Section: 101 (10am-11am) 102 (11am-12pm) 103 (1pm-2pm) 104 (1pm-2pm)

Transcription

1 Stat 0 Midterm Exam Instructor: Tessa Childers-Day 1 May 014 Please write your name and student ID below, and circle your section. With your signature, you certify that you have not observed poor or dishonest conduct on the part of your classmates. You also certify that you have not been a party to poor or dishonest conduct, and that the work on this exam is solely your own. Please show your work ONLY in the area provided. Work outside of this area will not be considered for credit. Name: Student ID: Signature: Date: Section: 101 (10am-11am) 10 (11am-1pm) 103 (1pm-pm) 104 (1pm-pm) 105 (pm-3pm) 106 (3pm-4pm) 107 (4pm-5pm) 108 (5pm-6pm) Answer the questions in the spaces provided. There are questions on the front and back of each page. This midterm covers the material from Lectures 1 through 13, and Homeworks 1 through 6. Show your work, including labeling quantities (such as z-scores). The clearer that your work is, the easier it is to award partial or full credit. If you do not show your work, you will not receive credit. You are welcome to leave your answers as fractions. If you use decimals, please round all answers to two significant figures, and hold your rounding until the final calculation. Please show your work ONLY in the area provided. Work outside of this area will not be considered for credit. Question Points Score Total: 60

2 Stat 0 Midterm Exam, Page of 9 1 May Pat the journalist sees a story in the media stating that the average mobile phone bill in America is $100 per month. Pat believes that the American subscribers of a major magazine pay less per month for their mobile phones than this. Pat seeks your help (as a statistician) to clarify this belief and aid in writing a story. (a) (3 points) State the null and alternative hypotheses in terms of the problem. What kind of test will you use, and why? Is your test one-sided or two-sided, and why? Solution: Null: Average mobile phone bill for all American subscribers is $100. Alternative: Average mobile phone bill for all American subscribers is less than $100. Use a one sample z-test because the data are quantitative and we only have one sample. The test is one-sided because the alternative is directed, due to the journalist having a particular belief/theory about the subscriber s bills. 1 point for the null and alternative hypotheses, 1 point for the type of test and reason, 1 point for the sidedness and reason. (b) (4 points) You and Pat decide to survey of,000 of the 1,47,8 subscribers of this magazine. Respondents are chosen via simple random sample and are asked the cost of their monthly mobile phone bill. The average of the sample is $98.75 with a sample SD of $ Draw (or describe) a box model representing this situation. Include information about the number and kind of tickets in the box, the number and kind of draws, the composition of the sample, and what is done with the tickets. If any of that information is unknown, indicate this. Solution: The box is filled with 1,47,8 tickets (one for each subscriber), and each ticket lists the monthly phone bill of the subscriber. We do not know the average or SD of the tickets in the box. A SRS (random draws without replacement) of,000 tickets is taken, and the average of those,000 tickets is calculated. The average of the sample is $98.75 with an SD of $ point is awarded for specifying each of the following parts: Composition of box Number and kind of draws What is being calculated (average of tickets) Composition of the sample (c) (3 points) Calculate your test statistic. Identify the SE, and explain why you calculated it this way.

3 Stat 0 Midterm Exam, Page 3 of 9 1 May 014 Solution: We find that: estimated SD of box = $41.35 estimated SE for avg = $ = $0.946 $98.75 $100 z = $0.946 = The SE is $0.946, because the SD of the average of the box is estimated by the data from the average. Note that there is no need to use the correction factor here since 1, 47, 8, 000 correction factor = 1, 478, 1 = SE (w/c.f.) = $0.946 = $98.75 $100 z = $0.939 = which doesn t really change anything. However, there is no penalty for using the correction factor. 1 point for using a z-score, 1 point for using the SE for the average, 1 point for the correct values. (d) ( points) Calculate the p-value for your test statistic. Use the context of the situation to make a conclusion about the null hypothesis. Solution: The z-score of corresponds to an area of 8.30%. Thus, the area below is 100% 8.30% = 17.7% = 8.85%. With a p-value of 8.85% there is not statistically significant evidence to reject the null hypothesis. We cannot reject the null hypothesis, that the average mobile phone bill of an American subscriber is $100 1 point for the p-value calculation, 1 point for the conclusion.. You are auditioning for a television game show where questions are drawn randomly from 6 categories. You are planning to spend time preparing for the show by studying each category for an equal amount of time. A friend points out that the questions are not necessarily equally distributed among each of the categories. Since you ve recently taken Stat 0, you do a statistical analysis to decide who is correct. (a) (3 points) State the null and alternative hypotheses in terms of the problem. What kind of test will you use, and why? Is your test one-sided or two-sided, and why? Solution: Null: There are equal numbers of questions from all 6 categories. Alternative: There are not equal numbers of questions from all 6 categories. Use a χ -test for distribution because the data are qualitative and from a single variable. The test is one-sided because a χ test cannot be two-sided. On the other hand, it could be viewed as two-sided, since the alternative hypothesis does not specify a direction. 1 point for the null and alternative hypotheses, 1 point for the type of test and reason, 1 point for the sidedness and reason.

