7.1 Generating Sampling Distributions Math 140 � Sampling Distributions. Distribution of summary statistics obtained from taking repeated random Introductory Statistics samples. � Steps for generating a sampling distribution: � I. Take a random sample of a fixed size n from a population. Professor Silvia Fernández � II. Compute a Summary Statistic for this sample. Chapter 7 � III. Repeat steps I and II many times. Based on the book Statistics in Action � IV. Display the distribution of the Summary Statistic. by A. Watkins, R. Scheaffer, and G. Cobb. Note : A way to remember these steps is, Random Sample, Summary Statistic, Repetition, Distribution. Example Shape, Center, and Spread � Westvaco case. � A good description of a 40 30 Randomly select three workers from the group of 10 with ages sampling distribution is 20 25, 33, 35, 38, 48, 55, 55, 55, 56, and 64, the trio shape, center, 10 and calculate the mean age of the three selected. and spread. 0 � Recall the rectangles 1 3 5 7 9 11 13 15 activity 4.2. (See Sample mean of the areas of 5 rectangles displays 5.7 and 5.8) 20 15 10 5 0 1 3 5 7 9 11 13 15 17 19 Population of Rectangle Areas 1
Shape, Center, and Spread Shape, Center, and Spread Notes. 20 40 40 � The standard deviation of 15 30 30 the sampling distribution is 10 20 20 often called the Standard 5 10 10 Error (SE) 0 0 0 1 3 5 7 9 11 13 15 1 3 5 7 9 11 13 15 1 3 5 7 9 11 13 15 17 19 � Most sample distributions are nearly normal, we’ll see Population of Rectangle Areas Sample mean of the areas of 5 rectangles Sample mean of the areas of 5 rectangles more about this later. � Shape: Irregular � Values that are in the middle � Shape: Normal with a hint of � Shape: Normal with a hint of skew to the right. skew to the right. 95% of a random distribution are called Reasonably Center= Mean = μ = 7.41 � � Center= Mean= x = 7.377 � Center= Mean= x = 7.377 Likely. � Values that are in the outer � Spread = � Spread = � Spread = 5% of a random distribution Standard Deviation = σ = 5.23 are called Rare Events . Standard Deviation = SE = Standard Deviation = SE = 2.23 2.23 7.2 Sampling distribution of the Sampling distribution of the sample mean sample mean. for sample sizes 1, 4, 10, 20, and 40. Example: Population 2
Properties of The Sampling Notation Distribution of The Sample Mean μ � The mean of the sampling distribution of equals the mean x x Population Sample Sampling of the population μ : μ = μ Distribution x σ � *The standard deviation of the sampling distribution of , x μ x μ also called the standard error of the mean, equals the standard x x deviation of the population σ divided by the square root of the Mean sample size n : σ Standard σ = σ σ or SE s x Deviation x n � The Shape of the sampling distribution will be approximately Size normal if the population is approximately normal; for other N n populations, the sampling distribution becomes more normal as n increases. This property is called the Central Limit Theorem . *This holds as long as you sample with replacement or your sample size is less than 10% of the population size. (See exercise E30.) Example 1 Example 1 � Example: Average Number Number of Children Proportion of families, � Problems usually involve a combination of the P ( x ) (per family), x of Children three properties of the Sampling Distribution What is the probability that a 0 0.524 of the Sample Mean, together with what we random sample of 20 1 0.201 families in the United States learned about the normal distribution. 2 0.179 will have an average of 1.5 3 0.070 children or fewer? 4 or more 0.026 � Example: Average Number of Children � Mean (of population) 0.6 What is the probability that a random sample μ = 0.873 0.5 of 20 families in the United States will have 0.4 � Standard Deviation 0.3 σ =1.095 an average of 1.5 children or fewer? 0.2 0.1 0 0 1 2 3 4 3
Example 1 Example 1 μ x = μ = μ x = μ = � Find z-score of the value 1.5 0 . 873 0 . 873 − − μ x mean x = = x z σ σ σ 1 . 095 1 . 095 SD σ = = = σ = = = x 0 . 2448 0 . 2448 x x − n 20 1 . 5 0 . 873 n 20 = ≈ 2 . 56 0 . 2448 − ≈ normalcdf ( 99999 , 2 . 56 ) . 9948 � So in a random sample of 20 families there is a 99.47% probability that the mean number of children per family 0.873 will be less than 1.5 0.873 − ≈ OR normalcdf ( 99999 , 1 . 5 , 0 . 873 , 0 . 2448 ) . 