Chapter 7 Sampling Distributions
Section 7.1 Sampling Distributions and the Central Limit Theorem
Sampling Distributions Sampling distribution The probability distribution of a sample statistic. Formed when samples of size n are repeatedly taken from a population. e.g. Sampling distribution of sample means
Sampling Distribution of Sample Means Population with μ, σ Sample 1 1 Sample 3 3 Sample 4 Sample 2 4 2 Sample 5 5 The sampling distribution consists of the values of the sample means, 1, 2, 3, 4, 5,...
Properties of Sampling Distributions of Sample Means 1. The mean of the sample means,, is equal to the population mean μ. 2. The standard deviation of the sample means,, is equal to the population standard deviation, σ, divided by the square root of the sample size, n. n Called the standard error of the mean.
Eample: Sampling Distribution of Sample Means The population values {1, 3, 5, 7} are written on slips of paper and put in a bo. Two slips of paper are randomly selected, with replacement. a. Find the mean, variance, and standard deviation of the population. Solution: Mean: 4 N 2 2 ( ) Varianc e: 5 N Standard Deviat ion: 5 2. 236
Probability Eample: Sampling Distribution of Sample Means b. Graph the probability histogram for the population values. Solution: 0.25 P() Probability Histogram of Population of 1 3 5 7 Population values All values have the same probability of being selected (uniform distribution)
Eample: Sampling Distribution of Sample Means c. List all the possible samples of size n = 2 and calculate the mean of each sample. Solution: Sample Sample 1, 1 1 5, 1 3 1, 3 2 5, 3 4 These means 1, 5 3 5, 5 5 form the 1, 7 4 5, 7 6 sampling 3, 1 2 7, 1 4 distribution of 3, 3 3 7, 3 5 sample means. 3, 5 4 7, 5 6 3, 7 5 7, 7 7
Eample: Sampling Distribution of Sample Means d. Construct the probability distribution of the sample means. Solution: f Probability f Probability 1 1 0.0625 2 2 0.1250 3 3 0.1875 4 4 0.2500 5 3 0.1875 6 2 0.1250 7 1 0.0625
Eample: Sampling Distribution of Sample Means e. Find the mean, variance, and standard deviation of the sampling distribution of the sample means. Solution: The mean, variance, and standard deviation of the 16 sample means are: 4 25. 2. 5 1. 581 2 2 5 These results satisfy the properties of sampling distributions of sample means. 4 5 2. 236 1. 581 n 2 2
Probability Eample: Sampling Distribution of Sample Means f. Graph the probability histogram for the sampling distribution of the sample means. Solution: 0.25 0.20 0.15 0.10 0.05 P() Probability Histogram of Sampling Distribution of 2 3 4 5 Sample mean 6 7 The shape of the graph is symmetric and bell shaped. It approimates a normal distribution.
The Central Limit Theorem 1. If samples of size n 30 are drawn from any population with mean = µ and standard deviation = σ, then the sampling distribution of sample means approimates a normal distribution. The greater the sample size, the better the approimation.
The Central Limit Theorem 2. If the population itself is normally distributed, then the sampling distribution of sample means is normally distribution for any sample size n.
The Central Limit Theorem In either case, the sampling distribution of sample means has a mean equal to the population mean. The sampling distribution of sample means has a variance equal to 1/n times the variance of the population and a standard deviation equal to the population standard deviation divided by the square root of n. 2 2 n n Mean Variance Standard deviation (standard error of the mean)
The Central Limit Theorem 1. Any Population Distribution 2. Normal Population Distribution Distribution of Sample Means, n 30 Distribution of Sample Means, (any n)
Eample: Interpreting the Central Limit Theorem Cellular phone bills for residents of a city have a mean of $63 and a standard deviation of $11. Random samples of 100 cellular phone bills are drawn from this population and the mean of each sample is determined. Find the mean and standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution of sample means.
