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Objectives 1. Describe the distribution of the sample mean: samples from normal populations 2. Describe the distribution of the sample mean: samples from a population that is not normal © 2010 Pearson Prentice Hall. All rights reserved 8-1 Statistics such as x are random variables since their value varies from sample to sample. As such, they have probability distributions associated with them. In this we focus on the shape, center and chapter spread of statistics such as x . © 2010 Pearson Prentice Hall. All rights reserved 8-2 The sampling distribution of a statistic is a probability distribution for all possible values of the statistic computed from a sample of size n. The sampling distribution of the sample mean x is the probability distribution of all possible values of the random variable x computed from a sample of size n from a population with mean and standard deviation . © 2010 Pearson Prentice Hall. All rights reserved 8-3 Illustrating Sampling Distributions Step 1: Obtain a simple random sample of size n. © 2010 Pearson Prentice Hall. All rights reserved 8-4 Illustrating Sampling Distributions Step 1: Obtain a simple random sample of size n. Step 2: Compute the sample mean. © 2010 Pearson Prentice Hall. All rights reserved 8-5 Illustrating Sampling Distributions Step 1: Obtain a simple random sample of size n. Step 2: Compute the sample mean. Step 3: Assuming we are sampling from a finite population, repeat Steps 1 and 2 until all simple random samples of size n have been obtained. © 2010 Pearson Prentice Hall. All rights reserved 8-6 Objective 1 • Describe the Distribution of the Sample MeanSamples from Normal Populations © 2010 Pearson Prentice Hall. All rights reserved 8-7 Parallel Example 1: Sampling Distribution of the Sample Mean-Normal Population The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams. Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 5 from this population. © 2010 Pearson Prentice Hall. All rights reserved 8-8 The data on the following slide represent the sample means for the 200 simple random samples of size n = 5. For example, the first sample of n = 5 had the following data: 2.493 2.466 2.473 2.492 Note: x=2.479 for this sample 2.471 © 2010 Pearson Prentice Hall. All rights reserved 8-9 Sample Means for Samples of Size n =5 © 2010 Pearson Prentice Hall. All rights reserved 8-10 The mean of the 200 sample means is 2.46, the same as the mean of the population. The standard deviation of the sample means is 0.0086, which is smaller than the standard deviation of the population. The next slide shows the histogram of the sample means. © 2010 Pearson Prentice Hall. All rights reserved 8-11 © 2010 Pearson Prentice Hall. All rights reserved 8-12 What role does n, the sample size, play in the standard deviation of the distribution of the sample mean? © 2010 Pearson Prentice Hall. All rights reserved 8-13 What role does n, the sample size, play in the standard deviation of the distribution of the sample mean? As the size of the sample gets larger, we do not expect as much spread in the sample means since larger observations will offset smaller observations. © 2010 Pearson Prentice Hall. All rights reserved 8-14 Parallel Example 2: The Impact of Sample Size on Sampling Variability • Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 20 from the population of weights of pennies minted after 1982 (=2.46 grams and =0.02 grams) © 2010 Pearson Prentice Hall. All rights reserved 8-15 The mean of the 200 sample means for n =20 is still 2.46, but the standard deviation is now 0.0045 (0.0086 for n = 5). As expected, there is less variability in the distribution of the sample mean with n =20 than with n =5. © 2010 Pearson Prentice Hall. All rights reserved 8-16 The Mean and Standard Deviation of the Sampling Distribution of x Suppose that a simple random sample of size n is drawn from a large population with mean and standard deviation . The sampling distribution of x willhave mean x and standard deviation . x n The standard deviation of the sampling distribution of x is called the standard error of the mean and is denoted x . © 2010 Pearson Prentice Hall. All rights reserved 8-17 The Shape of the Sampling Distribution of x If X is Normal If a random variable X is normally distributed, the distribution of the sample mean x is normally distributed. © 2010 Pearson Prentice Hall. All rights reserved 8-18 Parallel Example 3: Describing the Distribution of the Sample Mean The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams. What is the probability that in a simple random sample of 10 pennies minted after 1982, we obtain a sample mean of at least 2.465 grams? © 2010 Pearson Prentice Hall. All rights reserved 8-19 Solution • x 0.02 is normally distributed with x =2.46 and x 0.0063 . 