4 Stat 0 Midterm Exam, Page 4 of 9 1 May 014 (b) (4 points) Since you want to spend your preparation time as efficiently as possible, you decide to analyze questions from past episodes. You view a total of 660 questions, and gather the data in the table below about the number of questions from each of the six categories. Draw (or describe) box models representing an equal distribution of questions as well as the distribution described here. Include information about the number and kind of tickets in each box, the number and kind of draws, the composition of the sample, and what is done with the tickets. If any of that information is unknown, indicate this. Indicate observed and expected values as appropriate. Category Geography Arts Entertainment Science History Sports Total Quantity Solution: The equal distribution has 6 tickets (one marked with each category title) in the box. In 660 draws, the expected number of each category is 110, but there is no observed number. With the observed distribution, there are an unknown number of tickets marked with each category title, so there is no expected number. In 660 draws, the observed number of each category is shown in the table above. In both cases we are counting the number of each kind of ticket. Points are awarded for as follows: Composition of boxes (equal distribution, this distribution) (1/ pt each, 1 point total) Number and kind of draws (equal distribution, this distribution) (1/ pt each, 1 point total) What is being calculated (the quantities of each type are totaled) (1/ pt each, 1 point total) Expected values (equal distribution), Observed values (1/ pt each, 1 point total) (c) ( points) Calculate your test statistic. Solution: Recall that the χ test statistic is given by (observed expected) expected In this case, the frequency table (and other parts) is given by: which leads us to Quantity Geog. Arts Entert. Science History Sports Total Obs Exp Obs - Exp (Obs - Exp) (Obs - Exp) /Exp χ = point for the correct formula, one point for the correct value. (d) (3 points) Calculate the p-value for your test statistic using the value you calculated in part (c). If you did not calculate a value in that section, use a value of 30 as your test statistic. Use the context of the situation to make a conclusion about the null hypothesis. Solution: Since there are 6 categories, we are interested in the χ distribution with d.f. = 6 1 = 5. Since our test statistic is greater than (whether or not it was calculated in part (c), we know that our p-value is less than 1%. Thus, we reject the null hypothesis. There is statistically significant evidence that the distribution of questions is not equal. 1 point for the d.f., 1 point for the p-value calculation, 1 point for the conclusion.

