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regression, correlation and hypothesis testing

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regression, correlation and hypothesis testing

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1 Regression, correlation and hypothesis testing: page 3

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1 Regression, correlation and hypothesis testing: page 3

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Neil Sheldon's talk on stats and probability

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Neil Sheldon's talk on stats and probability

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The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975 ( Selvin 1975a), ( Selvin 1975b). It became famous as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in magazine in 1990 ( vos Savant 1990a):

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Vos Savant's response was that the contestant should switch to the other door ( vos Savant 1990a). Under the standard assumptions, contestants who switch have a chance of winning the car, while contestants who stick to their initial choice have only a chance.

The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. Yet another insight is that your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.

Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong ( Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy ( vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result ( Vazsonyi 1999).

The problem is a paradox of the type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to the much older Bertrand's box paradox.

Steve Selvin wrote a letter to the American Statistician in 1975 describing a problem based on the game show Let's Make a Deal, ( Selvin 1975a), dubbing it the "Monty Hall problem" in a subsequent letter ( Selvin 1975b). The problem is mathematically equivalent to the Three Prisoners problem described in Martin Gardner's "Mathematical Games" column in Scientific American in 1959 ( Gardner 1959a) and the Three Shells Problem described in Gardner's book Aha Gotcha ( Gardner 1982).

The same problem was restated in a 1990 letter by Craig Whitaker to Marilyn vos Savant's "Ask Marilyn" column in :

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? ( Whitaker, 1990, as quoted by vos Savant 1990a)

Under the standard assumptions, the probability of winning the car after switching is . The key to this solution is the behavior of the host. Ambiguities in the Parade version do not explicitly define the protocol of the host. However, Marilyn vos Savant's solution ( vos Savant 1990a) printed alongside Whitaker's question implies, and both Selvin (1975a) and vos Savant (1991a) explicitly define, the role of the host as follows:

1. The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).

2. The host must always open a door to reveal a goat and never the car.

3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

When any of these assumptions is varied, it can change the probability of winning by switching doors as detailed in the section below. It is also typically presumed that the car is initially hidden randomly behind the doors and that, if the player initially picks the car, then the host's choice of which goat-hiding door to open is random. ( Krauss and Wang, 2003:9) Some authors, independently or inclusively, assume that the player's initial choice is random as well. Selvin (1975a)

The solution presented by vos Savant (1990b) in Parade shows the three possible arrangements of one car and two goats behind three doors and the result of staying or switching after initially picking door 1 in each case:

A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three.

An intuitive explanation is that, if the contestant initially picks a goat (2 of 3 doors), the contestant will win the car by switching because the other goat can no longer be picked, whereas if the contestant initially picks the car (1 of 3 doors), the contestant will not win the car by switching ( Carlton 2005, concluding remarks). The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability.

Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence does not hold. Before the host opens a door there is a probability the car is behind each door. If the car is behind door 1 the host can open either door 2 or door 3, so the probability the car is behind door 1 AND the host opens door 3 is × = . If the car is behind door 2 (and the player has picked door 1) the host must open door 3, so the probability the car is behind door 2 AND the host opens door 3 is × 1 = . These are the only cases where the host opens door 3, so if the player has picked door 1 and the host opens door 3 the car is twice as likely to be behind door 2. The key is that if the car is behind door 2 the host must open door 3, but if the car is behind door 1 the host can open either door.

Another way to understand the solution is to consider the two original unchosen doors together ( Adams 1990; Devlin 2003, 2005; Williams 2004; Stibel et al., 2008). As Cecil Adams puts it ( Adams 1990), "Monty is saying in effect: you can keep your one door or you can have the other two doors." The chance of finding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is certain to reveal a goat. So the player's choice after the host opens a door is no different than if the host offered the player the option to switch from the original chosen door to the set of both remaining doors. The switch in this case clearly gives the player a probability of choosing the car.

As Keith Devlin says ( Devlin 2003), "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is . I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"

Vos Savant suggests that the solution will be more intuitive with 1,000,000 doors rather than 3. ( vos Savant 1990a) In this case, there are 999,999 doors with goats behind them and one door with a prize. After the player picks a door, the host opens 999,998 of the remaining doors. On average, in 999,999 times out of 1,000,000, the remaining door will contain the prize. Intuitively, the player should ask how likely it is that, given a million doors, he or she managed to pick the right one initially. Stibel et al. ( 2008) proposed that working memory demand is taxed during the Monty Hall problem and that this forces people to "collapse" their choices into two equally probable options. They report that when the number of options is increased to more than 7 choices (7 doors), people tend to switch more often; however, most contestants still incorrectly judge the probability of success at 50:50.

Vos Savant wrote in her first column on the Monty Hall problem that the player should switch ( vos Savant 1990a). She received thousands of letters from her readers - the vast majority of which, including many from readers with PhDs, disagreed with her answer. During 1990-1991, three more of her columns in Parade were devoted to the paradox ( vos Savant 1990-1991). Numerous examples of letters from readers of Vos Savant's columns are presented and discussed in The Monty Hall Dilemma: A Cognitive Illusion Par Excellence ( Granberg 2014).

The discussion was replayed in other venues (e.g., in Cecil Adams' " The Straight Dope" newspaper column, ( Adams 1990)), and reported in major newspapers such as The New York Times ( Tierney 1991).

In an attempt to clarify her answer, she proposed a shell game ( Gardner 1982) to illustrate: "You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are , agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger." She also proposed a similar simulation with three playing cards.

Vos Savant commented that, though some confusion was caused by some readers' not realizing they were supposed to assume that the host must always reveal a goat, almost all her numerous correspondents had correctly understood the problem assumptions, and were still initially convinced that vos Savant's answer ("switch") was wrong.

When first presented with the Monty Hall problem, an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter ( Mueser and Granberg, 1999). Out of 228 subjects in one study, only 13% chose to switch ( Granberg and Brown, 1995:713). In her book The Power of Logical Thinking, vos Savant (1996, p. 15) quotes cognitive psychologist Massimo Piattelli Palmarini: "No other statistical puzzle comes so close to fooling all the people all the time [and] even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans ( Herbranson and Schroeder, 2010).

Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show ( Krauss and Wang, 2003:9) and do not fully specify the host's behavior or that the car's location is randomly selected.( Granberg and Brown, 1995:712; Tierney 1991; VerBruggen 2015) Krauss and Wang ( 2003:10) conjecture that people make the standard assumptions even if they are not explicitly stated.

Although these issues are mathematically significant, even when controlling for these factors, nearly all people still think each of the two unopened doors has an equal probability and conclude that switching does not matter ( Mueser and Granberg, 1999). This "equal probability" assumption is a deeply rooted intuition ( Falk 1992:202). People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not ( Fox and Levav, 2004:637). The problem continues to attract the attention of cognitive psychologists. The typical behavior of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as 1) the endowment effect ( Kahneman et al., 1991), in which people tend to overvalue the winning probability of the already chosen - already "owned" - door; 2) the status quo bias ( Samuelson and Zeckhauser, 1988), in which people prefer to stick with the choice of door they have already made; and 3) the errors of omission vs. errors of commission effect ( Gilovich et al., 1995), in which, ceteris paribus, people prefer any errors for which they are responsible to have occurred through 'omission' of taking action, rather than through having taken an explicit action that later becomes known to have been erroneous. Experimental evidence confirms that these are plausible explanations that do not depend on probability intuition ( Kaivanto et al., 2014; Morone and Fiore, 2007). Another possibility is that people's intuition simply does not deal with the textbook version of the problem, but with a real game show setting ( Enßlin and Westerkamp, 2018). There, the possibility exists that the show master plays deceitfully by opening other doors only if a door with the car was initially chosen. A show master playing deceitfully half of the times modifies the winning chances in case one is offered to switch to "equal probability".

As already remarked, most sources in the field of probability, including many introductory probability textbooks, solve the problem by showing the conditional probabilities that the car is behind door 1 and door 2 are and (not and ) given that the contestant initially picks door 1 and the host opens door 3; various ways to derive and understand this result were given in the previous subsections. Among these sources are several that explicitly criticize the popularly presented "simple" solutions, saying these solutions are "correct but ... shaky" ( Rosenthal 2005a), or do not "address the problem posed" ( Gillman 1992), or are "incomplete" ( Lucas et al. 2009), or are "unconvincing and misleading" ( Eisenhauer 2001) or are (most bluntly) "false" ( Morgan et al. 1991).

Sasha Volokh (2015) wrote that "any explanation that says something like 'the probability of door 1 was 1/3, and nothing can change that...' is automatically fishy: probabilities are expressions of our ignorance about the world, and new information can change the extent of our ignorance."

Some say that these solutions answer a slightly different question - one phrasing is "you have to announce before a door has been opened whether you plan to switch" ( Gillman 1992, emphasis in the original).

The simple solutions show in various ways that a contestant who is determined to switch will win the car with probability , and hence that switching is the winning strategy, if the player has to choose in advance between "always switching", and "always staying". However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given that the player has picked door 1 and the host has opened door 3. As one source says, "the distinction between [these questions] seems to confound many" ( Morgan et al. 1991). The fact that these are different can be shown by varying the problem so that these two probabilities have different numeric values. For example, assume the contestant knows that Monty does not pick the second door randomly among all legal alternatives but instead, when given an opportunity to pick between two losing doors, Monty will open the one on the right. In this situation, the following two questions have different answers:

1. What is the probability of winning the car by always switching?

2. What is the probability of winning the car given the player has picked door 1 and the host has opened door 3?

The answer to the first question is , as is correctly shown by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is . This is because Monty's preference for rightmost doors means that he opens door 3 if the car is behind door 1 (which it is originally with probability ) or if the car is behind door 2 (also originally with probability ). For this variation, the two questions yield different answers. However, as long as the initial probability the car is behind each door is , it is never to the contestant's disadvantage to switch, as the conditional probability of winning by switching is always at least . ( Morgan et al. 1991)

Four university professors published an article (Morgan et al., 1991) in The American Statistician claiming that vos Savant gave the correct advice but the wrong argument. They believed the question asked for the chance of the car behind door 2 given the player's initial pick for door 1 and the opened door 3, and they showed this chance was anything between and 1 depending on the host's decision process given the choice. Only when the decision is completely randomized is the chance .

In an invited comment ( Seymann 1991) and in subsequent letters to the editor, ( vos Savant 1991c; Rao 1992; Bell, 1992; Hogbin and Nijdam, 2010) Morgan et al. were supported by some writers, criticized by others; in each case a response by Morgan et al. is published alongside the letter or comment in The American Statistician. In particular, vos Savant defended herself vigorously. Morgan et al. complained in their response to vos Savant (1991c) that vos Savant still had not actually responded to their own main point. Later in their response to Hogbin and Nijdam (2011), they did agree that it was natural to suppose that the host chooses a door to open completely at random, when he does have a choice, and hence that the conditional probability of winning by switching (i.e., conditional given the situation the player is in when he has to make his choice) has the same value, , as the unconditional probability of winning by switching (i.e., averaged over all possible situations). This equality was already emphasized by Bell (1992), who suggested that Morgan et al.'s mathematically involved solution would appeal only to statisticians, whereas the equivalence of the conditional and unconditional solutions in the case of symmetry was intuitively obvious.

There is disagreement in the literature regarding whether vos Savant's formulation of the problem, as presented in Parade magazine, is asking the first or second question, and whether this difference is significant ( Rosenhouse 2009). Behrends ( 2008) concludes that "One must consider the matter with care to see that both analyses are correct"; which is not to say that they are the same. One analysis for one question, another analysis for the other question. Several critics of the paper by Morgan et al. 1991, whose contributions were published alongside the original paper, criticized the authors for altering vos Savant's wording and misinterpreting her intention ( Rosenhouse 2009). One discussant (William Bell) considered it a matter of taste whether or not one explicitly mentions that (under the standard conditions), which door is opened by the host is independent of whether or not one should want to switch.

