Part 13: The Central Limit Theorem
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1 Part 13: The Central Limit Theorem As discussed in Part 12, the normal distribution is a very good model for a wide variety of real world data. And in this section we will give even further evidence of the normal distribution s utility, even in situations where you might not expect it. Suppose that we have a population of size N with mean and standard deviation. Now suppose that we devise an experiment that consists of randomly selecting a sample of size n (with n N of course) from the population and computing its mean. Then, we can now consider a new population of s that consists of all possible values for. That is, we are now thinking of the sample mean as a random variable: (now denoting it using a capital letter). We then ask: What is the mean of this random variable, denoted, and what is the standard deviation of this random variable, denoted here.. Let us look at a very simple example to illustrate what we mean Example 1: Suppose our population consists of the set {1, 2}. We take samples of size n=2, with replacement by choosing 1, replacing it, then choosing 2. So, by the Fundamental Principle of Counting there will be 4 samples of size 2: 2 ways to choose the first data unit, 2 ways to choose the second data unit; thus, 2*2 ways to choose both, with replacement of course. The samples are as follows: (1,1), (1,2), (2,1), (2,2). Accordingly, their means are listed in the distribution table below: Sample Random (1,1) 1 Variable Frequency (1,2) 1.5 Frequency Probability 2.5 (2,1) /4 2 (2,2) / / What are and as well as and? Example 2: Note that the above original population was symmetric about the mean. What if it is not? Again, we will chose a small original population in order to keep the work more manageable. 1
2 Frequency {1,2,7,10} Samples 8 (1,2,7) 3.33 (2,1,7) 3.33 (7,1,2) 3.33 (10,1,2) (1,2,10) 4.33 (2,1,10) 4.33 (7,1,10) 6.00 (10,1,7) (1,7,2) 3.33 (2,7,1) 3.33 (7,2,1) 3.33 (10,2,1) (1,7,10) 6.00 (2,7,10) 6.33 (7,2,10) 6.33 (10,2,7) (1,10,2) 4.33 (2,10,1) 4.33 (7,10,1) 6.00 (10,7,1) /3 13/3 6 19/3 (1,10,7) 6.00 (2,10,7) 6.33 (7,10,2) 6.33 (10,7,2) 6.33 Frequency 10/3 6 13/ /3 6 Note the distributions of means is symmetric even though the population is not and again, the mean of the means is the population mean. But, it doesn t yield a roughly, bell-shape curve here because our population is very small and the sample size is almost the same as the population size. In the first example, we sampled with replacement. In this case, we did not. By not replacing, we are changing the nature of the population because it is so small. When applying the Central Limit Theorem we should be conscious that the population is large in comparison to the sample size and that the sample is large as well. If the original population is normally distributed, then there are no concerns about size. Example 3: Let us look at another case with a bit larger population than the first and again use replacement: Samples Freq Distr Table {3,4,5} Frequency Prob (3,3) /9 (3,4) /9 (3,5) /9 (4,3) /9 (4,4) /9 (4,5) 4.5 (5,3) 4 (5,4) 4.5 (5,5) 5 What are and as well as and? Even before seeing the answers to our investigation questions, we probably could have taken an educated guess at them. Certainly some values of will lie below the value of and some will lie above the value of But, on average we would expect the values of to be close to. In other words, a good guess is =. As for, it seems from our examples that our statistic is 2
3 not matching our parameter. This seems reasonable since we would expect that would be affected by both and n. That is, if is large, then the values for are quite spread out and so the values for should likewise be more spread out. On the other hand, if n is quite large, then there just isn t much room for to vary (think of the extreme case n=n). Thus, larger values of n should correspond to smaller values of. To answer our questions about precisely, we must first specify two possibilities: Case 1: We select n separate values for our calculation of. That is, there are no repeats among our sample values unless there were also repeats among the values. (Most likely this is the scenario you imagined while reading the above description of the random variable.) Case 2: We select the values one at a time, replacing them back to the population each time. Here, there is a possibility of selecting a particular value twice. In Case 1, we will have = and ; and in Case 2, we will have = and. Note, however, that when n is much smaller than N (as is almost always the case in any practical setting), then for case 1, the formula is approximately equal. Thus, it s reasonable to use the formula for for either Case 1 or 2. n Now, not only do we have nice formulas for µ and σ. The most amazing part is that the distribution for the random variable is a very familiar one, regardless of what distribution the underlying population follows! The Central Limit Theorem: The distribution of the sample means, drawn from essentially any population with mean and standard deviation, is approximately normal with mean = and standard deviation, provided the sample size n is large enough. A keen observer may have noted the phrase n is large enough and wondered, How large is large enough?. The answer depends. If the initial population is normally distributed, then any n will do. If the population is at least symmetric, then n 10 or so should suffice. But if the distribution of is unknown, n 30 is probably enough to guarantee that is at least reasonably normally distributed. In general, larger sample sizes are better. How does this apply to our previous discussions? Well, we know that the sample means are approximately normally distributed with large populations and large enough samples; therefore, we can ask a question about probabilities using the normal curve. Example 1: The annual rainfall in a region has a distribution with a mean of 100 inches and a standard deviation of 12 inches. What is the approximate probability that the mean annual rainfall during 36 randomly picked years will exceed inches? Note that we do not know that the distribution is normal. However, we do know that the population is very large and so we can 3