5 Stat 0 Midterm Exam, Page 5 of 9 1 May Cameron has made up a game to play with friends. The game consists of putting six tickets in a box. Each ticket has a single letter on it, and the letters are C, A, M, E, R, O, and N. One ticket is then drawn out at random. If a vowel ( A, E, I, O, or U ) is pulled out, Person 1 wins. If any of the other 1 letters in the alphabet are pulled out, Person wins. The ticket drawn is then put back in the box, so the game can be played again. (a) ( points) With only this information, if you are playing this game, do you want to be Person 1 or Person? Does the number of draws made (games played) matter in your choice of being Person 1 or Person? Explain your answer. Solution: You want to be Person because they have a better chance of winning (4/7 vs 3/7). The number of draws/plays doesn t matter because the draws are independent, the chances don t change from game to game. 1 point for choice of person and reason, 1 point for number and reason. (b) (4 points) Suppose the game is slightly changed so that when Person 1 wins (a vowel is drawn), Person pays them $, but when Person wins (a non-vowel is drawn), Person 1 pays them $1. Draw (or describe) two box models to represent the final outcome of 50 repetitions of this game from each player s point of view. Include information about the number and kind of tickets in each box, the number and kind of draws from each box, the composition of the sample, and what is done with the tickets. If any of that information is unknown, indicate this. Solution: boxes, 1 for each person Box for Person 1: The box contains 4 tickets labeled $1 and 3 tickets labeled + $. 50 draws are made with replacement, and the tickets are summed together. The composition of the sample is unknown. Box for Person : The box contains 4 tickets labeled + $1 and 3 tickets labeled $. 50 draws are made with replacement, and the tickets are summed together. The composition of the sample is unknown. 1/ point is awarded for specifying each of the following parts for each person: Composition of box Number and kind of draws What is being calculated Composition of the sample (c) ( points) Keeping in mind the box models from (b), if you are playing this game, do you want to be Person 1 or Person? Does the number of draws made (games played) matter here? Explain your answer. Solution: You want to be Person 1 because they have a positive box avg (and thus EV) while Person has a negative box avg (and thus EV) (+/7 vs -/7). The number of draws/plays doesn t matter because the draws are independent, the chances don t change from game to game. 1 point for choice of person and reason, 1 point for number and reason. (d) (4 points) Use the appropriate box model from (b) to find the approximate chance that Person 1 wins $10 or more. If this is not possible, explain why not, giving as much detail as possible.

6 Stat 0 Midterm Exam, Page 6 of 9 1 May 014 Solution: We find that box avg = $ 7 4 box sd = ($ ( $1)) = $1.48 EV for sum = 50 $ 7 = $14.9 SE for sum = 50 $1.48 = $10.50 Note that we are interested in the chance that Person 1 wins $10 or more. In the probability histogram, the bar over 10 begins at 9.5. So, we use this (the continuity correction) to calculate a z-score, as z = = Looking up 0.45 in the z-table, it corresponds to an area of 34.73%. The area above is 34.73% + 50% = %. There is an approximately % chance that in 50 draws, Person 1 wins $10 or more. The continuity correction should absolutely be used in this case. However, if it is not used, the z-score is z = = Looking up 0.40 in the z-table, it corresponds to an area of 31.08%. The area above is 31.08% + 50% = 65.54%. There is an approximately 65.54% chance that in 50 draws, Person 1 wins $10 or more. 1 point for each of the following parts: Box average and sd EV for sum and SE for sum Usage of z-score Correct chance (must include continuity correction) 4. Imagine that you work for the Nevada Gaming Commission, checking that games of chance played in casinos actually have the advertised chances of winning and losing. You are tasked with evaluating a game which has two outcomes winning $1.50 or losing $1. The usual procedure for evaluating games is to perform a hypothesis test, where the null hypothesis is that the game s outcomes occur with the chances that the casino claims, and the alternative hypothesis is that the game s outcomes do not occur with the chances that the casino claims. If the p-value is statistically significant, the game is considered discredited, and must be closed. If it is not statistically significant, the game may remain open. When you arrive at the casino to test the game, you have forgotten the paperwork which tells you what the chances for the two outcomes should be, so you are unable to perform a test. (a) (3 points) Your assistant suggests that since you cannot complete a hypothesis test, you can instead calculate a confidence interval for the chance of winning. What do you think of this plan? Explain. If you agree with this plan, be sure to specify what confidence level the interval should have and why.

7 Stat 0 Midterm Exam, Page 7 of 9 1 May 014 Solution: This should be fine, since tests and confidence intervals are similar both make inferences about an unknown box using a sample. A two sided hypothesis test simply says whether or not a particular population value is plausible, while a confidence interval provides an entire range of plausible values. The confidence level should be 95%, since the test is usually rejected if the p-value is statistically significant (less than 5%). A confidence interval with level 95% is equivalent to all the values for which a test would produce non-significant (greater than 5%) p-values. 3 points for a correct explanation (yes to CI with good reason, 95% with good reason). points for a somewhat flawed answer, 1 point for a very flawed answer. 0 points for no answer, or an answer with no explanations. (b) (4 points) Since you are unable to contact your boss for instructions, you decide to go ahead with your assistant s plan, and calculate a confidence interval for the chance of winning the game. You play the game 1,000 times, winning 360 times and losing 640 times. Draw (or describe) a box model to represent this situation. Include information about the number and kind of tickets in the box, the number and kind of draws, the composition of the sample, and what is done with the tickets. If any of that information is unknown, indicate this. Solution: There are two kinds of tickets, and each ticket is either +$1.5 or -$1. There is an unknown percentage of each kind of ticket. 1,000 tickets are drawn with replacement, and we calculate the percentage of +$1.5. The sample percentage of the winning tickets is 360/1000, or 36%. 1 point is awarded for specifying each of the following parts: Composition of box Number and kind of draws What is being calculated (the percentage of winning tickets) Composition of the sample (c) (5 points) Using the information from the box model in part (b) and the level from part (a), calculate a confidence interval for the chance of winning and interpret it. If you did not state a level in part (a), then please calculate a 90% confidence level for the chance of winning and interpret it.

