: Theory of Statistics 2 Dr. Samah Alghamdi 10. Let π1 , π2 , … . . ππ be independent random variables have π(π, π 2 ). Find the best critical region of size πΌ for testing: (i) π»0 : π = π0 against π»1 : π = π1 when π0 > π1 and π 2 is known. (ii) π»0 : π 2 = π02 against π»1 : π 2 = π12 when π02 < π12 and π is known. 8 Stat 319 Exercises: Theory of Statistics 2 Dr. Samah Alghamdi Case 2: Simple Hypothesis against One-Sided Hypothesis 1. Consider a normal distribution of the form π(π, 4). Test the simple hypothesis π»0 : π = 0 against the alternative composite hypothesis π»1 : π > 0 and find the best rejection region. 2. Let π1 , π2 , … . , ππ be a random sample from a distribution with pdf π(π₯; π) = ππ₯ π−1 , 0 < π₯ < 1, zero elsewhere, where π > 0. Calculate the best critical region to reject π»0 : π = 6 versus π»1 : π < 6. 1 3. Let π have the pdf π(π₯; π) = π π₯ (1 − π)1−π₯ , π₯ = 0, 1, zero elsewhere. We test π»0 : π = 2 1 against π»1 : π < 2 by taking a random sample π1 , π2 , … . , π5 of size π = 5. Find the best critical region of size πΌ. 4. Suppose that π1 , π2 , … . , ππ is a random sample of size n from exponential population with 1 parameter π. Determine the best rejection region of size πΌ for π»0 : π = π0 against π»1 : π > π0 . 1 5. If π~πΊππππ (2, π). Find the best critical region for π»0 : π = 1 against π»1 : π < 1. 6. Let X be a random sample whose probability mass function is binomial distribution with parameters π = 10 and p. Test the hypotheses π»0 : π = 0.25 against π»1 : π > 0.25 and find the best critical region of size πΌ of this test. 9