9948 Finding Probabilities for Example 2 Sample Totals � Sometimes situations are stated in terms of the total number in � Example: Reasonably Likely Averages the sample rather than the average number: “What is the What average numbers of children are probability that there are 30 or fewer children in a random sample of 20 families in the United States?” You have the reasonably likely in a random sample of 20 choice of two equivalent ways to do this problem. families? � Method I: Find the equivalent average number of children, x , by dividing the total number of children, 30, by the sample size, 20: 30 = = x 1 . 5 � Recall that the values that are in the middle 20 95% of a random distribution are called Then you can use the same formulas and procedure as in the Reasonably Likely. previous examples. � Method II: Convert the formulas from the previous examples to Note that by calculating the z-scores of 2.5% and 97.5% we find that equivalent formulas for the sum, then proceed as in the next the Reasonably Likely values are those values within 1.96 standard example. deviations from the mean. That is, between μ – 1.96 σ and μ + 1.96 σ 4
Sampling Distribution Examples 3 and 4 of the Sum of a Sample � If a random sample of size n is selected with mean μ � Ex3: The Probability of 25 or fewer Children and standard deviation σ , then What is the probability that a random sample � the mean of the sampling distribution of the sum is μ sum = μ of 20 families in the United States will have a n total of 25 children or fewer? � the standard error of the sampling distribution of the sum is σ = ⋅ σ n � Ex4: Reasonably Likely Totals sum � the shape of the sampling distribution will be In a random sample of 20 families, what total approximately normal if the population is approximately numbers of children are reasonably likely? normal; for other populations, the sampling distribution becomes more normal as n increases. Note: To get the “sum” formulas just multiply by n 7.3 Sampling Distribution Sample Size vs. Population Size of the Sample Proportion � As long as the sample size is a small � You often hear reports of percentages or percentage (around 10% or less) of the proportions: About 60% of automobile drivers population size, it doesn’t matter much if you in Mississippi use seat belts. (The national sample with or without replacement, and, in average is about 82%.) fact, the population size will have little effect � To make intelligent decisions based on data on the statistical analysis. that is reported this way, you must � If the sample size is more than 10% of the understand the behavior of proportions that population size then a more complex formula arise from random samples. needs to be used for the standard error, we � The properties of sample proportions are will not do this here since it rarely happens in similar to the properties of sample means. practice. (See Exercise E30.) 5
Simulation The Sample Proportion p-hat (Activity 7.3a) � In a certain population we say that p is the proportion � Choose 10 random 25 numbers between 1 and of the population having a certain property. We say 10. (use your calculator that p is the proportion of “success” according to our 20 or a random row in the property. (e.g. using a seat belt) Table D on page 828) � Note that p is always a number between 0 and 1. 15 � Count the number of � When we select a sample of size n , we calculate the successes the following 10 proportion of successes in our sample by dividing the way: Numbers between number of successes by the sample size. We call this 1 and 6 are successes, 5 the sample proportion and we denote it by p-hat. 7 to 10 (or 0) are not. number of successes number of successes � Calculate 0 = = 0.00 0.20 0.40 0.60 0.80 1.00 p ˆ number of successes sample size n ˆ = p 10 Center and Spread for Computer Simulations Sample Proportions � � � The following diagram shows the exact sampling We can assign successes as Then we can calculate the follows: mean of the population as distributions of the sample proportion for samples of follows � Non-user of seatbelt 0 ∑ size 10,20, and 40; when p = 0.6 = 0.6 μ = ⋅ = + � User of seat-belt 1 x P ( x ) 0 (. 4 ) 1 ( 0 . 6 ) � We then have the following relative frequency table: � And in general = ∑ μ ⋅ Use Seat Relative In general x P ( x ) Belts Frequency = − + = 0 ( 1 p ) 1 ( p ) p 0 0.4 1- p 1- � On the other hand we know that 1 0.6 p the mean of the sampling distribution of the sample mean � And by adding the ones we get is equal to the mean of the population, that is sum of values = = ˆ p x μ ˆ = μ = μ = sample size p 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0.00 0.10 0.20 0.30 0.40 0.50 0.6 0.7 0.8 0.9 1 0.00 0.10 0.20 0.30 0.40 0.50 0.6 0.7 0.8 0.9 1 x p 6
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