8 Stat 0 Midterm Exam, Page 8 of 9 1 May 014 Solution: We estimate EV for percent = 36% SE for percent = (1 0) , % = , % = 1.518%. Since we are sampling with replacement, there is no need for the correction factor. The 95% confidence interval is given by: 36% ± (1.518%) 36% ± 3.036% (3.964%, %). We are about 95% confident that the interval between 3.964% and % covers the true chance of winning the game. If the 90% confidence interval is instead calculated, it should be given by: 36% ± 1.65(1.518%) 36% ±.5047% ( %, %). We are about 90% confident that the interval between % and % covers the true chance of winning the game. 1 point for each of the following parts: Observed value estimating EV for percent and bootstrap SE estimating SE for percent Format of CI (EV ± value SE) Correct numerical interval (either 95% or 90%, small amount of rounding error permissible). Additionally, points for a good interpretation, featuring the randomness in the right place (with the interval, not the average), 1 point for an okay interpretation, with possibly minor flaws, 0 points for no interpretation or a totally incorrect statement. 5. Jamie and Riley are reviewing material for their statistics midterm. (a) (6 points) Jamie explains the Law of Averages this way: The Law of Averages has to do with the frequency of an event. As the number of repetitions of an experiment increases, the Law of Averages says that the frequency of an event gets closer and closer to the expected frequency, eventually reaching it. That s why we use observed values to estimate expected values when we see only a sample from a box, but not the box itself. It s also why a big sample is always bigger than a small sample. Is this the statement Riley should know for the exam? Please describe two ways in which Jamie is right or wrong, correcting the statements if necessary. Solution: Jamie is way off. The Law of Averages says that as the number of repetitions of an experiment increases the relative frequency of an event gets closer and closer to the probability of that event, but never actually reaches it. Jamie is correct that we do use observed values to estimate expected values when we see only a sample from a box, but not the box itself, but this is because the center of the probability histogram of observed values is the expected value. Furthermore, while a big sample is better than a small sample in cases of simple random sampling or random sampling with replacement, increasing sample size is not helpful if a sample is biased to begin with. Also, a big sample can be better because it is more representative, and because it can result in a smaller standard error (for quantities other than the sum). To earn full credit the statement must be corrected in at least two ways, and no part of the explanation may be incorrect. (b) (6 points) Riley explains the Central Limit Theorem (CLT) this way: The CLT has to do with the probability histogram for any population parameter (e.g., the average of the box or the product of the tickets in the box). It tells us that with any number of draws, the probability histogram for a population parameter is approximately normal. It also says that as the number of draws increases, the standard error (or spread) of that probability

9 Stat 0 Midterm Exam, Page 9 of 9 1 May 014 histogram always decreases. Past a certain number of draws, the probability histogram for the population parameter fits a normal curve exactly. Is this the statement Jamie should know for the exam? Please describe two ways in which Riley is right or wrong, correcting the statements if necessary. Solution: Riley is also way off. The CLT has to do with the probability distribution for a sample statistic the population parameter is a single value with no distribution. The CLT does not work for any sample statistic for instance, the product of sample draws is definitely not approximately normally distributed. It is untrue that approximate normality applies with any sample size usually we need many samples to achieve approximate normality. With more draws, it is not true that the spread always decreases for the sum of sample draws, the spread increases with more draws. Finally, the probability histogram is never exactly normal, the normal curve is always an approximation to the probability histogram. To earn full credit the statement must be corrected in at least two ways, and no part of the explanation may be incorrect.