Solution: Interpreting the Central Limit Theorem The mean of the sampling distribution is equal to the population mean 63 The standard error of the mean is equal to the population standard deviation divided by the square root of n. 11 1.1 n 100
Solution: Interpreting the Central Limit Theorem Since the sample size is greater than 30, the sampling distribution can be approimated by a normal distribution with $63 $1.10
Eample: Interpreting the Central Limit Theorem Suppose the training heart rates of all 20-year-old athletes are normally distributed, with a mean of 135 beats per minute and standard deviation of 18 beats per minute. Random samples of size 4 are drawn from this population, and the mean of each sample is determined. Find the mean and standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution of sample means.
Solution: Interpreting the Central Limit Theorem The mean of the sampling distribution is equal to the population mean 135 The standard error of the mean is equal to the population standard deviation divided by the square root of n. 18 9 n 4
Solution: Interpreting the Central Limit Theorem Since the population is normally distributed, the sampling distribution of the sample means is also normally distributed. 135 9
Probability and the Central Limit Theorem To transform to a z-score z Value Mean Standard error n
Eample: Probabilities for Sampling Distributions The graph shows the length of time people spend driving each day. You randomly select 50 drivers ages 15 to 19. What is the probability that the mean time they spend driving each day is between 24.7 and 25.5 minutes? Assume that σ = 1.5 minutes.
Solution: Probabilities for Sampling Distributions From the Central Limit Theorem (sample size is greater than 30), the sampling distribution of sample means is approimately normal with 25 1.5 0.21213 n 50
Solution: Probabilities for Sampling Distributions Normal Distribution μ = 25 σ = 0.21213 P(24.7 < < 25.5) 24.7 25 z z 1 2 24. 7-25 1. 41 15. n 50 25. 5 25 2. 36 15. n 50 25.5 1.41 Standard Normal Distribution μ = 0 σ = 1 0.9909 0.0793 P( 1.41 < z < 2.36) 0 2.36 z P(24 < < 54) = P( 1.41 < z < 2.36) = 0.9909 0.0793 = 0.9116
Eample: Probabilities for and An education finance corporation claims that the average credit card debts carried by undergraduates are normally distributed, with a mean of $3173 and a standard deviation of $1120. (Adapted from Sallie Mae) 1. What is the probability that a randomly selected undergraduate, who is a credit card holder, has a credit card balance less than $2700? Solution: You are asked to find the probability associated with a certain value of the random variable.
Solution: Probabilities for and Normal Distribution μ = 3173 σ = 1120 P( < 2700) z = - m 2700-3173 = s 1120 Standard Normal Distribution μ = 0 σ = 1» -0.42 P(z < 0.42) 0.3372 2700 3173 0.42 0 z P( < 2700) = P(z < 0.42) = 0.3372
Eample: Probabilities for and 2. You randomly select 25 undergraduates who are credit card holders. What is the probability that their mean credit card balance is less than $2700? Solution: You are asked to find the probability associated with a sample mean. 3173 1120 224 n 25
Solution: Probabilities for and Normal Distribution μ = 3173 σ = 1120 Standard Normal Distribution μ = 0 σ = 1 P( < 2700) z = - m s n = 2700-3173 1120 25 = -473 224» -2.11 P(z < 2.11) 0.0174 2700 3173 2.11 0 z P( < 2700) = P(z < 2.11) = 0.0174
Solution: Probabilities for and There is about a 34% chance that an undergraduate will have a balance less than $2700. There is only about a 2% chance that the mean of a sample of 25 will have a balance less than $2700 (unusual event). It is possible that the sample is unusual or it is possible that the corporation s claim that the mean is $3173 is incorrect.
Chapter 8 Confidence Intervals
Section 8.1 Confidence Intervals for the Mean (Large Samples)
Point Estimate for Population μ Point Estimate A single value estimate for a population parameter Most unbiased point estimate of the population mean μ is the sample mean Estimate Population with Sample Parameter Statistic Mean: μ
Eample: Point Estimate for Population μ A social networking website allows its users to add friends, send messages, and update their personal profiles. The following represents a random sample of the number of friends for 40 users of the website. Find a point estimate of the population mean, µ. (Source: Facebook) 140 105 130 97 80 165 232 110 214 201 122 98 65 88 154 133 121 82 130 211 153 114 58 77 51 247 236 109 126 132 125 149 122 74 59 218 192 90 117 105
Solution: Point Estimate for Population μ The sample mean of the data is 5232 130.8 n 40 Your point estimate for the mean number of friends for all users of the website is 130.8 friends.
