 Suppose we take a sample of size n from a population. For example, we might have a set of quizzes, and we'll take a sample of size 3. We can compute the mean of this sample, but now consider this as the result of a random experiment. So, we'll select three quizzes, compute the mean, repeat the experiment. The mean of our data values is random. So, the mean data value will have a probability distribution. Now, while there are many different possibilities for this probability distribution, and our original data could have any probability distribution whatsoever, remarkably enough, this sample mean has a very familiar distribution. It's normally distributed, and this leads to what's known as the central limit theorem. So, for us, we'll use the following. Let a sample of n data values be drawn from a population with mean mu and standard deviation sigma. The sample mean will be normally distributed with the mean equal to the population mean, and the standard deviation equal to the population standard deviation, divided by the square root of the size of the sample. So, for example, suppose we have a sample of 100 light bulbs drawn from a population of light bulbs with a mean lifetime of, say, 10,000 hours and standard deviation of 1,000 hours. Let's find the probability a single light bulb has a lifespan of less than 9,800 hours. We'll assume the lifespans are normally distributed, then find the mean and standard deviation of the sample mean for the sample of 100 light bulbs, and the probability the sample mean is less than 9,800 hours. Phew, that's a lot of words. Good thing we don't have the actual data, because we'd have even more to deal with. So, let's take a look at this. Since we're assuming that the lifespan of light bulbs is normally distributed, the probability that a lifespan is less than 9,800 hours, that can be easily calculated using our favorite software tool, and it works out to be 0.4207. By the central limit theorem, the mean and standard deviation of the sample means for a sample of size n equals 100 will be normally distributed with a mean equal to the population mean and a standard deviation equal to the standard deviation divided by the square root of the sample size. So, substituting in those values we find, and so the sample mean will be normally distributed with a mean of 10,000 and a standard deviation of 100, and so we can find the probability that our sample mean is less than 9,800, which will be, and note that while the probability a single light bulb has a lifespan of less than 9,800 hours is rather high, the probability that our sample of 100 will have a mean of 9,800 hours or less is much lower. How does this relate to the fundamental problem of statistics? Give an information about a sample, infer information about the population. So the central limit theorem allows us to compute the probability a sample mean is close to the population mean. So can we use this to find the probability the sample mean is close to the population mean? No. But we can use the central limit theorem to form what we call a confidence interval. To understand confidence intervals, consider the following problem. Suppose you're six feet away from somebody. How far are they from you? And the answer is six feet. And this leads to the following idea. Suppose we find the sample mean and for now assume we have the population standard deviation but don't know the population mean. If the population mean is in fact much greater or much smaller than the sample mean, the probability of obtaining the sample mean we found is very low. But if the population mean is close to the sample mean, the probability of obtaining the observed sample mean is much greater. So we want to choose a confidence level. To understand confidence levels, consider the following. Suppose you want to be more confident a job will be done by an end date. What end date gives you more confidence? The job will be finished by tomorrow or the job will be finished by next year. To be more confident, use a larger interval. Suppose we want to find a C% confidence interval. We can find an interval that includes C% of the values. Now if we didn't know anything about the values, we'd use Chebyshevs. But since our sample means are normally distributed, we can use the normal distribution. Also, since we're expressing the interval in terms of the number of standard deviations, we can use any mean and any standard deviation. So here's what that looks like. Suppose we want to find a 75% confidence interval. We want to express our answer in terms of standard deviations rounded to two decimal places. Since we're expressing our confidence interval in terms of the standard deviations, we can use any value for our standard deviation. And so we'll use... Well, no reason to use anything other than one. And since our confidence interval is centered at the mean, we can use any value for the mean. And so we'll use our mean of zero. So using a standard deviation of one and a mean of zero, we find the probability of being within one standard deviation of the mean is 68%. Which, since we're looking for a 75% confidence interval, isn't high enough. So we'll widen our interval to... How about 1.5 standard deviations? And we find... Which is too much. So we'll narrow our interval to... How about 1.1 standard deviations? And we get... 73%, which is pretty close. Let's widen that just a little bit more. How about to 1.15 standard deviations? And we find... So the 75% confidence interval will be plus or minus 1.15 standard deviations. So we can calculate any confidence interval this way. The common ones are 90%, which is plus or minus 1.64 standard deviations. 95%, which is plus or minus 1.96 standard deviations. 99%, which is plus or minus 2.58 standard deviations. But again, don't memorize formulas. Understand concepts. In general, the C% confidence interval will be where C% of the values reside.