Among the simple solutions, the "combined doors solution" comes closest to a conditional solution, as we saw in the discussion of approaches using the concept of odds and Bayes theorem. It is based on the deeply rooted intuition that revealing information that is already known does not affect probabilities. But, knowing that the host can open one of the two unchosen doors to show a goat does not mean that opening a specific door would not affect the probability that the car is behind the initially chosen door. The point is, though we know in advance that the host will open a door and reveal a goat, we do not know which door he will open. If the host chooses uniformly at random between doors hiding a goat (as is the case in the standard interpretation), this probability indeed remains unchanged, but if the host can choose non-randomly between such doors, then the specific door that the host opens reveals additional information. The host can always open a door revealing a goat and (in the standard interpretation of the problem) the probability that the car is behind the initially chosen door does not change, but it is not because of the former that the latter is true. Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially chosen appear persuasive, but the assertion is simply untrue unless each of the host's two choices are equally likely, if he has a choice ( Falk 1992:207,213). The assertion therefore needs to be justified; without justification being given, the solution is at best incomplete. The answer can be correct but the reasoning used to justify it is defective.

The simple solutions above show that a player with a strategy of switching wins the car with overall probability , i.e., without taking account of which door was opened by the host ( Grinstead and Snell 2006:137-138 Carlton 2005). In contrast most sources in the field of probability calculate the conditional probabilities that the car is behind door 1 and door 2 are and given the contestant initially picks door 1 and the host opens door 3 ( Selvin (1975b), Morgan et al. 1991, Chun 1991, Gillman 1992, Carlton 2005, Grinstead and Snell 2006:137-138, Lucas et al. 2009). The solutions in this section consider just those cases in which the player picked door 1 and the host opened door 3.

If we assume that the host opens a door at random, when given a choice, then which door the host opens gives us no information at all as to whether or not the car is behind door 1. In the simple solutions, we have already observed that the probability that the car is behind door 1, the door initially chosen by the player, is initially . Moreover, the host is certainly going to open a (different) door, so opening a door ( which door unspecified) does not change this. must be the average probability that the car is behind door 1 given the host picked door 2 and given the host picked door 3 because these are the only two possibilities. But, these two probabilities are the same. Therefore, they are both equal to ( Morgan et al. 1991). This shows that the chance that the car is behind door 1, given that the player initially chose this door and given that the host opened door 3, is , and it follows that the chance that the car is behind door 2, given that the player initially chose door 1 and the host opened door 3, is . The analysis also shows that the overall success rate of , achieved by always switching, cannot be improved, and underlines what already may well have been intuitively obvious: the choice facing the player is that between the door initially chosen, and the other door left closed by the host, the specific numbers on these doors are irrelevant.

By definition, the conditional probability of winning by switching given the contestant initially picks door 1 and the host opens door 3 is the probability for the event "car is behind door 2 and host opens door 3" divided by the probability for "host opens door 3". These probabilities can be determined referring to the conditional probability table below, or to an equivalent decision tree as shown to the right ( Chun 1991; Carlton 2005; Grinstead and Snell 2006:137-138). The conditional probability of winning by switching is , which is ( Selvin 1975b).

The conditional probability table below shows how 300 cases, in all of which the player initially chooses door 1, would be split up, on average, according to the location of the car and the choice of door to open by the host.

Car hidden behind Door 3
(on average, 100 cases out of 300)

Car hidden behind Door 1
(on average, 100 cases out of 300)

Car hidden behind Door 2
(on average, 100 cases out of 300)

Player initially picks Door 1, 300 repetitions

Host must open Door 2 (100 cases)

Host randomly opens Door 2
(on average, 50 cases)

Host randomly opens Door 3
(on average, 50 cases)

Host must open Door 3 (100 cases)

Switching wins

Switching loses

Switching loses

Switching wins

On those occasions when the host opens Door 2,
switching wins twice as often as staying (100 cases versus 50)

On those occasions when the host opens Door 3,
switching wins twice as often as staying (100 cases versus 50)

Bayes' theorem

Many probability text books and articles in the field of probability theory derive the conditional probability solution through a formal application of Bayes' theorem; among them Gill, 2002 and Henze, 1997. Use of the odds form of Bayes' theorem, often called Bayes' rule, makes such a derivation more transparent ( Rosenthal, 2005a), ( Rosenthal, 2005b).

Initially, the car is equally likely to be behind any of the three doors: the odds on door 1, door 2, and door 3 are 1 : 1 : 1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given that the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is, by definition, the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance that the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus, the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given that the host opened door 3, the probability that the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1.

Richard Gill ( 2011) analyzes the likelihood for the host to open door 3 as follows. Given that the car is not behind door 1, it is equally likely that it is behind door 2 or 3. Therefore, the chance that the host opens door 3 is 50%. Given that the car is behind door 1, the chance that the host opens door 3 is also 50%, because, when the host has a choice, either choice is equally likely. Therefore, whether or not the car is behind door 1, the chance that the host opens door 3 is 50%. The information "host opens door 3" contributes a Bayes factor or likelihood ratio of 1 : 1, on whether or not the car is behind door 1. Initially, the odds against door 1 hiding the car were 2 : 1. Therefore, the posterior odds against door 1 hiding the car remain the same as the prior odds, 2 : 1.

In words, the information which door is opened by the host (door 2 or door 3?) reveals no information at all about whether or not the car is behind door 1, and this is precisely what is alleged to be intuitively obvious by supporters of simple solutions, or using the idioms of mathematical proofs, "obviously true, by symmetry" ( Bell 1992).

Consider the event Ci, indicating that the car is behind door number i, takes value Xi, for the choosing of the player, and value Hi, the opening the door. The player initially chooses door i = 1, C = X1 and the host opens door i = 3, C = H3.

In this case, we have:

P(H3|X1) = 1/2 because this expression only depends on X1, not on any Ci. So, in this particular expression, the choosing of the host doesn't depend on where the car is, and there's only two remaining doors once X1 is chosen (for instance, P(H1|X1) = 0); and P(Ci,Xi) = P(Ci)P(Xi) because Ci and Xi are independent events (the player doesn't know where is the car in order to make a choice).