Interval estimate Interval Estimate An interval, or range of values, used to estimate a population parameter. Left endpoint 115.1 Point estimate 130.8 ( ) 115 120 125 130 135 140 145 150 Interval estimate Right endpoint 146.5 How confident do we want to be that the interval estimate contains the population mean μ?
Level of confidence c Level of Confidence The probability that the interval estimate contains the population parameter. ½(1 c) ½(1 c) z c z = 0 z c Critical values The remaining area in the tails is 1 c. c c is the area under the standard normal curve between the critical values. Use the Standard Normal Table to find the corresponding z-scores. z
Level of Confidence If the level of confidence is 90%, this means that we are 90% confident that the interval contains the population mean μ. c = 0.90 ½(1 c) = 0.05 ½(1 c) = 0.05 z c = 1.645 z c z = 0 z c = 1.645 z c z The corresponding z-scores are ±1.645.
Sampling error Sampling Error The difference between the point estimate and the actual population parameter value. For μ: the sampling error is the difference μ is generally unknown varies from sample to sample μ
Margin of error Margin of Error The greatest possible distance between the point estimate and the value of the parameter it is estimating for a given level of confidence, c. Denoted by E. E z σ z c c σ n When n 30, the sample standard deviation, s, can be used for σ. Sometimes called the maimum error of estimate or error tolerance.
Eample: Finding the Margin of Error Use the social networking website data and a 95% confidence level to find the margin of error for the mean number of friends for all users of the website. Assume the sample standard deviation is about 53.0.
Solution: Finding the Margin of Error First find the critical values 0.95 0.025 0.025 z c = z 1.96 c z = 0 z c = 1.96 95% of the area under the standard normal curve falls within 1.96 standard deviations of the mean. (You can approimate the distribution of the sample means with a normal curve by the Central Limit Theorem, because n = 40 30.) z c z
Solution: Finding the Margin of Error E zc z n 53.0 1.96 40 16.4 c s n You don t know σ, but since n 30, you can use s in place of σ. You are 95% confident that the margin of error for the population mean is about 16.4 friends.
Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ E E where E z c n The probability that the confidence interval contains μ is c.
Constructing Confidence Intervals for μ Finding a Confidence Interval for a Population Mean (n 30 or σ known with a normally distributed population) In Words 1. Find the sample statistics n and. In Symbols n 2. Specify σ, if known. Otherwise, if n 30, find the sample standard deviation s and use it as an estimate for σ. s ( ) n 1 2
Constructing Confidence Intervals for μ In Words 3. Find the critical value z c that corresponds to the given level of confidence. 4. Find the margin of error E. 5. Find the left and right endpoints and form the confidence interval. In Symbols Use the Standard Normal Table or technology. E z c n Left endpoint: E Right endpoint: E Interval: E E
Eample: Constructing a Confidence Interval Construct a 95% confidence interval for the mean number of friends for all users of the website. Solution: Recall 130.8 and E 16.4 Left Endpoint: E 130.8 16.4 114.4 114.4 < μ < 147.2 Right Endpoint: E 130.8 16.4 147.2
Solution: Constructing a Confidence Interval 114.4 < μ < 147.2 With 95% confidence, you can say that the population mean number of friends is between 114.4 and 147.2.
Eample: Constructing a Confidence Interval σ Known A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. From past studies, the standard deviation is known to be 1.5 years, and the population is normally distributed. Construct a 90% confidence interval of the population mean age.
