Introduction to the Practice of Statistics Fifth Edition Moore, McCabe

Introduction to the Practice of Statistics Fifth Edition Moore, McCabe Section 5.1 Homework Answers 5.7 In the proofreading setting if Exercise 5.3, what is the smallest number of misses m with P(X m) no larger than 0.05? You might consider m or more misses as evidence that a proofreader actually catches fewer than 70% of word errors. Answer To me this question provides evidence of why it is so important to first identify your random variable. Notice that the constant m is the number of misses, this means that the variable X, is not the same as in question 7.3. Let us define the random variable Y as the number of misses out of 20 attempts. So this means that p = 0.3. We want the sum of the individual probabilities to be at most 0.5 = P(Y = m) + P(Y = m+1) + P(Y = 20). Let us try and see what P(Y = 15) gives us. P(Y = 15) = 0.000037 which is very small, thus m must be smaller than 15. P(Y = 11) = 0.0120 which is still less than 0.05. But the question is what value of m produces a probability sum equal to at most 0.05 when we consider values greater than or equal to m. Below is the distribution of Y. I can see by inspecting the table and estimating the sums, that the value of m that would give P(Y m) 0.05, is for m to equal 10. Let me verify by calculating P(Y 10) = 1 P(Y 9) = 1 binomdist(9,20,0.3, true) = 0.047962. Thus P(Y 10) = 0.047962 < 0.05. Y P(Y = y) Y P(Y = y) 0 0.000798 11 0.0120 1 0.006839 12 0.0039 2 0.027846 13 0.0010 3 0.071604 14 0.0002 4 0.130421 15 0.0000 5 0.178863 16 0.0000 6 0.191639 17 0.0000 7 0.164262 18 0.0000 8 0.114397 19 0.0000 9 0.06537 20 0.0000 10 0.030817

5.11 In 1998, Mark McGwire of the St. Louis Cardinals hit 70 home runs, a new Major League record. Was this feat as surprising as most of us thought? In the three seasons before 1998, McGwire hit a home run in 11.6% of his times at bat. He went to bat 509 times in 1998. If he continues his past performance, McGwire's home run count in 509 times at bat has approximately the binomial distribution with n = 509 and p = 0.116. (a) What is the mean number of home runs McGwire will hit in 509 times at bat? Assuming that he comes up 509 at bats, n = 509, and the probability of hitting a homerun is 0.116, p = 0.116, then the mean number of homeruns is µ = 509(0.116) = 59.044. What does this number mean? It depicts a person who comes up to the plate 509 times with a probability of hitting a homerun equal to 0.116. Suppose we can observe many people with the same abilities as mentioned. Some would hit 45 homeruns for the season, others 65, and so on. If we took all those numbers and average them, the average is about 59.044 homeruns. (b) What is the probability that he hits 70 or more home runs? To do this problem we need to arrange what we know, and determine what we want. Afterwards we will worry about how we will calculate the probability. In part (a) I described our scenario. We understand that this is like sampling from a population that either results in a homerun being hit or not. The assumptions of the situation will satisfy the binomial setting. The question involves a count of homeruns out of the sample size of 509. Thus we have a question that will require the use of the binomial distribution, with the parameter p= 0.116 and n = 506. So now we know what we are dealing with. The next question is how are we going to calculate this probability? Let the random variable Y denote the number of homeruns out of 509 attempts. So the random variable Y can be any number between 0 and 509. The question is this P(Y 70) =? OK, now if I did this by hand, and I tried to use the formula n! r PX ( = r) = p(1 p) r! n r! complement, ( ) n r I would have to use it 509 69 = 440 times. If I consider the P(Y 70) = 1 - P(Y < 70) I would still need to calculate the above formula 71 times. So, I have two choices. I notice that np> 10 and n(1- p) > 10 so I could use a normal approximation. The other choice is to let the computer crank out calculation after calculation until it comes up with the result. On the exam I want you to do both!!! Not necessarily on the same problem.