Then, if the player initially selects door 1, and the host opens door 3, we prove that the conditional probability of winning by switching is:

From the Bayes' rule, we know that P(A,B) = P(A|B)P(B) = P(B|A)P(A). Extending this logic to multiple events, for example A, B and C, we get that we can play with the different subsets of {A, B, C} to calculate the probability of the intersection, as a tool to simplify the calculation of our conditional probability:

In our case, since we know that P(H3|C2,X1) = 1, we are in luck:

Going back to Nalebuff (1987), the Monty Hall problem is also much studied in the literature on game theory and decision theory, and also some popular solutions correspond to this point of view. Vos Savant asks for a decision, not a chance. And the chance aspects of how the car is hidden and how an unchosen door is opened are unknown. From this point of view, one has to remember that the player has two opportunities to make choices: first of all, which door to choose initially; and secondly, whether or not to switch. Since he does not know how the car is hidden nor how the host makes choices, he may be able to make use of his first choice opportunity, as it were to neutralize the actions of the team running the quiz show, including the host.

Following Gill, 2011 a strategy of contestant involves two actions: the initial choice of a door and the decision to switch (or to stick) which may depend on both the door initially chosen and the door to which the host offers switching. For instance, one contestant's strategy is "choose door 1, then switch to door 2 when offered, and do not switch to door 3 when offered". Twelve such deterministic strategies of the contestant exist.

Elementary comparison of contestant's strategies shows that, for every strategy A, there is another strategy B "pick a door then switch no matter what happens" that dominates it ( Gnedin, 2011). No matter how the car is hidden and no matter which rule the host uses when he has a choice between two goats, if A wins the car then B also does. For example, strategy A "pick door 1 then always stick with it" is dominated by the strategy B "pick door 1 then always switch after the host reveals a door": A wins when door 1 conceals the car, while B wins when one of the doors 2 and 3 conceals the car. Similarly, strategy A "pick door 1 then switch to door 2 (if offered), but do not switch to door 3 (if offered)" is dominated by strategy B "pick door 3 then always switch".

Dominance is a strong reason to seek for a solution among always-switching strategies, under fairly general assumptions on the environment in which the contestant is making decisions. In particular, if the car is hidden by means of some randomization device - like tossing symmetric or asymmetric three-sided die - the dominance implies that a strategy maximizing the probability of winning the car will be among three always-switching strategies, namely it will be the strategy that initially picks the least likely door then switches no matter which door to switch is offered by the host.

Strategic dominance links the Monty Hall problem to the game theory. In the zero-sum game setting of Gill, 2011, discarding the non-switching strategies reduces the game to the following simple variant: the host (or the TV-team) decides on the door to hide the car, and the contestant chooses two doors (i.e., the two doors remaining after the player's first, nominal, choice). The contestant wins (and her opponent loses) if the car is behind one of the two doors she chose.

A simple way to demonstrate that a switching strategy really does win two out of three times with the standard assumptions is to simulate the game with playing cards ( Gardner 1959b; vos Savant 1996, p. 8). Three cards from an ordinary deck are used to represent the three doors; one 'special' card represents the door with the car and two other cards represent the goat doors.

The simulation can be repeated several times to simulate multiple rounds of the game. The player picks one of the three cards, then, looking at the remaining two cards the 'host' discards a goat card. If the card remaining in the host's hand is the car card, this is recorded as a switching win; if the host is holding a goat card, the round is recorded as a staying win. As this experiment is repeated over several rounds, the observed win rate for each strategy is likely to approximate its theoretical win probability, in line with the law of large numbers.

Repeated plays also make it clearer why switching is the better strategy. After the player picks his card, it is already determined whether switching will win the round for the player. If this is not convincing, the simulation can be done with the entire deck. ( Gardner 1959b; Adams 1990). In this variant, the car card goes to the host 51 times out of 52, and stays with the host no matter how many non-car cards are discarded.

A common variant of the problem, assumed by several academic authors as the canonical problem, does not make the simplifying assumption that the host must uniformly choose the door to open, but instead that he uses some other strategy. The confusion as to which formalization is authoritative has led to considerable acrimony, particularly because this variant makes proofs more involved without altering the optimality of the always-switch strategy for the player. In this variant, the player can have different probabilities of winning depending on the observed choice of the host, but in any case the probability of winning by switching is at least (and can be as high as 1), while the overall probability of winning by switching is still exactly . The variants are sometimes presented in succession in textbooks and articles intended to teach the basics of probability theory and game theory. A considerable number of other generalizations have also been studied.

The version of the Monty Hall problem published in Parade in 1990 did not specifically state that the host would always open another door, or always offer a choice to switch, or even never open the door revealing the car. However, vos Savant made it clear in her second follow-up column that the intended host's behavior could only be what led to the probability she gave as her original answer. "Anything else is a different question." ( vos Savant 1991a) "Virtually all of my critics understood the intended scenario. I personally read nearly three thousand letters (out of the many additional thousands that arrived) and found nearly every one insisting simply that because two options remained (or an equivalent error), the chances were even. Very few raised questions about ambiguity, and the letters actually published in the column were not among those few." ( vos Savant 1996) The answer follows if the car is placed randomly behind any door, the host must open a door revealing a goat regardless of the player's initial choice and, if two doors are available, chooses which one to open randomly ( Mueser and Granberg, 1999). The table below shows a variety of other possible host behaviors and the impact on the success of switching.

Determining the player's best strategy within a given set of other rules the host must follow is the type of problem studied in game theory. For example, if the host is not required to make the offer to switch the player may suspect the host is malicious and makes the offers more often if the player has initially selected the car. In general, the answer to this sort of question depends on the specific assumptions made about the host's behavior, and might range from "ignore the host completely" to "toss a coin and switch if it comes up heads"; see the last row of the table below.

