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## The standard error of the sample mean and confidence intervals

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**The standard error of the sample mean and confidence**intervals How far is the average sample mean from the population mean? In what interval around mu can we expect to find 95% or 99% or sample means**An introduction to random samples**• When we speak about samples in statistics, we are talking about random samples. • Random samples are samples that are obtained in line with very specific rules. • If those rules are followed, the sample will be representative of the population from which it is drawn. • One way that it will be representative of the population is that the sample mean will be close to the population mean. • Specifically, on the average, sample means are closer to mu than are individual scores.**Random samples: Some principles**• In a random sample, each and every score must have an equal chance of being chosen each time you add a score to the sample. • Thus, the same score can be selected more than once, simply by chance. (This is called sampling with replacement.) • The number of scores in a sample is called “n.” • Sample statistics based on random samples provide least squared, unbiased estimates of their population parameters.**The variance and the standard deviation are the basis for**the rest of this chapter. • In Chapter 1 you learned to compute the average squared distance of individual scores from mu. We called it the variance. • Taking a square root, you got the standard deviation. • Now we are going to ask a slightly different question and transform the variance and standard deviation in another way.**As you add scores to a random sample**• Each randomly selected score tends to correct the sample mean back toward mu • If we have several samples, as we add scores the sample means get closer to each other and closer to mu • The larger the samples, the closer they will be to mu, on the average.**Let’s see how that happens**Population is 1320 students taking a test. is 72.00, = 12 Let’s randomly sample one student at a time and see what happens.We’ll create a random sample with 8 students’ scores in the sample.**Scores**Mean Standard deviations 3 2 1 0 1 2 3 102 72 66 76 66 78 69 63 Test Scores F r e q u e n c y score 36 48 60 96 108 72 84 Sample scores: Means: 87 80 79 76.4 76.7 75.6 74.0**How much closer to mu does the sample mean get when you**increase n, the size of the sample? (1) • The average squared distance of individual scores is called the variance. You learned to compute it in Chapter 1. • The symbol for the mean of a sample is the letter X with a bar over it.We will write that as X-bar.**How much closer to mu does the sample mean get when you**increase n, the size of the sample? (2) • The average squared distance of sample means from mu is the average squared distance of individual scores from mu divided by n, the size of the sample. • Let’s put that in a formula • sigma2X-bar = sigma2/n**Let’s take that one step further**• As you know, the square root of the variance is called the standard deviation. It is the average unsquared distance of individual scores from mu. • The average unsquared distance of sample means from mu is the square root of sigma2X-bar = sigmaX-bar. sigmaX-bar is called the standard error of the sample mean or, more briefly, the standard error of the mean. Let’s look at the formulae: sigma2X-bar = sigma2/n sigmaX-bar = sigma/**The standard error of the mean**• Let’s translate the formula into English, just to be sure you understand it. Here is the formula again: sigmaX-bar = sigma/ • In English: The standard error of the sample mean equals the ordinary standard deviation divided by the square root of the sample size. • Another way to say that: The average unsquared distance of the means of random samples from mu equals the average unsquared distance of individual scores from the population mean divided by the square root of the sample size.**The standard error of the mean is the standard deviation of**the sample means around mu. • We could compute the average unsquared distance of sample means from mu by 1. subtracting mu from each sample mean. 2. squaring the differences, 3. getting a sum of squares 4. dividing by the number of sample means and 5. taking the square root. • We would need to do that for all possible samples of a particular size from a population. That’s a lot of calculations. (A real lot.)**Example: Start with a tiny population N=5**• The scores in this population form a perfectly rectangular distribution. • Mu = 5.00 • Sigma = 2.83 • We are going to list all the possible samples of size 2 (n=2) • First see the population, then the list of samples**If we did compute a standard deviation of sample means from**mu, it should give the same result as the formula • Let’s see if it does. • We can only do all the computations if we have a very small population and an even tinier sample. • Let’s use an example with N=5 and n, the size of each sample = 2.