Computer does the work. Even if the computer is let to do the work, I will still calculate 1 - P(Y < 70) to get P(Y 70). So here is what I will type into Excel: =1-binomdist(69,509,0.116,true) Which gives the result of 0.076374. This is result is correct till the last digit. A normal approximation Since 509(.116) > 10 and 509(1 0.116) > 10 a normal approximation will give good results. 70-59.044 P(Y 70) P Z > 509(0.116)(1 0.116) = P(Z > 1.516) = 0.06476 =P(Z > 1.52) = 0.06426 Probability Partial Distribution of the Binomial Distribution n = 509, p = 0.116 0.06 0.04 0.02 0 27 32 37 42 47 52 57 62 67 72 77 82 87 Number of Homeruns A normal Approximation with Continuity Correction 69.5-59.044 P(Y 70) P Z > 509(0.116)(1 0.116) = P(Z > 1.447) = 0.07395 =P(Z > 1.45) = 0.07353 The lesson I want you learn is that a normal approximation is just what it sounds like an approximation. Here is a case where we can glimpse at the actual error. A second lesson is that continuity correction generally, in the long run, will provide better results. 5.11c. In 2001, Barry Bonds of the San Francisco Giants hit 73 home runs, breaking McGwire's record. This was surprising. In the three previous seasons, Bonds hit a home run in 8.65% of his times at bat. He batted 476 times in 2001. Considering his home run count as a binomial random, variable with n = 476 and p = 0.0865, what is the probability of 73 or more home runs? Computer does the work. Even if the computer is let to do the work I will still calculate 1 - P(Y < 73) to get P(Y 73). So here is what I will type into Excel: =1-binomdist(72,476,0.0865,true) Which gives the result of 0.0000014481. This is result is correct till the last digit.

A normal approximation Since 476(0.0865) > 10 and 509(1 0.0865) > 10 a normal approximation will give good results. 73-476(0.0865) P(Y 73) P Z > 476(0.0865)(1 0.0865) = P(Z > 5.189) = 0.0000001059 A normal Approximation with Continuity Correction There is really no point in using continuity correction since the basic concept has been discovered, that the probability was nearly zero; a very rare occurrence. I am just doing the first set up in the calculation to give you another example. 72.5-59.044 P(Y 70) P Z > 509(0.116)(1 0.116) 5.15. A Gallup poll finds that 30% of adults visited a casino in the past 12 months, and that 6% bet on college sports. These results come from a random sample of 1011 adults. For an SRS of size n = 1011. a) What is the probability that the sample proportion ˆp is between 0.29 and 0.31 if the population proportion is p = 0.30? We can change the proportions into counts by the formula X = np. The question proportion ˆp is between 0.29 and 0.31 can be viewed as count X is between 0.29(1011) = 293.19 and 0.31(1011) = 313.41 Likewise p = 0.30 is now 0.30(1011) = 303.3 Obviously if you are going to use Excel s Binomdist you will have to enter whole values, which is not the case with a normal approximation. 0.3(1 0.3) We are assuming that µ = p = 0.3, and σ =. 1011 Since np> 10 and n(1 p) > 10 we can use a normal approximation. I want to calculate P(0.29 < p ˆ < 0.31). ˆ ˆ ˆ P(0.29 < p < 0.31) = P(p < 0.29) - P(p < 0.31) 0.31.3 0.29.3 = PZ < PZ < 0.3(0.7) 0.3(0.7) 1011 1011 =P(Z < 0.69) - P(Z < -0.69) = 0.7549-0.2451 = 0.5098

b) What is the probability that the sample proportion ˆp is between 0.05 and 0.07 if the population proportion is p = 0.06? 0.06(1 0.06) We are assuming that µ = p = 0.06, and σ =. 1011 Since np> 10 and n(1 p) > 10 we can use a normal approximation. I want to calculate P(0.05 < p ˆ < 0.07). ˆ ˆ ˆ P(0.05 < p < 0.07) = P(p < 0.07) - P(p < 0.05) 0.07 0.06 0.05 0.06 = PZ < PZ < 0.06(0.94) 0.06(0.94) 1011 1011 =P(Z < 1.34) - P(Z < -1.34) = 0.9099-0.0901 = 0.8198 (c) How does the probability that pˆ falls within ±0.01 of the true p change as p gets closer to 0? In problem 5.5 I displayed a graph which shows that as p gets closer to zero or 1 the standard deviation of counts gets closer to zero. Here is the graph for the standard deviations of proportions. 0.02 0.01 σ ˆp σ ˆp = p(1 p) 1011 p 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1. The two examples show that as we get closer to zero, if we consider the likelihood of being within 1% of the center that value is increasing. Why? Because the graph shows that the variability reduces as p gets closer to 0 or 1. So if the variability is reduced what are we getting closer to? The center! In problem (a) the center is p = 0.3 and in problem (b) the center is p = 0.06.