Morgan et al. ( 1991) and Gillman ( 1992) both show a more general solution where the car is (uniformly) randomly placed but the host is not constrained to pick uniformly randomly if the player has initially selected the car, which is how they both interpret the statement of the problem in Parade despite the author's disclaimers. Both changed the wording of the Parade version to emphasize that point when they restated the problem. They consider a scenario where the host chooses between revealing two goats with a preference expressed as a probability q, having a value between 0 and 1. If the host picks randomly q would be and switching wins with probability regardless of which door the host opens. If the player picks door 1 and the host's preference for door 3 is q, then the probability the host opens door 3 and the car is behind door 2 is while the probability the host opens door 3 and the car is behind door 1 is . These are the only cases where the host opens door 3, so the conditional probability of winning by switching given the host opens door 3 is which simplifies to . Since q can vary between 0 and 1 this conditional probability can vary between and 1. This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching. However neither source suggests the player knows what the value of q is so the player cannot attribute a probability other than the that vos Savant assumed was implicit.

Possible host behaviors in unspecified problem

The host acts as noted in the specific version of the problem.

Switching wins the car two-thirds of the time.
(Specific case of the generalized form below with p = q = )

The host always reveals a goat and always offers a switch. If he has a choice, he chooses the leftmost goat with probability p (which may depend on the player's initial choice) and the rightmost door with probability q = 1 − p. ( Morgan et al. 1991) ( Rosenthal, 2005a) ( Rosenthal, 2005b).

If the host opens the rightmost door, switching wins with probability 1/(1+ q).

"Monty from Hell": The host offers the option to switch only when the player's initial choice is the winning door. ( Tierney 1991)

Switching always yields a goat.

"Mind-reading Monty": The host offers the option to switch in case the guest is determined to stay anyway or in case the guest will switch to a goat. ( Enßlin and Westerkamp, 2018)

Switching always yields a goat.

"Angelic Monty": The host offers the option to switch only when the player has chosen incorrectly ( Granberg 1996:185).

Switching always wins the car.

"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car ( Granberg and Brown, 1995:712) ( Rosenthal, 2005a) ( Rosenthal, 2005b).

Switching wins the car half of the time.

The host knows what lies behind the doors, and (before the player's choice) chooses at random which goat to reveal. He offers the option to switch only when the player's choice happens to differ from his.

Switching wins the car half of the time.

The host opens a door and makes the offer to switch 100% of the time if the contestant initially picked the car, and 50% the time otherwise. ( Mueser and Granberg 1999)

Switching wins the time at the Nash equilibrium.

Four-stage two-player game-theoretic ( Gill, 2010, Gill, 2011). The player is playing against the show organizers (TV station) which includes the host. First stage: organizers choose a door (choice kept secret from player). Second stage: player makes a preliminary choice of door. Third stage: host opens a door. Fourth stage: player makes a final choice. The player wants to win the car, the TV station wants to keep it. This is a zero-sum two-person game. By von Neumann's theorem from game theory, if we allow both parties fully randomized strategies there exists a minimax solution or Nash equilibrium ( Mueser and Granberg 1999).

Minimax solution ( Nash equilibrium): car is first hidden uniformly at random and host later chooses uniform random door to open without revealing the car and different from player's door; player first chooses uniform random door and later always switches to other closed door. With his strategy, the player has a win-chance of at least , however the TV station plays; with the TV station's strategy, the TV station will lose with probability at most , however the player plays. The fact that these two strategies match (at least , at most ) proves that they form the minimax solution.

As previous, but now host has option not to open a door at all.

Minimax solution ( Nash equilibrium): car is first hidden uniformly at random and host later never opens a door; player first chooses a door uniformly at random and later never switches. Player's strategy guarantees a win-chance of at least . TV station's strategy guarantees a lose-chance of at most .

Deal or No Deal case: the host asks the player to open a door, then offers a switch in case the car hasn't been revealed.

Switching wins the car half of the time.

D. L. Ferguson (1975 in a letter to Selvin cited in Selvin 1975b) suggests an N-door generalization of the original problem in which the host opens p losing doors and then offers the player the opportunity to switch; in this variant switching wins with probability . This probability is always greater than , therefore switching always brings an advantage.

Even if the host opens only a single door ( ), the player is better off switching in every case. As N grows larger, the advantage decreases and approaches zero ( Granberg 1996:188).

At the other extreme, if the host opens all losing doors but one ( p = N − 2) the advantage increases as N grows large (the probability of winning by switching is , which approaches 1 as N grows very large).

A quantum version of the paradox illustrates some points about the relation between classical or non-quantum information and quantum information, as encoded in the states of quantum mechanical systems. The formulation is loosely based on quantum game theory. The three doors are replaced by a quantum system allowing three alternatives; opening a door and looking behind it is translated as making a particular measurement. The rules can be stated in this language, and once again the choice for the player is to stick with the initial choice, or change to another "orthogonal" option. The latter strategy turns out to double the chances, just as in the classical case. However, if the show host has not randomized the position of the prize in a fully quantum mechanical way, the player can do even better, and can sometimes even win the prize with certainty ( Flitney and Abbott 2002, D'Ariano et al. 2002).

The earliest of several probability puzzles related to the Monty Hall problem is Bertrand's box paradox, posed by Joseph Bertrand in 1889 in his Calcul des probabilités ( Barbeau 1993). In this puzzle, there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, the question is what is the probability that the other coin is gold. As in the Monty Hall problem, the intuitive answer is , but the probability is actually .

The Three Prisoners problem, published in Martin Gardner's Mathematical Games column in Scientific American in 1959 ( 1959a, 1959b), is equivalent to the Monty Hall problem. This problem involves three condemned prisoners, a random one of whom has been secretly chosen to be pardoned. One of the prisoners begs the warden to tell him the name of one of the others to be executed, arguing that this reveals no information about his own fate but increases his chances of being pardoned from to . The warden obliges, (secretly) flipping a coin to decide which name to provide if the prisoner who is asking is the one being pardoned. The question is whether knowing the warden's answer changes the prisoner's chances of being pardoned. This problem is equivalent to the Monty Hall problem; the prisoner asking the question still has a chance of being pardoned but his unnamed colleague has a chance.

