Review Test Submission: Chapter 3 Blackboard Homework H Review Test Submission: Chapter 3 Blackboard Homework User Angel Tafoya Course IDS 270 Introduction to Business Statistics(31167) 2013 Fall Test Chapter 3 Blackboard Homework Started 9/20/13 10:16 PM Submitted 9/20/13 10:55 PM Status Completed Score 60 out of 70 points Time Elapsed 38 minutes. Instructions Question 1 3.1 (3.1) A market researcher interviews a large sample of consumers, both men and women. She asks each consumer which of two health plans he or she prefers. Is this study an experiment? Why or why not? Selected Answer: This is not an experiment because we are only asking people for an opinion that they already have. Correct Answer: This is not an experiment because we are only asking people for an opinion that they already have. Question 2 3.1 (3.1) Which is the explanatory and which the response variable? Selected Answer: Gender is the explanatory; Health plan is the response. Correct Answer: Gender is the explanatory; Health plan is the response. Question 3 3.14 (3.14) Which of the two questions received 40% favorable and which 80% and why. Selected Answer: Question A (Question 1 in the 2nd edition of the textbook) got 40% agreement and Question B (Question 2) got 80% agreement due to the use of phrases like 7 out of 7 points 3 out of 3 points 0 out of 10 points
| If event A can occur in m possible ways and event B can occur in n possible ways,|
there are m+n possible ways for either event A or event B to occur,
but only if there are no events in common between them.
n(A or B)=n(A)+n(B)-n(AB)
Because often one works with non-overlapping events, you will find that the last term is commonly omitted, but added later. It is better to learn the formula correctly the first time and make a special case when the intersection is indeed empty. An empty intersection might occur due to happenstance or it might occur because the events cannot occur simultaneously, i.e. the events are mutually exclusive. In the M&M® example above, the color selections were mutually exclusive.
In the homework you will look at an example of overlapping events when you calculate the probability of the green die having a 2 or the red die of having a 5. A careful inspecation of the diagram in the prior lesson indicates that although there are six outcomes where the green die has a 2 and six outcomes where the red die has a 5, we must be careful not to double count the event where both the green die has a 2 and the red die has a 5. There are thus only 11 not 12 corresponding outcomes and the probability was 11/36 or about 0.306.
Factorial RuleThe factorial rule is used when you want to find the number of arrangements for ALL objects.
Example: Suppose you have four candles you wish to arrange from left to right on your dinner table. The four candles are vanilla, mulberry, orange, and raspberry fragrances (shorthand: V, M, O, R). How many options do you have?
Solution: If you select V first then you still have three options remaining. If you then pick O, you have two candles to choose from. You can compute the number of ways to decorate your table by the factoral rule: for the first choice (event) you have 4 choices; for the second, 3; for the third, 2; and for the last, only 1. The total ways then to select the four candles are: 4!=4321 = 24.
These types of problems occur frequently and can be summarized as follows.
|Factorial Rule: For n different items, there are n! arrangements.|
PermutationsAnother word for arrangements is permutations. Please recall that the symbol ! is mathematical shorthand for factorial. n!=n(n-1)! and 1!=1. Please also note that by definition and because it makes these types of problems easier, 0!=1. 5! = 54321 = 120, 4! = 24, 3! = 321 = 6, and 2!=2.
Try solving this exercise on your own: You need to study, practice football, fix dinner, phone a friend, and go buy a notebook. How many different ways can you arrange your schedule?
Permutation is another name for possible arrangements with SOME items from a given set. It is important to remember that order chosen or position arranged is taken into account. Hence permutations are similar to anagrams. Given below is the necessary equation.
|nPr = n! / (n - r)!|
where r is the number of items arranged from n elements.
More information on permutations, permutations with repeated elements, and permutations on a circle can be found at this location.
CombinationsCombinations are arrangements of elements without regard to their order or position.
|nCr = n! / (r!(n - r)!)|
where r is the number of items taken from n elements.
Note that these numbers are the same as those in Pascal's Triangle, the binomial formula, and the binomial distribution. Those less than about four digits become very familiar.
Example: You have five places left for stamps in your stamp book and you have eight stamps. How many different ways can you select five?
Answer: 8!/(5!3!) = 876/(32)=56.
Think of putting them in slots, the first has eight choices, the next slot has seven choices and so forth as demonstrated.
Each combination of choosing 5 out of the 8 has permutations of its own. The five can be arranged in the following ways:
Thus there are (8!÷3!)÷5! = 8!÷(5!3!)=56 ways to select five of eight, but 6720 ways to arrange five of eight.
