#### Transcript LGM 8

Presentation 2 Sampling Distributions • Sampling Distribution of sample proportions • Sampling Distribution of sample means 1 Statistics VS Parameters Statistic – is a numerical value computed from a sample. Parameter – is a numerical value associated with a population. Essentially, we would like to know the parameter. But in most cases it is hard to know the parameter since the population is too large. So we have to estimate the parameter by some proper statistics computed from the sample. 2 Some Notation p = population proportion pˆ = sample proportion μ = population mean x = sample mean σ = standard deviation s = sample standard deviation 3 A. Sampling Distribution of the Sample Proportion Situation 1: A survey is undertaken to determine the proportion of PSU students who engage in under-age drinking. The survey asks 200 random under-age students (assume no problems with bias). Suppose the true population proportion of those who drink is 60%. Thus, p = 0.6 and drink. pˆ is the proportion in the sample who 4 Repeated Samples Imagine repeating this survey many times, and each time we record the sample proportion of those who have engaged in under-age drinking. What would the ˆ look like? sampling distribution of p Sample (n=200) Sample Proportion pˆ1 pˆ2 1 2 pˆ 3 pˆ4 pˆ5 3 4 5 … 150,000 pˆ is a random variable assigning a value to each sample! … pˆ 150,000 5 pˆ for 150 000 samples. 0 2 4 6 8 10 Histogram of 0.4 0.5 pˆ 0.6 0.7 0.8 6 Sampling Distribution of pˆ Let X be the number of respondents who say they engage in under age drinking. X is binomial with n =200 and p =0.6. So, we can calculate the probability of X for each possible outcome (0-200). The PDF is plotted below: 0.06 0.05 Probability 0.04 0.03 0.02 0.01 0.00 69 74 79 84 89 94 99 104 109 114 119 124 129 134 139 144 149 154 159 164 169 X 7 Sampling Distribution of pˆ pˆ number of students in the sample that drink X X totalnumber of students in the sample n 200 Since X ~Bin (n =200, p =0.6), the sampling distribution ˆ is the same as that of the binomial distribution of p divided by n. Therefore we have E ( X ) np p n n np (1 p ) sd ( X ) Std Dev. for pˆ n n Mean for pˆ p (1 p ) n 8 Sampling Distribution of pˆ - Cont. Using the Normal approximation to the binomial distribution pˆ we have that the sampling distribution of is approximately Normal with mean p and std. dev. p (1 p ) / n i.e. approx . p (1 p ) pˆ ~ N p , n The conditions for this approximation to be valid are: 1. The sample selected from the population is random. 2. The sample must be large enough, np and n(1-p) MUST be greater than 5, and should be greater than 10. 9 Example: Recent studies have shown that about 20% of American adults fit the medical definition of being obese. A large medical clinic would like to estimate what percent of their patients are obese, so they take a random sample of 100 patients and find that 18 percent are obese. Suppose in truth, the same percentage holds for the patients of the medical clinic as for the general population, 20%. Give notation and the numerical value for the following. 10 Problem - Cont. a. The population proportion of obese patients in the medical clinic: b. The proportion of obese patients in the sample of 100 patients: c. The mean of the sampling distribution of pˆ : d. The standard deviation of the sampling distribution of pˆ : e. The variance of the sampling distribution of pˆ : 11 B. Sampling Distribution of the Sample Mean Situation 2: The mean height of women age 20 to 30, X , is normally distributed (bell-shaped) with a mean of 65 inches and a standard deviation of 3 inches. i.e. X ~N(65,9) A random sample of 200 women was taken and the sample mean X recorded. Now IMAGINE taking MANY samples of size 200 from the population of women. For each sample we record the X . What is the sampling distribution of X ? 12 Histograms for the Distribution of X and X -Bar Distribution of Sample Means: X-bar = mean of random sample of size 200. 0.0 0.00 0.02 0.5 0.04 0.06 1.0 0.08 0.10 1.5 0.12 Original Population of Women: X= height of random woman 50 55 60 65 X 70 75 80 62 63 64 65 x 66 67 68 13 Normal Data Consider a Normal random variable X with mean μ and standard deviation σ, X ~N( μ , σ2 ). The sampling distribution of the sample mean of X for a sample of size n is Normal with Mean or ExpectedValue of X E ( X ) n 2 Variance of X Var ( X ) n i.e. Std Dev. of X s .d .