Steve Selvin posed the Monty Hall problem in a pair of letters to the American Statistician in 1975 ( Selvin 1975a), ( Selvin 1975b). The first letter presented the problem in a version close to its presentation in Parade 15 years later. The second appears to be the first use of the term "Monty Hall problem". The problem is actually an extrapolation from the game show. Monty Hall did open a wrong door to build excitement, but offered a known lesser prize - such as $100 cash - rather than a choice to switch doors. As Monty Hall wrote to Selvin:

And if you ever get on my show, the rules hold fast for you - no trading boxes after the selection.

A version of the problem very similar to the one that appeared three years later in Parade was published in 1987 in the Puzzles section of The Journal of Economic Perspectives ( Nalebuff 1987). Nalebuff, as later writers in mathematical economics, sees the problem as a simple and amusing exercise in game theory.

Phillip Martin's article in a 1989 issue of Bridge Today magazine titled "The Monty Hall Trap" ( Martin 1989) presented Selvin's problem as an example of what Martin calls the probability trap of treating non-random information as if it were random, and relates this to concepts in the game of bridge.

A restated version of Selvin's problem appeared in Marilyn vos Savant's Ask Marilyn question-and-answer column of Parade in September 1990. ( vos Savant 1990a) Though vos Savant gave the correct answer that switching would win two-thirds of the time, she estimates the magazine received 10,000 letters including close to 1,000 signed by PhDs, many on letterheads of mathematics and science departments, declaring that her solution was wrong. ( Tierney 1991) Due to the overwhelming response, Parade published an unprecedented four columns on the problem. ( vos Savant 1996, p. xv) As a result of the publicity the problem earned the alternative name Marilyn and the Goats.

In November 1990, an equally contentious discussion of vos Savant's article took place in Cecil Adams's column " The Straight Dope" ( Adams 1990). Adams initially answered, incorrectly, that the chances for the two remaining doors must each be one in two. After a reader wrote in to correct the mathematics of Adams's analysis, Adams agreed that mathematically he had been wrong. "You pick door #1. Now you're offered this choice: open door #1, or open door #2 and door #3. In the latter case you keep the prize if it's behind either door. You'd rather have a two-in-three shot at the prize than one-in-three, wouldn't you? If you think about it, the original problem offers you basically the same choice. Monty is saying in effect: you can keep your one door or you can have the other two doors, one of which (a non-prize door) I'll open for you." Adams did say the Parade version left critical constraints unstated, and without those constraints, the chances of winning by switching were not necessarily two out of three (e.g., it was not reasonable to assume the host always opens a door). Numerous readers, however, wrote in to claim that Adams had been "right the first time" and that the correct chances were one in two.

The Parade column and its response received considerable attention in the press, including a front-page story in the New York Times in which Monty Hall himself was interviewed. ( Tierney 1991) Hall understood the problem, giving the reporter a demonstration with car keys and explaining how actual game play on Let's Make a Deal differed from the rules of the puzzle. In the article, Hall pointed out that because he had control over the way the game progressed, playing on the psychology of the contestant, the theoretical solution did not apply to the show's actual gameplay. He said he was not surprised at the experts' insistence that the probability was 1 out of 2. "That's the same assumption contestants would make on the show after I showed them there was nothing behind one door," he said. "They'd think the odds on their door had now gone up to 1 in 2, so they hated to give up the door no matter how much money I offered. By opening that door we were applying pressure. We called it the Henry James treatment. It was ' The Turn of the Screw.'" Hall clarified that as a game show host he did not have to follow the rules of the puzzle in the vos Savant column and did not always have to allow a person the opportunity to switch (e.g., he might open their door immediately if it was a losing door, might offer them money to not switch from a losing door to a winning door, or might allow them the opportunity to switch only if they had a winning door). "If the host is required to open a door all the time and offer you a switch, then you should take the switch," he said. "But if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood."

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The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975 ( Selvin 1975a), ( Selvin 1975b).

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The original motivation behind this investigation was an attempt to save my Statistics students a few precious seconds in their upcoming S1 module paper.

The mean or expectation of a Binomial Distribution is always very close to mode, (the value of X that has greatest probability). I want to know if you can use the mean to reliably predict the mode.

Binomial Distributions come up all over the place. A classic example would be where you try to score, say, a 5 with an ordinary dice. You perform n trials, and the probability of success on a particular go is 1/6. On each trial you'll either succeed, i.e. score a 5 (the probability of which is 1/6) or you'll fail, i.e. not score 5 (the probability of which is 5/6). X is simply the number of times you score 5 out of those n trials. Binomial literally means 'two numbers': the probability of success and the probability of failure; and the two numbers must add up to 1.

But we can be very general. In the following example we are not even given the context of the experiment; we're just given n which is the number of trials, and p which is the probability of success on each separate trial.

Example Problem
Let X be the number of successes and X~B(20,0.42)
i.e. X is binomially distributed across 20 trials with probability of success in each trial 0.42

Find:
(i) The expected number of successes(the mean).
(ii) The most likely number of successes(the mode).

Part (i) is straightforward. Just use the formula np.
20×0.42 = 8.4

Part (ii) is only a little fiddlier. Textbooks recommend using the expectation np as a guide, and to calculate the probabilities that X takes each of the values either side of np. The answer is the value of X that yields the greatest probability:

Is there a quicker way that doesn't involve using the formula nCrpr(1-p)n-r ?
In this example, the expectation, np, rounded to the nearest integer, is 8. Is it a coincidence that this value is also the most likely number of successes?

Most textbook problems like this one are such that np rounded to the nearest integer gives the most likely number of successes. If this were a reliable fact for any n and p, then we could take all calculations away from the solution to part (ii) and simply write down the answer, saving a minute or more in an exam. So let's see if we can prove or disprove it...

Hypothesis
"If X~B(n,p) then the mode is equal to the expectation rounded to the nearest integer."
.

