Survey

Survey

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Sampling Distribution of a Sample Proportion Lecture 26 Sections 8.1 – 8.2 Wed, Mar 8, 2006 Preview of the Central Limit Theorem We looked at the distribution of the sum of 1, 2, and 3 uniform random variables U(0, 1). We saw that the shapes of their distributions was moving towards the shape of the normal distribution. If we replace “sum” with “average,” we will obtain the same phenomenon, but on the scale from 0 to 1 each time. Preview of the Central Limit Theorem 2 1 0 1 Preview of the Central Limit Theorem 2 1 0 1 Preview of the Central Limit Theorem 2 1 0 1 Preview of the Central Limit Theorem Some observations: Each distribution is centered at the same place, ½. The distributions are being “drawn in” towards the center. That means that their standard deviation is decreasing. Can we quantify this? Preview of the Central Limit Theorem = ½ 2 = 1/12 2 1 0 1 Preview of the Central Limit Theorem = ½ 2 = 1/24 2 1 0 1 Preview of the Central Limit Theorem = ½ 2 = 1/36 2 1 0 1 Preview of the Central Limit Theorem This tells us that a mean based on three observations is much more likely to be close to the population mean than is a mean based on only one or two observations. Parameters and Statistics THE PURPOSE OF A STATISTIC IS TO ESTIMATE A POPULATION PARAMETER. A sample mean is used to estimate the population mean. A sample proportion is used to estimate the population proportion. Sample statistics, by their very nature, are variable. Population parameters are fixed. Some Questions We hope that the sample proportion is close to the population proportion. How close can we expect it to be? Would it be worth it to collect a larger sample? If the sample were larger, would we expect the sample proportion to be closer to the population proportion? How much closer? The Sampling Distribution of a Statistic Sampling Distribution of a Statistic – The distribution of values of the statistic over all possible samples of size n from that population. The Sample Proportion Let p be the population proportion. Then p is a fixed value (for a given population). Let p^ (“p-hat”) be the sample proportion. Then p^ is a random variable; it takes on a new value every time a sample is collected. The sampling distribution of p^ is the probability distribution of all the possible values of p^. Example Suppose that this class is 3/4 freshmen. Suppose that we take a sample of 2 students, selected with replacement. Find the sampling distribution of p^. Example 3/4 3/4 F P(FF) = 9/16 N P(FN) = 3/16 F P(NF) = 3/16 N P(NN) = 1/16 1/4 1/4 3/4 N F 1/4 Example Let X be the number of freshmen in the sample. The probability distribution of X is x P(x) 0 1/16 1 6/16 2 9/16 Example Let p^ be the proportion of freshmen in the sample. (p^ = X/n.) The sampling distribution of p^ is x P(p^ = x) 0 1/16 1/2 6/16 1 9/16 Samples of Size n = 3 If we sample 3 people (with replacement) from a population that is 3/4 freshmen, then the proportion of freshmen in the sample has the following distribution. x 0 1/3 2/3 1 P(p^ = x) 1/64 = .02 9/64 = .14 27/64 = .42 27/64 = .42 Samples of Size n = 4 If we sample 4 people (with replacement) from a population that is 3/4 freshmen, then the proportion of freshmen in the sample has the following distribution. x 0 1/4 P(p^ = x) 1/256 = .004 12/256 = .05 2/4 3/4 1 54/256 = .21 108/256 = .42 81/256 = .32 The Parameters of the Sampling Distributions When n = 1, the sampling distribution is p^ 0 1 P(p^) 1/4 3/4 The mean and standard deviation are = 3/4 = 0.75 2 = 3/16 = 0.1875 The Parameters of the Sampling Distributions When n = 2, the sampling distribution is p^ P(p^) 0 1/16 1/2 6/16 1 9/16 The mean and standard deviation are = 3/4 = 0.75 2 = 3/32 = 0.09375 The Parameters of the Sampling Distributions When n = 3, the sampling distribution is p^ P(p^) 0 1/64 = .02 1/3 9/64 = .14 2/3 27/64 = .42 1 27/64 = .42 The mean and standard deviation are = 3/4 = 0.75 2 = 3/48 = 0.0625 The Parameters of the Sampling Distributions When n = 4, the sampling distribution is p^ P(p^) 0 1/256 = .004 1/4 12/256 = .05 2/4 54/256 = .21 3/4 108/256 = .42 1 81/256 = .32 The mean and standard deviation are = 3/4 = 0.75 2 = 3/64 = 0.046875 Sampling Distributions Run the program Central Limit Theorem for Proportions.exe. Use n = 30 and p = 0.75; generate 100 samples. 100 Samples of Size n = 30 = 0.75 = 0.079 Observations and Conclusions Observation #1: The values of p^ are clustered around p. Conclusion #1: p^ is probably close to p. Larger Sample Size Now we will select 100 samples of size 120 instead of size 30. Run the program Central Limit Theorem for Proportions.exe. Pay attention to the spread (standard deviation) of the distribution. 100 Samples of Size n = 120 = 0.75 = 0.0395 Observations and Conclusions Observation #2: As the sample size increases, the clustering is tighter. Conclusion #2A: Larger samples give more reliable estimates. Conclusion #2B: For sample sizes that are large enough, we can make very good estimates of the value of p. Larger Sample Size Now we will select 10000 samples of size 120 instead of only 100 samples. Run the program Central Limit Theorem for Proportions.exe. Pay attention to the shape of the distribution. 10,000 Samples of Size n = 120 = 0.75 = 0.0395 10,000 Samples of Size n = 126 More Observations and Conclusions Observation #3: The distribution of p^ appears to be approximately normal. One More Conclusion Conclusion #3: We can use the normal distribution to calculate just how close to p we can expect p^ to be. However, we must know the values of and for the distribution of p^. That is, we have to quantify the sampling distribution of p^. The Sampling Distribution of It turns out that the sampling distribution of p^ is approximately normal with the following parameters. Mean of pˆ p Variance of pˆ p1 p n Standard deviation of pˆ ^ p p1 p n This is the Central Limit Theorem for Proportions, summarized on page 519. The Sampling Distribution of ^ p The approximation to the normal distribution is excellent if np 5 and n1 p 5. Why Surveys Work Suppose 51% of the population plan to vote for candidate X, i.e., p = 0.51. What is the probability that an exit survey of 1000 people would show candidate X with less than 45% support, i.e., p^ < .45? Why Surveys Work First, describe the sampling distribution of p^ if the sample size is n = 1000 and p = 0.51. Check: np = 510 5 and n(1 – p) = 490 5. p^ is approximately normal. pˆ 0.51 pˆ 0.510.49 0.01581 1000 Why Surveys Work The z-score of 0.45 is z = (0.45 – 0.51)/.01581 = -3.795. P(p^ < 0.45) = P(Z < -3.795) = 0.00007385 (not likely!) Or use normalcdf(-E99, 0.45, 0.51, 0.01581). Why Surveys Work Perform the same calculation, but with a smaller sample size, say n = 50. The probability turns out to be 0.1980, nearly a 20% chance. By symmetry, there is also a 20% chance that the sample proportion is greater than 57%. Thus, there is a 40% chance that the sample proportion is off by at least 6 percentage points.