10 2.465 2.46 0.79 . • Z 0.0063 • P(Z>0.79)=1-0.7852 =0.2148. © 2010 Pearson Prentice Hall. All rights reserved 8-20 Objective 2 • Describe the Distribution of the Sample MeanSamples from a Population That is Not Normal © 2010 Pearson Prentice Hall. All rights reserved 8-21 Parallel Example 4: Sampling from a Population that is Not Normal The following table and histogram give the probability distribution for rolling a fair die: Face on Die Relative Frequency 1 0.1667 2 0.1667 3 0.1667 4 0.1667 5 0.1667 6 0.1667 =3.5, =1.708 Note that the population distribution is NOT normal © 2010 Pearson Prentice Hall. All rights reserved 8-22 Estimate the sampling distribution of x by obtaining 200 simple random samples of size n=4 and calculating the sample mean for each of the 200 samples. Repeat for n = 10 and 30. distribution of the sample Histograms of the sampling mean for each sample size are given on the next slide. © 2010 Pearson Prentice Hall. All rights reserved 8-23 © 2010 Pearson Prentice Hall. All rights reserved 8-24 © 2010 Pearson Prentice Hall. All rights reserved 8-25 © 2010 Pearson Prentice Hall. All rights reserved 8-26 Key Points from Example 4 • The mean of the sampling distribution is equal to the mean of the parent population and the standard deviation of the sampling distribution of the sample mean is regardless of the sample n size. • The Central Limit Theorem: the shape of the distribution of the sample mean becomes approximately normal as the sample size n increases, regardless of the shape of the population. © 2010 Pearson Prentice Hall. All rights reserved 8-27 Parallel Example 5: Using the Central Limit Theorem Suppose that the mean time for an oil change at a “10-minute oil change joint” is 11.4 minutes with a standard deviation of 3.2 minutes. (a) If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean. (b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes? © 2010 Pearson Prentice Hall. All rights reserved 8-28 Parallel Example 5: Using the Central Limit Theorem Suppose that the mean time for an oil change at a “10-minute oil change joint” is 11.4 minutes with a standard deviation of 3.2 minutes. (a) If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean. Solution: x is approximately normally distributed with mean=11.4 and std. dev. = 3.2 0.5409. 35 (b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes? 1111.4 0.74 , P(Z<-0.74)=0.23. Solution: Z 0.5409 © 2010 Pearson Prentice Hall. All rights reserved 8-29 Section 8.2 Distribution of the Sample Proportion © 2010 Pearson Prentice Hall. All rights reserved Objectives 1. Describe the sampling distribution of a sample proportion 2. Compute probabilities of a sample proportion © 2010 Pearson Prentice Hall. All rights reserved 8-31 Objective 1 • Describe the Sampling Distribution of a Sample Proportion © 2010 Pearson Prentice Hall. All rights reserved 8-32 Point Estimate of a Population Proportion Suppose that a random sample of size n is obtained from a population in which each individual either does or does not have a certain characteristic. The sample proportion, denoted (read “p-hat”) is given by pˆ x pˆ n where x is the number of individuals in the sample with the specified characteristic. The sample proportion is a statistic that estimates the population proportion, p. © 2010 Pearson Prentice Hall. All rights reserved 8-33 Parallel Example 1: Computing a Sample Proportion In a Quinnipiac University Poll conducted in May of 2008, 1,745 registered voters nationwide were asked whether they approved of the way George W. Bush is handling the economy. 349 responded “yes”. Obtain a point estimate for the proportion of registered voters who approve of the way George W. Bush is handling the economy. © 2010 Pearson Prentice Hall. All rights reserved 8-34 Parallel Example 1: Computing a Sample Proportion In a Quinnipiac University Poll conducted in May of 2008, 1,745 registered voters nationwide were asked whether they approved of the way George W. Bush is handling the economy. 