.Consider the possible shapes of a binomial distribution:

Symmetrical (p=0.5)
When n is even, the hypothesis clearly holds:

When n is odd, np will be exactly halfway between two integers, and will round to the greater of them. This is ok because the distribution is bimodal and we're still ending up with one of the two modes:

Asymmetrical or 'skew' (p≠0.5)
Usually there is one value of X that has greater probability than any of the others. Consider these two similar distributions (notice the value of p is slightly different in each):

.

The value of p changes only slightly from 0.44 to 0.45, but the most likely value of X has now jumped from X= 3 to X= 4. So there must be a value of p between 0.44 and 0.45 where P(X=3) and P(X=4) are exactly equal.

We can find this in the general case, for any n and for any pair P(X=r) and P(X=r-1):

This can be derived in either direction, so: if there exists a positive integer r ≤n such that p=r/(n+1), then P(X=r) = P(X=r-1).

But, also, there is a tipping point at which np stops being rounded up and starts being rounded down: this is halfway between the two integers, at np = r-½. Divide both sides of this equation by n, to give p=(r-½)/n. Now we have two expressions for p.

If the value of p can fall between r/(n+1) and ( r-½)/n then the hypothesis does not hold, i.e. we cannot confidently round the expectation to the nearest integer to calculate the mode.

Example: Given n=20 and r=8, then r/(n+1) = 0.381 and (r-½)/n = 0.375
If p takes a value between these figures then the hypothesis does not hold. Given how close these figures are to each other, this explains why a counterexample seemed so elusive.

Formally, rounding the expectation to the nearest integer does not give the correct mode if and only if there exists some positive integer r ≤n such that,

To paraphrase (or 'para-equate'?), if the expectation np has already been calculated, it may make more sense to consider whether np falls within certain bounds, so rounding the expectation to the nearest integer does not give the correct mode if and only if there exists some positive integer r ≤n such that

Final considerations
These compound inequalities provide only small windows into which p must fall in order to negate the hypothesis. We will now add up the sizes of these small windows to determine the likelihood that the hypothesis does not hold. Effectively we're working out: if you blindly round the expectation to the nearest integer, what's the probability you get the correct (or incorrect) mode? Since the binomial problem could come from any source and in any context, it seems reasonable to assume a uniform distribution for p between 0 and 1.

Consider the size of these small windows - that size is the difference between the two expressions we have for p. We can sum over this expression for all values of r from 1 to n. This gives the probability that, for a randomly chosen value of p, the hypothesis does not hold.

For large values of n, this is approximately equal to ¼. So, for a randomly chosen probability of success from a single trial, the probability that the hypothesis holds is at least ¾. That means you're likely but far from certain to get it right. It's probably not worth just guessing. Shame!

This interactive diagram shows the regions where the value of p negates the hypothesis (the red regions show the values of p that provide counterexamples). Drag the slider to modify the value of n:

The graph below shows how the probability that the hypothesis holds varies with increasing n. See how it tends to ¾ from above.

Conclusion
The recommendation by the textbooks, to separately calculate the probabilities that X takes each of the values either side of np, then choosing the larger, is the best approach to finding the mode.
But there is at least a ¾ chance that it will be the integer value closest to np, so if (and only if) you're really short of time in an exam, guess!

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The original motivation behind this investigation was an attempt to save my Statistics students a few precious seconds in their upcoming S1 module paper. The mean or expectation of a Binomial Distribution is always very close to mode, (the value of X that has greatest probability).

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MSV 11: Binomial Reverse

Sometimes a problem becomes a lot more interesting when turned around the other way.
Here is a standard (and maybe a little dull?) Binomial question that livens up when reversed.
msv-11.ppt - The Task with Answers

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MSV 11: Binomial Reverse Sometimes a problem becomes a lot more interesting when turned around the other way. Here is a standard (and maybe a little dull?) Binomial question that livens up when reversed. msv-11.ppt - The Task with Answers

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proof of variance of binomial distribution

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proof of variance of binomial distribution

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Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.

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Type 1 and 2 errors for binomial distribution

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Type 1 and 2 errors for binomial distribution

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Statistical hypothesis testing implies that no test is ever 100% certain: that's because we rely on probabilities to experiment.

When online marketers and scientists run hypothesis tests, both seek out statistically relevant results. This means that the results of their test have to be true within a range of probabilities (typically 95%).

Even though hypothesis tests are meant to be reliable, there are two types of errors that can occur.
These errors are known as type 1 and type 2 errors.
Understanding Type 1 errors

Type 1 errors - often assimilated with false positives - happen in hypothesis testing when the null hypothesis is true but rejected. The null hypothesis is a general statement or default position that there is no relationship between two measured phenomena.

Simply put, type 1 errors are "false positives" - they happen when the tester validates a statistically significant difference even though there isn't one.

Type 1 errors have a probability of "α" correlated to the level of confidence that you set. A test with a 95% confidence level means that there is a 5% chance of getting a type 1 error.

Consequences of a type 1 Error

Type 1 errors can happen due to bad luck (the 5% chance has played against you) or because you didn't respect the test duration and sample size initially set for your experiment.

Consequently, a type 1 error will bring in a false positive. This means that you will wrongfully assume that your hypothesis testing has worked even though it hasn't.

In real life situations, this could potentially mean losing possible sales due to a faulty assumption caused by the test.
Related: Sample Size Calculator for A/B Testing
A real-life example of a type 1 error

Let's say that you want to increase conversions on a banner displayed on your website. For that to work out, you've planned on adding an image to see if it increases conversions or not.

You start your A/B test running a control version (A) against your variation (B) that contains the image. After 5 days, the variation (B) outperforms the control version by a staggering 25% increase in conversions with an 85% level of confidence.

You stop the test and implement the image in your banner. However, after a month, you noticed that your month-to-month conversions have actually decreased.

That's because you've encountered a type 1 error: your variation didn't actually beat your control version in the long run.

Understanding type 2 errors

If type 1 errors are commonly referred to as "false positives", type 2 errors are referred to as "false negatives".

Type 2 errors happen when you inaccurately assume that no winner has been declared between a control version and a variation although there actually is a winner.

In more statistically accurate terms, type 2 errors happen when the null hypothesis is false and you subsequently fail to reject it.

