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# INTRO TO STATISTICAL ANALYSIS ECON 160

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This 5 page Class Notes was uploaded by Adriel Kub on Monday October 5, 2015. The Class Notes belongs to ECON 160 at Clark University taught by Myles Callan in Fall. Since its upload, it has received 33 views. For similar materials see /class/219542/econ-160-clark-university in Economcs at Clark University.

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Date Created: 10/05/15

Econ 160 Introduction to Statistics Myles J Callan Spring 2009 1 The mean medanand mode ofthed t but on 2 The normat curve 39ts symmetrtcat about the mean 3 The tatts of the curve do not touch the honzontat axts and Wt not touch 39tt no matter how far away they may be extended 4 The totat area underthe normat curve eduats 1 The area to the tett of the mean 39ts 5 or 50 and the area to the 39ght of the mean 395 atso 5 50 That 39s we expect 50 of the data to be oetow the mean and 50 above the mean 5 Data that te bevond three standard dev39tat39tons from the mean are very rare atthough the range of a true normat dtstr39tbutton t5 tnttntte e In that 39tntervat G39tven any normat d39tstr39tbut39ton we can obta39tn a more exact statement concern39tng the proport39ton of the 39ttems w39tth39tn spe cstandard de a onsfromthe mean The emp cat rute g39tvenasfo ows 1 About 68 of the ttems wt te w39tth39tn the tntervat from one standard dev39tat39ton below the mean to one standard dev39tat39ton abovethe meanthat 39ts w39tth39tn the 39tntervat 1 T v M 1 a 2 About 95 of the 39ttems w e wtthtn the 39tntervat from two standard devtat39tons below the mean to two standard dev39tat39tons abovethe meanthat 39ts w39tth39tn the 39tntervat I T 2quot t0 ll 2H 3 Atmost aH 997 ofthe tems w e n the ntervat from three standard de a ons below the mean to three standard dev39tat39tons abovethe mean that w39tth39tn the39tntervat u a 30 w u 30 The emptrtcat rute can be tustrated on a normat d39tstr39tbut39ton Propemesotthenormatcwye EmptvtcatRute pageng urE p of pm 135 t t Rama Z scores and the standard normal curve The I score orstandmd units 39ts a measure of the retat39tve pos39tt39ton of a data 39ttem In terms ofthe number of standard sfromthe mean We can transform each org nat yatue nto standard units 1 bvus ngthetormuta angina Valuei mean I standard deviation 1 7 7 I a t z s forsampledata or z 039 forpopulatlondata one standard deviatton 0 00 002 00a 004 005 006 007 008 009 0 0008 0016 0024 0032 00398 00478 00558 00638 00718 00796 00876 00956 01034 01114 01192 01272 0135 01428 01506 01586 01664 01742 0182 01896 01974 02052 02128 02206 02282 02358 02434 0251 02586 02662 02736 02812 02886 0296 03034 03108 03182 03256 03328 034 03472 03544 03616 03688 03758 0383 039 0397 04038 04108 04176 04246 04314 0438 04448 04514 04582 04648 04714 04778 04844 04908 04972 05034 0 5098 0516 05222 05284 05346 05408 05468 05528 05588 05646 05704 05762 0582 05878 05934 0599 06046 06102 06156 06212 06266 06318 06372 06424 06476 06528 06578 0663 0668 0673 06778 06826 06876 06922 0697 07016 07062 07108 07154 07198 07242 07286 0733 07372 07416 07458 07498 0754 0758 0762 0766 07698 07738 07776 07814 0785 07888 07924 0796 07994 0803 08064 08098 08132 08164 08198 0823 08262 08294 08324 08354 08384 08414 08444 08472 08502 0853 08558 08584 08612 08638 08664 0869 08714 0874 08764 08788 08812 08836 08858 08882 08904 08926 08948 08968 0899 0901 0903 0905 0907 0909 09108 09128 09146 09164 09182 09198 09216 09232 0925 09266 09282 09298 09312 09328 09342 09356 09372 09386 09398 09412 09426 09438 09452 09464 09476 09488 095 09512 09522 09534 09544 09556 09566 09576 09586 09596 09606 09616 09624 09634 09642 09652 0966 09668 09676 09684 09692 097 09708 09714 09722 09728 09736 09742 0975 09756 09762 09768 09774 0978 09786 09792 09796 09802 09808 09812 09818 09822 09826 09832 09836 0984 09844 0985 09854 09858 09862 09864 09868 09872 09876 0988 09882 09886 0989 09892 09896 09898 09902 09904 09906 0991 09912 09914 09918 0992 09922 09924 09926 09928 0993 09932 09934 09936 09938 0994 09942 09944 09946 09948 09948 0995 09952 09954 09954 09956 09958 09958 0996 09962 09962 09964 09964 09966 09968 09968 0997 0997 09972 09972 09974 09974 09974 09976 09976 09978 09978 09978 0998 0998 Example 1 Suppose a sample of scotes yields a mean of 100 and a standatd deviation of 15 Assume that the distribution is bellshape symmetrical about the mean Use the Empirical mle to answer the questions below Prapemesannenarmatewe EmptvtcalRule gram M E at wze new 4 55 7D 85 100 115 130145 1 1 1 Panza R 35 a 1 d aa7 1 What percent ofthe scores should fall between 85 and 115 68 2 What percent ofthe scores should fa between 70 and 130 95 3 What percent ofthe scores should fa between 55 and 145 Almost e 1 997 Example 2 F391nd the area underthe standard normal curve wh39tch es between 2 113 and z 186 Solution F39trst draw the standard normal curve Table 0 m 1 86 4686 From T 0 Ana U 4685 Reqmed U 1 13 ZTUS Arena 13 l86 4686 3708 30976 a z 0 1 13 186 Next we w have to read 1 