349 responded “yes”. Obtain a point estimate for the proportion of registered voters who approve of the way George W. Bush is handling the economy. Solution: pˆ 349 0.2 1745 © 2010 Pearson Prentice Hall. All rights reserved 8-35 Parallel Example 2: Using Simulation to Describe the Distribution of the Sample Proportion According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry. Describe the sampling distribution of the sample proportion for samples of size n=10, 50, 100. © 2010 Pearson Prentice Hall. All rights reserved 8-36 © 2010 Pearson Prentice Hall. All rights reserved 8-37 © 2010 Pearson Prentice Hall. All rights reserved 8-38 © 2010 Pearson Prentice Hall. All rights reserved 8-39 Key Points from Example 2 • Shape: As the size of the sample, n, increases, the shape of the sampling distribution of the sample proportion becomes approximately normal. • Center: The mean of the sampling distribution of the sample proportion equals the population proportion, p. • Spread: The standard deviation of the sampling distribution of the sample proportion decreases as the sample size, n, increases. © 2010 Pearson Prentice Hall. All rights reserved 8-40 Sampling Distribution of pˆ For a simple random sample of size n with population proportion p: ˆ is • The shape of the sampling distribution of p approximately normal provided np(1-p)≥10. ˆ is • The mean of the sampling distribution of p pˆ p. • The standard deviation of the sampling distribution ˆ is of p p(1 p) pˆ n © 2010 Pearson Prentice Hall. All rights reserved 8-41 Sampling Distribution of pˆ • The model on the previous slide requires that the sampled values are independent. When sampling from finite populations, this assumption is verified by checking that the sample size n is no more than 5% of the population size N (n ≤ 0.05N). • Regardless of whether np(1-p) ≥10 or not, the ˆ is p, and mean of the sampling distribution of p the standard deviation is pˆ p(1 p) n © 2010 Pearson Prentice Hall. All rights reserved 8-42 Parallel Example 3: Describing the Sampling Distribution of the Sample Proportion According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry. Suppose that we obtain a simple random sample of 50 voters and determine which believe that gay and lesbian couples should be allowed to marry. Describe the sampling distribution of the sample proportion for registered voters who believe that gay and lesbian couples should be allowed to marry. © 2010 Pearson Prentice Hall. All rights reserved 8-43 Solution The sample of n=50 is smaller than 5% of the population size (all registered voters in the U.S.). Also, np(1-p)=50(0.42)(0.58)=12.18≥10. The sampling distribution of the sample proportion is therefore approximately normal with mean=0.42 and standard deviation= 0.42(1 0.42) 0.0698. 50 (Note: this is very close to the standard deviation of 0.072 found using simulation in Example 2.) © 2010 Pearson Prentice Hall. All rights reserved 8-44 Objective 2 • Compute Probabilities of a Sample Proportion © 2010 Pearson Prentice Hall. All rights reserved 8-45 Parallel Example 4: Compute Probabilities of a Sample Proportion According to the Centers for Disease Control and Prevention, 18.8% of school-aged children, aged 6-11 years, were overweight in 2004. (a) In a random sample of 90 school-aged children, aged 6-11 years, what is the probability that at least 19% are overweight? (b) Suppose a random sample of 90 school-aged children, aged 6-11 years, results in 24 overweight children. What might you conclude? © 2010 Pearson Prentice Hall. All rights reserved 8-46 Solution • • • n=90 is less than 5% of the population size np(1-p)=90(.188)(1-.188)≈13.7≥10 pˆ is approximately normal with mean=0.188 and standard deviation = (0.188)(1 0.188) 0.0412 90 (a) In a random sample of 90 school-aged children, aged 6-11 years, what is the probability that at least 19% are overweight? 0.19 0.188 Z 0.0485 , P(Z>0.05)=1-0.5199=0.4801 0.0412 © 2010 Pearson Prentice Hall. All rights reserved 8-47 Solution • pˆ is approximately normal with mean=0.188 and standard deviation = 0.0412 (b) Suppose a random sample of 90 school-aged children, aged 6-11 years, results in 24 overweight children. What might you conclude? 24 0.2667 0.188 pˆ 0.2667, Z 1.91 90 0.0412 P(Z>1.91)=1-0.9719=0.028. We would only expect to see about 3 samples in 100 resulting in a sample proportion of 0.2667 or more. This is an unusual sample if the true population proportion is 0.188. © 2010 Pearson Prentice Hall. All rights reserved 8-48

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