(X ) 2 X ~ N , n 14 Skewed or Non-Normal Data 0 10 20 30 40 Situation 3: In a college survey, students were asked to report the number of cd’s they own. Clearly CDs is a right skewed data set. Suppose our population looked something like this, let us take repeated samples from this population and see what the sample mean looks like. 0 100 200 300 400 500 600 CDs 15 1200 n=8 0 0 200 500 400 600 1000 800 1500 n=4 1000 2000 Suppose we take repeated samples of size n = 4, 8, 16, 32 0 100 200 0 300 50 100 150 200 250 Sample Mean for n=8 800 Sample Mean for n=4 800 n = 32 0 0 200 200 400 400 600 600 n = 16 50 100 150 Sample Mean for n16 200 40 60 80 100 120 140 160 180 Sample Mean for n=32 16 Statistics From Skewed Data Using that CD sample as the population, µ = 87.6, σ = 87.8 The sample means from the previous slide had the following summary statistics: Sample Size Mean of X-bar Std. Dev. of X-bar n=4 86.6 43.2 n=8 86.8 30.9 n = 16 86.7 21.9 n = 32 86.6 15.6 Note: that the mean remains constant, and the std. deviation decreases as the sample size increases! 17 Central Limit Theorem For non-normal data coming from a population with mean µ and standard deviation σ the sampling distribution of the sample mean is approximately normal with Mean or ExpectedValue of X E ( X ) n 2 Variance of X Var ( X ) n approx. 2 i.e. X ~ N , n Std Dev. of X s .d .(X ) Conditions: The above is true if the sample size is large enough, usually n > 30 is sufficient. 18 What next? We have shown that both the sampling distribution of the sample proportion, and the sampling distribution of the sample mean are both normal under certain conditions. Now we can use what we know about normal distributions ˆ and X ! to make conclusions about p In the following we will see how to use the values of the statistics (p-hat, x-bar) to make inferences about the parameters (p, µ). 19 Exercise 1 The population proportion is 0.30. Consider the following questions. 1. Find the sampling distribution of p-hat for each of the following sample sizes n=100, n=200, n=1000 2. What is the probability that a sample proportion will be within ±.04 of the population proportion for each of these sample sizes? 3. What is the advantage of larger sample size? 20 Exercise 2 A certain antibiotic in known to cure 85% of strep bacteria infections. A scientist wants to make sure the drug does not lose its potency over time. He treats 100 strep patients with a 1 year old supply of the antibiotic. Let pˆ be the proportion of individuals who are cured. ASSUME the drug has NOT lost potency, answer the following questions… 1. 2. 3. What is the sampling distribution of pˆ ? Draw a picture If we repeated this study many times we would expect 95% of pˆ to fall within what interval? What is the probability that more than 90% in the sample are cured? 21 Exercise 3 A newspaper conducts a poll to determine the proportion of adults who favor a certain candidate. They ask a random sample of 800 people whether or not they favor that candidate (Assume no bias!). Suppose the true proportion of adults who favor the candidate is 58%. 1. 2. 3. 4. The newspaper records the sample proportion who favor the candidate. What is the sampling distribution of the sample proportion? Draw a picture of its PDF (center it correctly and include the appropriate scale). What is the probability that the newspaper would have recorded a sample proportion greater than 62%? What is the probability that less than 50% of the newspaper respondents would support this candidate? What is the probability that a randomly selected individual favors this candidate? 22 Exercise 4 Suppose the number of calories FIT students consume in a day is normally distributed with mean 2000 and standard deviation 300. 1. 2. 3. 4. 5. About 95% of PSU students have a daily caloric intake between what two values? What is the probability that a randomly selected individual consumed between 1800 and 2100 calories yesterday? Suppose I take a random sample of 36 students and recorded the number of calories each consumed on a given day. Describe the sampling distribution of the sample mean. Draw a picture of the sampling distribution of the sample mean (center it correctly and include the appropriate scale). If I take a sample of size 36 from the student body, what is the probability that the sample mean will be less than 2050? 23 Exercise 5 Assume the length of trout living in the Susquehanna River is normally distributed with mean of 14 inches and standard deviation of 2 inches. A random sample of 16 trout is taken from the river. 1. 2. 3. 4. 5. What is the sampling distribution of the average trout length (i) in a sample of size 16 (ii) in a sample of size 100? What happens to the sampling distribution of the sample mean as the sample size increases? (Draw a picture) What is the probability that a random sample of 16 trout will provide a sample mean within one in of the population mean? What is the probability that a random sample of 100 trout will provide a sample mean within one in of the population mean? What is the advantage of a larger sample size when one is attempting to estimate the population mean? 24