4 assume the distribution of sample means is normal distributed and that the mean of the sample means is the population mean and the sample standard deviation is the population standard deviation divided by the square root of n. Example 2: An airplane with room for 100 passengers has a total baggage limit of 6000 pounds. Suppose the total weight of the baggage checked by an individual passenger is a random variable with mean of 56 pounds and standard deviation of 23 pounds. (a) If a flight is full, what is the approximate probability that the total weight of the passengers baggage will exceed the limit? (b) Find a new baggage limit so that the total weight of the passengers baggage on a full flight will only exceed the limit about once every 100 flights. Example 3: Suppose 100 dice are tossed. Estimate the probability that the sum of the numbers on the top faces of the dice exceeds 370. (Note: What is the expected value of 6 rolls, namely, what is the expected value of the random variable = 1,2,3,4,5,6? In this case, the empirical mean value from 100 rolls is 370/100.) 4
5 Now let us turn our attention to a portion of a population having a certain property. For example, we might want to know the proportion of a population exposed to the H1N1 virus, or the percentage of classmates with black hair. Again, we are interested in knowing about the entire population by examining a sample. Is the population proportion statistic from a sample a good approximation for the parameter? Let us look at an example from above with the population {3,4,5}. What is the proportion of odd numbers? Obviously, it is 2/3. But, what if we can only sample the population. Looking at our samples from before we see the mean of the sample proportions is the population proportion: Sample Proportion of odds (3,3) 1 (3,4) 1/2 (3,5) 1 (4,3) 1/2 (4,4) 0 (4,5) 1/2 (5,3) 1 (5,4) 1/2 (5,5) 1 2/3 Mean of Proportions It appears there is a connection to the Central Limit Theorem! Well, let us form that connection. Now consider a very different looking situation. Suppose the proportion of a population having some property (call it Property A) is p. Now we form a new experiment that consists of randomly selecting n units from the population and computing the proportion in that sample having Property A. We denote this calculation by x/n and call it the sample proportion. Then, since our sample proportions can vary over all possible choices of n units from the population, we again arrive at a random variable, denoted /n, and we again may ask: What is the mean of this variable, denoted µ(/n,) and what is the standard deviation of this variable, denoted σ(/n )? But rather than try to answer these questions by working from scratch, we instead make a shrewd observation. Computing a sample proportion is equivalent to the following: For each of the units in our sample, first record a 1 for each unit that has Property A and record a 0 otherwise. Then add up all these values and divide by n. In other words, we re calculating the sample mean of 1s and 0s. So, if we relabel the initial population so that each unit having Property A is replaced with a 1 and the rest are replaced with a 0, then /n is simply a special case of. Therefore, we can simply apply the Central Limit Theorem once we calculate µ and σ for the initial population, suitably relabeled with 1s and 0s. This should look familiar. Our set is made of only 1 s and 0 s. So, the P( = 1) = p and using the complement of set P( = 0) = 1 p. Thus, direct calculation yields µ = p and p p 2 p1 p. Now applying the Central Limit Theorem we can say: The sample proportion /n is approximately normally distributed with mean p and standard deviation, provided n is suitable large. 5
6 As before, we should ask what suitably large means. Here, the initial population is not symmetric unless p = 0.5. Moreover, the further p is from 0.5 the more lopsided the distribution will be. Taking both of these into account, to see if our sample size is suitably large we can check that both of the following hold: np 10 and n(1 p) 10. In general we will assume our requirements are satisfied. Nonetheless, we should keep a list of such requirements (an upcoming assignment) so that we don t waste our time on useless data. Example 4: The College wishes to estimate student reaction to having all course text books be available online. The college takes a random sample of 50 students and 20 said they are in favor of the idea. Assuming the sample is representative of the student population, what is the proportion of all student who are in favor of the idea? What is the standard deviation for this estimate? Example 5: If 30% of all people have blood type O, what is the approximate probability that, for a random sample of 560 people, the percentage having blood type O will be less than 25% or greater than 35%? 6
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