186 and z 113 from the standard norma1tab1e Area from z 186 to z 186 09372 wn39tcn 1mp11es Area from z 0 to z 186 04686 Area fromz 113 to z 113 07416wh391ch391mp es Areafromz 0 to z 113 03708 Answer The redu39tred Area 04686 03708 00976 or976 Example 3 F391nd the area underthe standard normal curve wh39tch es between 2 O93 and z 159 Solution F39trst draw the standard normal curve Rammed From To Area 0 093 4441 Rammed 0 159 3238 M31193 159 444132387679 Next we w have to read 1 093 and z 159 from the standard norma1tab1e Area fromz 093 to 2093 06476 mm 1mp es Area from 20 to 2093 03238 Areafromz159 to 2159 O8882wh1ch391mphesAreafromz0 to 2159 04441 Answer The redu39tred Area 093 159 07679 or 7679 Example 4 Suppose x 1s normany d391str391buted random variable w39tth 40 and a 15 F391nd the probab ty that x 1s between 35 and 57 Requued names 3 5 4 I assures g 33 3 From To Area 0 33 1293 0 113 3708 R d Xiana 23 113 1293 3708 5001 5qu on Transform 35 and 57391nto standard un39tts o39 15 40 35 5 X It 740 i 0 33 x35 235 039 15 15 5 5 1 13 x57 15 15 15 Next we read 1 033 and z 113 fromthe standard norma1tab1e Area fromz 033 to z 033 02586 which 391rn sAreafromzO to 2033 01293 Area from z 113 to z 113 07416 wntcn 1mp11es Area from z 0 to z 113 03708 Answer The redu39tred Area 01293 03708 05001 or 5001 ECON 160 PROBABILITY PUTTING STRUCTURE ON UNCERTAI39NTY We begin putting structure on uncertainty by finding the simplest assumptions that we can about certainty and uncertainty 7 that is we define what mathematicians call axioms The Simplest Assumptions Axioms The chance of any event occurring lies somewhere between certainty and impossibility 0 5 PE 5 1 An event either happens or it doesn tl PEC 1 PE If two events are independent then the occurrence of one event doesn t affect the chance of the other occurring PBA PB if A and B are independent If two events are mutually exclusive then the chance of them happening together is zero it is impossible PA 0 B 0 ifA and B are mutually exclusivez The probability of A and B occurring PA m B PA PBA The probability of either A or B occurring3 PA U B PA PB PA W B Using These Simple Assumptions The next step is to use these simple assumptions to derive rules of uncertainty that are applicable in particular circumstances To understand this step you must remember that the approach that we are following is the Frequentist Approach That is when calculating the chance of an event A occurring we are trying to determine PA lim NA N total the number of times the event A occurs divided by the number of trials That is we count the number of occurrences of A and get the relative frequency of the number of occurrences by dividing by the total number of trails In other words the rules that we derive for probability are shortcuts that allow us to avoid this tedious method of counting This means that the rules of probabilitymicertainty are based on counting rules and the simple assumptions that we have outlined above Independent Events If events are independent then the chance of multiple events occurring is just the product of their individual probabilities we combine the two simple rules above one relating to independence and the other relating to the probability of two events 1 Where Ec is the complement of E 2 Note the last two axioms are simply a translation of what we mean in English into the notation that we use in probabilit 3 See classnotes for an explanation ofwhy these are simple assumptions MYLES J CALLAN Example Tossing three coins what is the probability of H T H We know PH 05 PT 05 3 PHTH PHPT PH0125 053 We can derive this from the frequentist approach by imagining an experiment where you toss 3 coins 100 times Of the hundred times how many times will the first coin land heads Obviously 50 times Of those 50 times how many times will the second coin land tails 25 times half of the 50 Finally of those 25 how many times will the third coin land heads 125 half of the 25 So the probability is 125 divided by 100 or 0125 In the above example we have determined the probability of a particular sequence H T H If we were interested in the probability of exactly 2 heads in 3 tosses of a coin the probability would not be the same This is because there are other ways that exactly 2 heads can arise for example in the sequence H H T Thus we have to find a method for counting all of these possible sequences if we are to determine the probability that we require For independent events this is a straightforward matter At first glance the method may look odd but it is a simple if tedious matter of finding out how many different orderings are possible As before once we determine the rule of counting the number of orderings we can use this rule to avoid having to actually count the number of orderings but to understand the rule you will have to at least once go through the process of counting We are going to use this method for all independent events To make the rule that we derive a general rule we think in terms of success and failure rather than thinking of a particular event like tossing a