5.17. The Harvard College Alcohol Study finds that 67% of college students support efforts to crack down on underage drinking. The study took a sample of almost 15,000 students, so the population proportion who support a crackdown is very close to p = 0.67. The administration of your college surveys an SRS of 100 students and finds that 62 support a crackdown on underage drinking. a) What is the sample proportion who support a crackdown on underage drinking? Notice the use of symbols in the set up of the problem. The 67% value is not considered a statistic, but rather a parameter, since the symbol p was used; p = 0.67. There is a mention of a second survey of 100 62 students. Then the sample proportion, ˆp, is 100 = 0.62 b) If in fact the proportion of all students on your campus who support a crackdown is the same as the national 67%, what is the probability that the proportion in an SRS of 100 students is as small or smaller than the result of the administration s sample? So I need to make sure I understand what they want. It looks like they are asking for the chance that I would see a sample proportion as low as or lower than 0.62. Using symbols: P( ˆp 0.62). Now I need to figure out HOW I will go about figuring this out. I know the situation fits a binomial setting, and a proportion contains the same information as if I had talked about counts. The computer using Excel The parameters for this binomial situation is p = 0.67 and n = 100. Let the random variable X count the number of students out of the 100 sampled that support efforts to crack down on underage drinking. I convert the proportion to a count, 0.62(100) = 62. P(X 62) = binomdist(62,100,0.67,true) = 0.1690 A normal approximation - First note that I meet the requirement that np > 10 and n(1 p) > 10. If I did not then I can not proceed with this approach. I need to find the mean of the distribution, µ = 100(0.67) = 67, and the standard deviation, σ = 100(0.67)(1 0.67) and use that in the basic formula that converts a measurement to a z-score, x µ z =, which we will use to calculate a probability. σ Partial Distribution of the Binomial Distribution n = 100, p = 0.67 62-67 P(X 62) P Z < 100(0.67)(1 0.67) = P(Z < -1.06) = 0.1446 A normal approximation with continuity correction. Probability 0.1 0.08 0.06 0.04 0.02 0 42 47 52 57 62 67 72 77 82 87 Number of Students 62.5-67 P(X 62) P Z < 100(0.67)(1 0.67) = P(Z < -0.96) = 0.1685 which is a better approximation than without the continuity correction.

c) A writer in the student paper says that support for a crackdown is lower on your campus than nationally. Write a short letter to the editor explaining why the survey does not support this conclusion. This question is indicative of the argument that will be used throughout MTH 244. Here is how the argument goes: If the real proportion is 0.67, p = 0.67, and we took a sample of 100, how likely is it to get a sample proportion as low or lower than the one we observed, 0.62? The answer was 16.90%. So while the value we got was lower than the target of 67%, it is not that unusual to arrive at the number of 62%. So there this is not strong evidence to suggest that the true proportion is something else than 67%. 5.19 In a test for ESP (extrasensory perception), the experimenter looks at cards that are hidden from the subject. Each card contains either a star, a circle, a wave, or a square. As the experimenter looks at each of 20 cards in turn, the subject names the shape on the card. (a) If a subject simply guesses the shape on each card, what is the probability of a successful guess on a single card? Because the cards are independent, the count of successes in 20 cards has a binomial distribution. This part is not quite clear from the description, but I have to assume this is what is occurring in order to have independence. The person will pick a card at random, look at it, then the person with ESP says what the card is. The person then puts the card back, shuffles, and picks again a card at random. The process is repeated 20 times. The probability of success is 0.25, p = 0.25.. (b) What is the probability that a subject correctly guesses at least 10 of the 20 shapes? P(X 10) =? If I use Excel then I need to calculate the compliment. P(X 10) = 1 P(X 9) I need to calculate the compliment because the function I am going to use =binomdist( has the feature of looping so that it can repeatedly calculate not just one named success but several. The problem is that is only adds or calculates in a particular direction, from the upper most number of successes entered to zero successes. =1 binomdist(9,20,0.25,true) = 0.013864 Notice, we can not use a normal approximation since np is not greater than 10..

