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The Normal Curve and Sampling A sample will always be different from the true population This is called “sampling error” The difference between a sample and the true population, regardless of how well the survey was designed or implemented Different from measurement error or sample bias.

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## The Normal Curve and Sampling A sample will always be different from the true population

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**The Normal Curve and Sampling**A sample will always be different from the true population This is called “sampling error” The difference between a sample and the true population, regardless of how well the survey was designed or implemented Different from measurement error or sample bias**Sampling distribution of Means**• The existence of sampling error means that if you take a 1000 random samples from a population and calculate a 1000 means and plot the distribution of those means you will get a consistent distribution that has the following characteristics:**Characteristics of a Sampling Distribution**• 1. the distribution approximates a normal curve • 2. the mean of a sampling distribution of means is equal to the true population • 3. the standard deviation of a sampling distribution is smaller than the standard deviation of the population. Less variation in the distribution because we are not dealing with raw scores but rather central tendencies.**Probability and the Normal Curve**In chapter 6 – we are not interested in the distribution of raw scores but rather the distribution of sample means and making probability statements about those sample means.**Probability and the Sampling Distribution**Why is making probabilistic statements about a central tendency important? • 1. it will allow us to engage in inferential statistics (later in ch. 7) • 2. it allows us to produce confidence intervals**Example of number 1:**• President of UNLV states that the average salary of a new UNLV graduate is $60,000. We are skeptical and test this by taking a random sample of a 100 UNLV students. We find that the average is only $55,000. Do we declare the President a liar?**Not Yet!!!!**We need to make a probabilistic statement regarding the likelihood of Harter’s statement. How do we do that? With the aid of the standard error of the mean we can calculate confidence intervals - the range of mean values within with our true population mean is likely to fall.**How do we do that?**• First, we need the sample mean • Second, we need the standard deviation of the sampling distribution of means (what’s another name for this?) • a.k.a standard error of the mean**What’s the Problem?**• The problem is… • We don’t have the standard deviation of the sampling distribution of means? • What do we do?**First – let’s pretend**• Let’s pretend that I know the Standard Deviation of the Sampling Distribution of Means (a.k.a. the standard error of the mean). It’s 3000 • For a 95% confidence interval we multiply the standard error of the mean by 1.96 and add & subtract that product to our sample mean • Why 1.96?**So is the President Lying?**CI = Mean + or – 1.96 (SE) = 55,000 +/- 1.96 (3000) = 55,000 +/- 5880 = $49,120 to 60,880**Estimating the SE**• We Can Estimate the Standard Error of the Mean. • Divide the standard deviation of the sample by √n-1 • For example a sample standard deviation of 29849 would produce a estimate of the SE of around 3000 [29849 divide by √n-1] [remember n = 100] • Then multiply this estimate by t rather than 1.96 and then add this product to our sample mean. Why t?**The t Distribution**• Empirical testing and models shows that a standard deviation from a sample underestimates the standard deviation of the true population • This is why we use N-1 not N when calculating the standard deviation and the standard error • So in reality, we are calculating t-scores, not z-scores since we are not using the true sd.**So when we are using a sample and calculating a 95%**confidence interval (CI) we need to multiply the standard error by t, not 1.96 • How do we know what t is? • Table in back of book (Appendix C; Table C) • Df = N – 1 • 100-1 = 99; Use the df of 60 and level of significance of .05 (why?) • T = 2**Confidence Intervals for Proportions**Calculate the standard error of the proportion: Sp = 95% conf. Interval = P +/- (1.96)Sp**Example**• National sample of 531 Democrats and Democratic-leaning independents, aged 18 and older, conducted Sept. 14-16, 2007 • Clinton 47%; Obama 25%; Edwards 11% • P(1-P) = .47(1-.47) = .47(.53) = .2491 • Divide by N = .2491/531 = .000469 • Take square root = .0217 • 95% CI = .47 +/- 1.96 (.0217) • .47 +/- .04116 or 0.429 to .511

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