Dependent vs. IndependentWhen working with the multiplication rule, keep in mind whether or not the events are independent. Independent events are those that do not affect each other. Otherwise the events are dependent. When sampling is done with replacement, the selected object is put back before the next object is selected. The events remain independent. When down without replacement the events become dependent. P(A|B) represents the probability of A occurring after B has already taken place. This is known as the conditional probability.
|P(A and B) = P(A)·P(B) if A and B are independent.|
P(A and B) = P(A)·P(B|A) if A and B are dependent.
Sometimes the probability of A and B occurring is given, but the question asks for the probability of B occurring after A. All that requires is solving the algebraic equation, P(A and B) = P(A)·P(B|A) for P(B|A), the conditional probability.
Tree diagrams are a method of double checking your work when the sample space is small.
Example: A couple plans on having 3 children. What is the probability of them having two boys and one girl?
|B ---- B ---B|
|---- G ---B|
|G ----- B ---B|
|---- G ---B|
|2 × 4 = 8|
In the chart, there are three different ways to have two boys and one girl. Thus the probability is 3/8 or 0.375. One can also think of the only girl being born first, second, or third. We can do it in a different way: P(GBB) + P(BGB) + P(BBG) = ½×½×½ + ½×½×½ + ½×½×½ = 1/8 + 1/8 + 1/8 = 3/8. Of course, those of us who have done this awhile immediately think in terms of Pascal's Triangle and nCr!
Example: What is the probability of rolling a die twice and getting two sixes?
Answer:P(6)·P(6) = 1/6 × 1/6 = 1/36 = 0.0278.
Complementary EventsIn Geometry, complementary angles summed to 90°these angles together complete a right angle. Another widely used meaning is that complement is opposite, or the negation of something. In probability, the complement of event A are the outcomes which do NOT have event A occurring. The notation of the complement of A is a horizontal bar over A (or for these webpages: Ã). Please note that this spelling and meaning for complement is distinct from compliment which means a formal expression of esteem, respect, affection, or admiration.
Example: A local theater group is planning to give away a season ticket via a raffle. Eighty women dropped their ticket stubs in the bucket while only 35 men did. What is the probability of the winning ticket not going to a woman?
Solution: Thirty-five men dropped their stubs of the 115 total tickets. P(not getting a woman) = P(man) = 35/115 = 7/23 = 0.304.
At Least OneUsing the complementary rule with the multiplication rule, one can find the probability of at least one event being what we want. At least one means the same as one or more. The complement of one or more is none. So the multiplication rule is used to find P(none) and then take the complement of it. P(at least one) = 1 - P(none)!!!
Example: A person deals you a new five card hand. What is the probability of having at least one heart?
Solution:P(at least one heart) = 1 - P(none) = 1 - 13C039C5 ÷ 52C5 = 1 - (39/52)(38/51)(37/50)(36/49)(35/48) = 1 - 0.222 = 0.778. Just think how long it would have taken if instead you calculated the probabilities for getting one heart, two hearts...!
Please note, the method used above for computing none is very general and not well nor widely documented. I'm referring specifically to the expression: 13C039C5 ÷ 52C5. This expression is saying of the 13 hearts we choose 0, whereas of the other 39 cards we choose 5. These two items are multipied together then divided by the number of ways to choose 5 cards from 52. Thus to calculate the probability for getting one heart would be: 13C139C4 ÷ 52C5.
Rules of ComplementAs we have seen before, the probability of something certain to occur (occurring 100% of the time) is one. Using the addition rule for P(A) and P(Ã), which are mutually exclusive because A and Ã cannot occur at the same time and knowing all that is not in A is in Ã, we end up with P(A) + P(Ã) = 1.
|P(A) + P(Ã) = 1|
P(Ã) = 1 - P(A)
P(A) = 1 - P(Ã)
Solution: 20% is destroyed by mice, drought or other means. Remember that percentages are equivalent to probabilities: 80% = 0.80 = P(A). Thus P(Ã) = 1 - 0.8 = 0.2 = 20%.
Bayes' TheoremThomas Bayes was a 18th century English Presbyterian minister (and statistician) who said that probabilities should be revised when we learn more about an event. Bayesian statistics is very much in vogue and is considered by some a different "flavor" of statistics. Bayes' Theorem, also known as Bayes' Rule gives the solution to what Rev. Bayes called the "converse problem". Many medical tests give what are known as false positives. Thus Bayes Theorem is commonly used in paternity suits to calculate the probability that a defendant really is the father of a child, given test results which support such a conclusion. Another example can be found here.
Probability DistributionsA casino operator generally isn't worried about distribution of a gambler winnings because he knows that over the long run the odds favor the casino. Occasionally, someone able to count cards well might distort the otherwise random nature of the outcomes. This random nature is dependent on the cards being well schuffled and multiple decks together makes counting cards impractical for most humans. Such randomness underlies the probability upon which inferential statistics depends.