**Computing sigma**• SS=(1-5)2+(3-5)2+(5-5)2+ (7-5)2+ (9-5)2=40 • sigma2=SS/N=40/5=8.00 • sigma = 2.83**The standard error = the standard deviation divided by the**square root of n, the sample size • In the example you just saw, sigma = 2.83. Divide that by the square root of n (1.414) and you get the standard error of the mean (2.00). • The formula works. And it works every time.**Let’s see what sigmaX-bar can tell us**• We know that the mean of SAT/GRE scores = 500 and sigma = 100 • So 68.26% of individuals will score between 400 and 600 and 95.44% will score between 300 and 700 • But if we take random samples of SAT scores, with 4 people in each sample, the standard error of the mean is sigma divided by the square root of the sample size = 100/2=50. • 68.26% of the sample means (n=4) will be within 1.00 standard error of the mean from mu and 95.44% will be within 2.00 standard errors of the mean from mu • So, 68.26% of the sample means (n=4) will be between 450 and 550 and 95.44% will fall between 400 and 600 • NOTE THAT SAMPLE MEANS FALL CLOSER TO MU, ON THE AVERAGE, THAN DO INDIVIDUAL SCORES.**Let’s make the samples larger**• Take random samples of SAT scores, with 400 people in each sample, the standard error of the mean is sigma divided by the square root of 400 = 100/20=5.00 • 68.26% of the sample means will be within 1.00 standard error of the mean from mu and 95.44% will be within 2.00 standard errors of the mean from mu. • So, 68.26% of the sample means (n=400) will be between 495 and 505 and 95.44% will fall between 490 and 510. • Take random samples of SAT scores, with 2500 people in each sample, the standard error of the mean is sigma divided by the square root of 2500 = 100/50=2.00. • 68.26% of the sample means will be within 1.00 standard error of the mean from mu and 95.44% will be within 2.00 standard errors of the mean from mu. • 68.26% of the sample means (n=2500) will be between 498 and 502 and 95.44% will fall between 496 and 504**What happens as n increases?**• The sample means get closer to each other and to mu. • Their average squared distance from mu equals the variance divided by the size of the sample. • Therefore, their average unsquared distance from mu equals the standard deviation divided by the square root of the size of the sample. • The sample means fall into a more and more perfect normal curve. • These facts are called “The Central Limit Theorem” and can be proven mathematically.**We want to define two intervals around mu:One interval into**which 95% of the sample means will fall. Another interval into which 99% of the sample means will fall.**95% of sample means will fall in a symmetrical interval**around mu that goes from 1.960 standard errors below mu to 1.960 standard errors above mu • A way to write that fact in statistical language is: CI.95: mu + 1.960 sigmaX-bar or CI.95: mu - 1.960 sigmaX-bar < X-bar < mu + 1.960 sigmaX-bar**As I said, 95% of sample means will fall in a symmetrical**interval around mu that goes from 1.960 standard errors below mu to 1.960 standard errors above mu • Take samples of SAT/GRE scores (n=400) • Standard error of the mean is sigma divided by the square root of n=100/ = 100/20.00=5.00 • 1.960 standard errors of the mean with such samples = 1.960 (5.00)= 9.80 • So 95% of the sample means can be expected to fall in the interval 500+9.80 • 500-9.80 = 490.20 and 500+9.80 =509.80 CI.95: mu + 1.960 sigmaX-bar = 500+9.80 or CI.95: 490.20 < X-bar < 509.20**99% of sample means will fall within 2.576 standard errors**from mu • Take the same samples of SAT/GRE scores (n=400) • The standard error of the mean is sigma divided by the square root of n=100/20.00=5.00 • 2.576 standard errors of the mean with such samples = 2.576 (5.00)= 12.88 • So 99% of the sample means can be expected to fall in the interval 500+12.88 • 500-12.88 = 487.12 and 500+12.88 =512.88 CI.99: mu + 2.576 sigmaX-bar = 500+12.88 or CI.99: 487.12 < the sample mean < 512.88**Let’s do another one.**• What are the 95% and 99% confidence intervals for samples of 25 randomly selected IQ scores • First compute the standard error for samples of 25 IQ scores • IQ: mu =100, sigma = 15 • Standard error for samples of size 25 is 15 divided by the square root of 25 = 15/5.00 =3.00**IQ scores – CI.95 and CI.99 for samples n=25**• The standard error of the mean is sigma divided by the square root of n=15/5.00=3.00 • 1.960 standard errors of the mean with such samples = 1.960 (3.00)= 5.88 points • So, 95% of the sample means (n=25) can be expected to fall in the interval 100 + 5.88 • 100-5.88 = 94.12 and 100+5.88 =105.88 CI.95: mu + 1.960 sigmaX-bar = 100+5.88 or CI.95: 94.12 < X-bar < 105.88 99% of the sample means (n=25) can be expected to fall in the interval 100 + (2.576)(3.00) = 100 + 7.73 CI.99: 100+7.73 or CI.99: 92.27 < X-bar < 107.73**Here is another example. This time we start with an even**smaller population (N=4) and take all possible samples of size 3. There are 64 of them. Let’s see that again the means form a normal curve around mu and the standard error equals sigma divided by the square root of the sample size (3).**Standard error of the mean - 2**• The standard deviation of the individual scores was 3.35 • Sample size was 3 • 3.35 divided by the square root of 3 = 1.94 • Computing the standard error directly from the sample means shows the standard error = 1.94