coin In this example we will use the case of 4 events each of which is independent and each of which can be a success or failure Thus possible sequences are S F F S or F S S S and so on From the simple rules it is easy to determine the probability of a particular sequence If we let p be the probability of success then we know that 1 p is the probability of failure events either happen or they don t Independence means that the probability of any sequence is simply the product of the individual events In the case of the first sequence above p1p1pp pz1pz This is the first step in answering the more general question of what is the probability of exactly 2 successes in 4 events Each of the sequences that contain exactly 2 successes5 will have the same probability prove this to yourself If we define n as the number of events and k as the number of successes then we have pquot1p k 4 The question here is how many possible sequences are there ie we have to s on into the actually number of orderings 5 Others include S S F F and F F S S and so on Each of the individual sequences that have exactly k successes will have this probability as they will have k success and therefore nk failures but how many of these sequences are there Learning How To Count We begin by counting the total possible number of orderings of all sequences ie not just for a particular number of successes We do this by determining the each of the possible sequences and then counting them If the first event is a success how many other ways are there for successes As seen in class the way that we determine this is by using a tree diagram 4 4 3 2 4 3 2 4 4 2 4 2 3 3 2 If the first event is a success then we are left with three other possibilities for successes ie that the 2quotd 3 or 43911 could be a success If we follow the first branch of this tree that is when l and 2 are a success then we are left with 2 options either 3 or 4 could be a success For the moment we don t concern ourselves about overcounting for example the ordering 123 on the first branch is the same as 132 on the second branch 4 3 1 4 3 1 4 4 1 4 1 3 3 We do the same for the other scenarios obviously we are magnifying the overcounting which makes it seem a strange way to proceed but we can adjust for that The remaining cases are 24 43 3lt2lt lt2 1 23 4 ltlt 32 1 That is we have 4 possible starting points Once we choose a starting point we have 3 other possibilities remaining If we follow the branches along we are reducing the number of possibilities by one each time This gives us a way of counting the number of possible orderings 4 x3 x 2 x1 24 This logic follows regardless of the number of events So if there are n events the number of possible orderings is n xn1 x n2 x x 1 n Which is called nfaclarial Now back to the initial question How many ways can you get 2 successes in 4 events or more generally how many ways can you get k successes in n events To answer this we return to the tree diagrams and limit the number of orderings that we look at 2 4 If we are trying to determine the number of ways that we can get two successes in four events then we will only be interested in the first two levels of our ordering tree As above if the first event is a success the second could be a success also giving 2 successes or the lst and the 3quoti could be the two successes and so on Thus as we have 4 possible starting points each of which leave 3 options the number of possible orderings is 4 x 3 The rule that we use is the total number of sequences including overcounting that contain 2 successes from 4 l 4 events is E or in general the number of sequences including overcounting that contain k successes in n events6 n is n k Now to the issue of overcounting The possible sequence are 2 6 Check to see if you understand this rule That is ensure that in this case it gives 4 x 3 what should it give in the case of3 successes in 4 events does it 3 Notice however that the lst being a success and the 2nd being a success is including in both the first tree and the second ie as l 7 2 and 2 7 1 each of the sequences is overcounted in this way How do we remove the overcounting The answer is simple How many ways are we goin to overcount We are going to overcount by the number of possible ways to order k successes 7 which is just I That is we just use the same counting rule that we used to count the total number of possible orderings to count the number of possible orderings of successes Thus we have derived the rule that we require the total number of sequences that contain 2 successes from 4 events is 4 2 2 In general the number of sequences that contain k successes in n events7 is n kn k 39 The Binomial Theorem Thus the answer to our original question what is the probability of exactly k successes in n events if the events are independent is k1 n7k kn k p p 7 Again check to see if you understand this rule That is ensure that in this case it gives 4 x 3 2 6 Thatis 12 13 14 2 3 24 34 What should it give in the case of3 successes in 4 events does it

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