Seven, eleven, or doubles may get you out of jail in Monopoly, but a close examination of the 36 possible outcomes when two dies are rolled and the pips summed indicates these have probabilities of and for a total of =0.389. The corresponding probability for getting out on the second turn is =0.151. The corresponding randomness gives variety to the game just as randomness gives variety to statistical results. However, there is an underlying distribution which can be analyzed.
The underlying distribution of possible outcomes is important when we consider the probability of any specific sample. We will briefly look at a few theoretical distributions which are commonly encountered.
Consider the combinations examined above and apply it specifically to the case of selecting six of ten when the ten are five males and five females. Any number, say x, between 1 and 5, inclusive, could occur with 6 - x of the other gender occurring. However, if we examine the distribution of the 10C6 = 210 different possibilities we discover 5C55C1 = 5 ways five women and one man might be selected, 5C45C2 = 50 ways four women and two men might be selected, 5C35C3 = 100 ways three women and three men might be selected, and the remaining results are symmetric hence given above. This is a very leptokurtic distribution (Hinkle Figure 7.4).
Another common underlying distribution is the binomial distribution which we already examined in homework 4 problem 1 with the flipping of four coins and gave some formulae in lesson 4. This is a special case in the family of binomial distributions for a given number of trials, where p=q=½. It is natural to ask what happens when p#q#½. The same formula as before applies, namely:
|P(x) = nCx px qn-x where x = 0, 1, 2,..., n|
Example: Find the probability of having five left-handed students in a class of twenty-five, given p=0.1 (n = 25, x = 5, p = 0.1).
Solution:P(5) = (25! ÷(20! · 5!)) (0.1)5 (0.9)20 = 0.064593.
Thus, the probability that 5 of the 25 students will be left-handed is about 6%. As usual, it is important to set up your solution logically. Carefully identify the important values (n, x, p, etc.) before cranking out the numbers and presenting your answer. The TI-83/84 series calculators have BINOMPDF which, if given the two arguments of n and p, in that order, will output a list of n+1 probabilities for each value of x, with the first one being for x=0. BINOMCDF is similar but gives cumulative frequency. Both are under the 2ndVARS or DISTR button (entries 0 and A, so you may need to scroll down). It can be shown that the mean, variance, and standard deviation of a binomial distribution can be expressed in simple formulae as follows:
Example: Suppose 20 biased coins are flipped and each coin has a probability of 75% of coming up heads. Find the mean and standard deviation for this binomial experiment.
Solution:n=20, p=0.75, so q=¼. =n · p = 20 · 0.75 = 15. This is as expected, we expect heads to come up about three quarters the time. = (n · p · q) = (20 · 0.75 · ¼) = 3.75 1.936.
Since the binomial distribution tends to become more like the normal distribution as sample size increases, especially when p and q are nearly equal, we can often approximate the binomial using the normal distribution. More information can be found here which we will summarize by saying np and nq must be greater than 10 (or 5 or 15) before this can be done.
The normal distribution is the most important underlying distribution due to is prevalence in such measurements as intelligence, aptitude, and achievement. In addition, many of the statistics generated through inferential statistics are normally distributed or close to normally distributed.
Central Limit TheoremThis lesson's discussion has established a foundation for the probability and reasonings behind the procedures known as inferential statistics. Specifically, once we have taken a sample and measured a corresponding statistic, we either estimate population parameters or test hypotheses about these unknown parameters. One of the most common parameters we wish to estimate is the population center and the mean is a good measure of such central tendancy. As it turns out, if the sample mean is from a random sample, it is a good estimator of the population mean. To establish how good we need to examine how the means from all possible samples are distributed.
The center, width, and variability of the sampling distribution of the mean is determined by the central limit theorem.
|Central Limit Theorem: As sample size increases, the sampling distribution of sample means approaches that of a normal distribution with a mean the same as the population and a standard deviation equal to the standard deviation of the population divided by the square root of n (the sample size).|
Stated another way, if you draw simple random samples (SRS) of size n from any population whatsoever with mean and finite standard deviation , when n is large, the sampling distribution of the sample means is close to a normal distribution with mean and standard deviation / (n). This standard deviation is often called the standard error of the mean.
It is important to recognize that this standard error of the mean decreases as sample size increases. This means increased precision with larger sample size. However, to improve the precision by a factor of 2 would require an increase in the sample size by a factor of 4.
A second result is that the shape of the sampling distribution of the mean resembles more closely a normal distribution as the sample size increases, even when the population is not normal.
The sampling distribution in the case above of sample means becomes the underlying distribution of the statistic. It is an important component in the chain of reasoning which underpins inferential statistics. Different sampling distributions will apply to different sample parameters. The study of inferential statistics is largely an examination of which distribution applies to which parameter and developing a familiarity with this distribution and how to apply an appropriate statistical test.