If the probability of making a type 1 error is determined by "α", the probability of a type 2 error is "β". Beta depends on the power of the test (i.e the probability of not committing a type 2 error, which is equal to 1-β).
There are 3 parameters that can affect the power of a test:

Your sample size (n)

The significance level of your test (α)

The "true" value of your tested parameter (read more here)

Consequences of a type 2 error

Similarly to type 1 errors, type 2 errors can lead to false assumptions and poor decision making that can result in lost sales or decreased profits.

Moreover, getting a false negative (without realizing it) can discredit your conversion optimization efforts even though you could have proven your hypothesis. This can be a discouraging turn of events that could happen to all CRO experts and digital marketers.

A real-life example of a type 2 error

Let's say that you run an e-commerce store that sells high-end, complicated hardware for tech-savvy customers. In an attempt to increase conversions, you have the idea to implement an FAQ below your product page.

You launch an A/B test to see if the variation (B) could outperform your control version (A).

After a week, you do not notice any difference in conversions: both versions seem to convert at the same rate and you start questioning your assumption. Three days later, you stop the test and keep your product page as it is.

At this point, you assume that adding an FAQ to your store didn't have any effect on conversions.

Two weeks later, you hear that a competitor has implemented an FAQ at the same time and observed tangible gains in conversions. You decide to re-run the test for a month in order to get more statistically relevant results based on an increased level of confidence (say 95%).

After a month - surprise - you discover positive gains in conversions for the variation (B). Adding an FAQ at the bottom of your product page has indeed brought your company more sales than the control version.

That's right - your first test encountered a type 2 error!

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Statistical hypothesis testing implies that no test is ever 100% certain: that's because we rely on probabilities to experiment. When online marketers and scientists run hypothesis tests, both seek out statistically relevant results. This means that the results of their test have to be true within a range of probabilities (typically 95%).

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The Art of Statistics with Professor Sir David Spiegelhalter

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David McCandless: The beauty of data visualization

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David McCandless turns complex data sets (like worldwide military spending, media buzz, Facebook status updates) into beautiful, simple diagrams that tease out unseen patterns and connections. Good design, he suggests, is the best way to navigate information glut -- and it may just change the way we see the world.

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https://t.co/TaGPnPvHWL

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by Jonny Griffiths

Activities for the A Level Statistics Classroom

www.making-statistics-vital.co.uk

This Edition: June 2014

Sister site 1 - www.risps.co.uk

40 open pure activities for the A level mathematics classroom

Sister site 2 - www.carom-maths.co.uk

40 pure mathematics active Powerpoints bridging between A Level and university

Sister site 3 - www.further-risps.co.uk

Risps for Further Maths A Level and equivalent

Jonny's email: hello@jonny-griffiths.net
Jonny's website:www.jonny-griffiths.net

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by Jonny Griffiths Activities for the A Level Statistics Classroom www.making-statistics-vital.co.uk This Edition: June 2014 Sister site 1 - www.risps.co.uk 40 open pure activities for the A level mathematics classroom Sister site 2 - www.carom-maths.co.uk 40 pure mathematics active Powerpoints bridging between A Level and university Sister site 3 - www.further-risps.co.uk Risps for Further Maths A Level and equivalent Jonny's email: hello@jonny-griffiths.net Jonny's website:www.jonny-griffiths.net

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Hypothesis testing resources

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integral mathematics resources: log in to the site

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Integral Mathematics Resources: Log in to the site

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Last updated June 16, 2013

This material is based upon work supported by the National Science Foundation under Award Number DMI-0216656. Any opinions, findings, and conclusions or recommendations expressed in this publication are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

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Last updated June 16, 2013 This material is based upon work supported by the National Science Foundation under Award Number DMI-0216656. Any opinions, findings, and conclusions or recommendations expressed in this publication are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

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This resource allows you to run simulations of two situations which can each be modelled by a simple probability distribution.

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The sets of data below have all got muddled up. Can you sort them out?

The sixteen sets of data below were generated by four different probability distributions. Four sets were taken from each of the four distributions.

By searching for patterns in the data, can you put the sets into groups of four which are likely to have come from the same distributions?

You may wish to use a computer to analyse the data. We have created a GeoGebra Worksheet which contains the data sets, so you can use the statistical tools in the spreadsheet function to examine each of the sixteen sets.

Alternatively, you can download this csv file: Data Matching, which can be opened by most spreadsheet software.

This resource is part of the collection Statistics - Maths of Real Life

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rt06 - problem inviting multiple approaches/representations

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The sets of data below have all got muddled up. Can you sort them out? The sixteen sets of data below were generated by four different probability distributions. Four sets were taken from each of the four distributions.

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This resource simulates the binomial distribution. On the first sheet, there is a context of tossing n coins. The second sheet is more general and the values of n and p can be varied to see the distribution.

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This resource simulates the binomial distribution. On the first sheet, there is a context of tossing n coins. The second sheet is more general and the values of n and p can be varied to see the distribution.

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Hypothesis testing using the binomial distribution

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Hypothesis testing using the binomial distribution

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Hypothesis testing using the binomial distribution

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Hypothesis testing using the binomial distribution

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with venn diagrams

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independence and biased dice

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ideas behind independence

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adding two normal distributions gives you another

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adding two normal distributions gives you another

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The normal approximation to the binomial

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This chart accurately shows the probability density functions for three normal distributions.

I estimate that the probability that a variable drawn from the blue distribution is negative is $0.25$. Can you suggest how I made this estimate? Is it likely to be an over- or under-estimate?

How about the probabilities that random variables drawn from the other distributions are negative?

My friend says that the means and variances of these normal distributions are whole numbers. Can you use your estimates to work out the values for these means and variances, using just tables and estimates? Can you specify the values exactly, or can you only specify a certain range of possibilities?

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This chart accurately shows the probability density functions for three normal distributions. I estimate that the probability that a variable drawn from the blue distribution is negative is $0.25$. Can you suggest how I made this estimate? Is it likely to be an over- or under-estimate?

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Hypothesis testing for the mean of a Normal distribution