(c) In many repetitions of this experiment with a subject who is guessing, how many cards will the subject guess correctly on the average? What is the standard deviation of the number of correct guesses? µ = 20(.25) = 4. (d) A standard ESP deck actually contains 25 cards. There, are five different shapes, each of which appears on 5 cards. The subject knows that the deck has this makeup. Is a binomial model still appropriate for the count of correct guesses in one pass through this deck? If so, what are n and p? If not, why not? As long as we are allowed, after each guess, to place the card back and not show the person the card, it is a binomial situation. If we however, after each guess, don t allow the card back into the deck then the situation is not binomial since the probability of success is changing after each guess. 5.20 A selective college would like to have an entering class of 1200 students. Because not all students who are offered admission accept, the college admits more than 1200 students. Past experience shows that about 70% of the students admitted will accept. The college decides to admit 1500 students. Assuming that students make their decisions independently, the number who accept has the B(1500, 0.7) distribution. If this number is less than 1200, the college will admit students from its waiting list. (a) What are the mean and the standard deviation of the number X of students who accept? 1500(0.7) = 1050 = µ x σ x = 1500(0.7)(0.3) 17.75 (b) Use the normal approximation to find the probability that at least 1000 students accept. P(X >1000)= P Z > = P(Z > -2.82) = 0.9976 1000 1050 1500(0.3)(0.7) Actual value P(X > 1000) = 1 binomdist(999,1500,0.7,true) =0.9976 (c) The college does not want more than 1200 students. What is the probability that more than 1200 will accept? P(X >1200)= P Z > = P(Z > 8.45) 0 is very unlikely. 1200 1050 1500(0.3)(0.7) Once you reach a z-score this high you realize that the result whatever it might be

(d) If the college-decides to increase the number of admission offers to 1700, what is the probability that more than 1200 will accept? P(X >1200)= 1200 1700(0.7) P Z > 1700(0.3)(0.7) = P(Z > 0.53) = 0.2981 With continuity correction. P(X >1200)= 1199.5 1700(0.7) P Z > 1700(0.3)(0.7) = P(Z > 0.50) = 0.3085 Actual value P(X > 1200) = 1 binomdist(1199,1700,0.7,true) =0.3085 5.21 When the ESP study of Exercise 5.19 discovers a subject whose performance appears to be better than guessing, the study continues at greater length. The experimenter looks at many cards bearing one of five shapes (star, square, circle, wave, and cross) in an order determined by random numbers. The subject cannot see the experimenter as he looks at each card in turn, in order to avoid any possible nonverbal clues. The answers of a subject who does not have ESP should be independent observations, each with probability 1/5 of success. We record 1000 attempts. (a) What are the mean and the standard deviation of the count of successes? µ = 0.2(1000) = 200; σ = 0.2(1 0.8)1000 = 12.6491 (b) What are the mean and standard deviation of the proportion of successes among the 1000 attempts? µ = p = 0.2; σ ˆp ˆp 0.2(0.8) = = 0.01265 1000 (c) What is the probability that a subject without ESP will be successful in at least 24% of 1000 attempts? Here we go again. We have the situation in which np > 10 and n(1 p) > 10, so a normal approximation is in order here. We can either go with counts or proportions. I personally prefer counts. If we want at least 24% success out of 1000 attempts then that means I want 0.24(1000) = 240 success or more. Let the random variable X count the number of successes out of 1000 attempts.

Using Excel and the Computer Now I will calculate the probability using Excel s =binomdist( function. P(X 240) = 1 P(X < 240) =1 binomdist(239,1000,0.2, true) = 0.001108 Normal Approximation 240 200 P(X 240) = P Z > 1000(0.2)(0.8) = P(Z > 3.162) = 0.0007835 =P(Z > 3.16) =0.0008 Normal Approximation with Continuity Correction P(X 239.5 200 240) = P Z > 1000(0.2)(0.8) = P(Z > 3.123) = 0.0008952 Notice there is not much difference between the two. But again the approximation with continuity correction wins. (d) The researcher considers evidence of ESP to be a proportion of successes so large that there is only probability 0.01 that a subject could do this well or better by guessing. What proportion of successes must a subject have to meet this standard? (Example 1.30, page 79, shows how to do a normal calculation of the type required here.) The random variable X counts the number of correct guesses out of 1000 attempts. We have a probability of success equal to 0.2 if the subject is guessing. The expected, mean, number of correct guesses is 0.2(1000) = 200. We clearly meet the goal of np 10 and n(1 p) 10. This mean I can think of the distribution as being normally distributed even thought this is not true since the random variable X is not continuous. You can find the corresponding z-score that pertains to 0.01 by using the excel command normsinv(

= normsinv(0.99) which gives the z-score for the given area to the left of that z-score. You must afterward use the formula Z = x - µ σ to convert to a X value. X 200 2.326 = solving this we get x 229.4, which is a sample proportion of 0.2294. So a person 12.6491 would need to guess 230 or more correct. 5.22 The mailing list of an agency that markets scuba-diving trips to the Florida Keys contains 70% males and 30% females. The agency calls 30 people chosen at random from its list. (a) What is the probability that 20 of the 30 are men? (Use the binomial probability formula.) The probability of success is 0.7, and n = 30. P(X = 20) = binomdist(20,30,0.7,false) = 0.1416 30! 20 10 P(X = 20) = (0.7) (0.3) 20!(30 20)! 30! 20 10 = (0.7) (0.3) 20!10! 30(29)(28)(27)..21 = (0.7) (0.3) 10(9)(8)(7)..(2)(1) Be patient cancel away, and you will get the count. 20 10 (b) What is the probability that; the first woman is reached on the fourth call? (That is, the first 4 calls give MMMR) P(M and M and M and F) = 0.7 3 (0.3) = 0.1029 5.25 Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher p than a weak student.) The correctness of an answer to a question is independent of the correctness of answers to other questions. Jodi is a good student for whom p = 0.75. (a) Use the normal approximation to find the probability that Jodi scores 70% or lower on a 100-question test. p = 0.75, n = 100. I want to find P( ˆp < 0.70). I will answer this question using counts instead of proportions. Just a personal preference, and since I want to compare this to normal approximation with continuity correction, it is much easier dealing with counts. Let the random variable X count the number of correct answers in a 100 question exam. P( ˆp < 0.70) = P(X < 70).

I verify that a normal approximation can be done. 100(0.70) > 10 and 100(0.30) > 10 so a normal approximation is possible. With Continuity Correction 70-75 P(X 70) = P Z < 100(0.75)(0.25) 70.5-75 P(X 70) = P Z < = P(Z < -1.155) 100(0.75)(0.25) = 0.12405 = P(Z < -1.039) = 0.1494 Actual value is 14.95% which I got by using Excel (=binomdist(70, 100, 0.75, true) ) (b) If the test contains 250 questions, what is the probability that Jodi will score 70% or lower? p = 0.75; n = 250. 0.70(250) = 175 175-0.75(250) P(X 175) = P Z < 250(0.75)(0.25) = P(Z < -1.826) = 0.03393 With Continuity Correction 175.5-0.75(250) P(X 175) = P Z < 250(0.75)(0.25) = P(Z < -1.753) = 0.03980 Actual value using Excel: 0.0418. Again you see that the continuity correction procedure gives a better approximation. (c) How many questions must the test contain in order to reduce the standard deviation of Jodi's proportion of correct answers to half its value for a 100-item test? The standard deviation for proportions (not counts) in the 100 question test is 0.043301 (I used the formula for proportions). Half of that is 0.021651. 0.25(0.75) 0.021651 = solve for n. n n = 400 (d) Laura is a weaker student for whom p = 0.6. Does the answer you gave in (c) for the standard deviation of Jodi's score apply to Laura's standard deviation also? Yes, in order for Laura to reduce her standard deviation (which is not the same as Jodi s) in half, she will also need to have a 400 question exam. Verify this on your own.