Solution: Constructing a Confidence Interval σ Known First find the critical values c = 0.90 ½(1 c) = 0.05 ½(1 c) = 0.05 z c = 1.645 z c z = 0 z c = 1.645 z c z z c = 1.645
Solution: Constructing a Confidence Interval σ Known Margin of error: E z c n Confidence interval: Left Endpoint: E 22.9 0.6 22.3 1.5 1.645 0.6 20 22.3 < μ < 23.5 Right Endpoint: E 22.9 0.6 23.5
Solution: Constructing a Confidence Interval σ Known 22.3 < μ < 23.5 Point estimate 22.3 22.9 23.5 ( ) E E With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years.
Interpreting the Results μ is a fied number. It is either in the confidence interval or not. Incorrect: There is a 90% probability that the actual mean is in the interval (22.3, 23.5). Correct: If a large number of samples is collected and a confidence interval is created for each sample, approimately 90% of these intervals will contain μ.
Interpreting the Results The horizontal segments represent 90% confidence intervals for different samples of the same size. In the long run, 9 of every 10 such intervals will contain μ. μ
Sample Size Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate the population mean µ is n z c E If σ is unknown, you can estimate it using s, provided you have a preliminary sample with at least 30 members. 2
Eample: Sample Size You want to estimate the mean number of friends for all users of the website. How many users must be included in the sample if you want to be 95% confident that the sample mean is within seven friends of the population mean? Assume the sample standard deviation is about 53.0.
Solution: Sample Size First find the critical values 0.95 0.025 0.025 z c = z 1.96 c z = 0 z c = 1.96 z c z z c = 1.96
Solution: Sample Size z c = 1.96 σ s 53.0 E = 7 n z c E 2 2 1.9653.0 7 220.23 When necessary, round up to obtain a whole number. You should include at least 221 users in your sample.
Section 8.2 Confidence Intervals for the Mean (Small Samples)
The t-distribution When the population standard deviation is unknown, the sample size is less than 30, and the random variable is approimately normally distributed, it follows a t-distribution. t = - m s Critical values of t are denoted by t c. n
Properties of the t-distribution 1. The t-distribution is bell shaped and symmetric about the mean. 2. The t-distribution is a family of curves, each determined by a parameter called the degrees of freedom. The degrees of freedom are the number of free choices left after a sample statistic such as is calculated. When you use a t-distribution to estimate a population mean, the degrees of freedom are equal to one less than the sample size. d.f. = n 1 Degrees of freedom
Properties of the t-distribution 3. The total area under a t-curve is 1 or 100%. 4. The mean, median, and mode of the t-distribution are equal to zero. 5. As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f., the t- distribution is very close to the standard normal z- distribution. d.f. = 2 d.f. = 5 Standard normal curve 0 t The tails in the t- distribution are thicker than those in the standard normal distribution.
Eample: Critical Values of t Find the critical value t c for a 95% confidence level when the sample size is 15. Solution: d.f. = n 1 = 15 1 = 14 Table 5: t-distribution t c = 2.145
Solution: Critical Values of t 95% of the area under the t-distribution curve with 14 degrees of freedom lies between t = ±2.145. c = 0.95 t c = 2.145 t c = 2.145 t
Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ s E E where E tc n The probability that the confidence interval contains μ is c.
Confidence Intervals and t-distributions In Words 1. Identify the sample statistics n,, and s. 2. Identify the degrees of freedom, the level of confidence c, and the critical value t c. In Symbols s n d.f. = n 1 ( ) n 1 2 3. Find the margin of error E. E t c s n
Confidence Intervals and t-distributions In Words 4. Find the left and right endpoints and form the confidence interval. In Symbols Left endpoint: Right endpoint: Interval: E E E E
Eample: Constructing a Confidence Interval You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 162.0ºF with a sample standard deviation of 10.0ºF. Find the 95% confidence interval for the population mean temperature. Assume the temperatures are approimately normally distributed. Solution: Use the t-distribution (n < 30, σ is unknown, temperatures are approimately normally distributed).
Solution: Constructing a Confidence Interval n =16, = 162.0 s = 10.0 c = 0.95 df = n 1 = 16 1 = 15 Critical Value Table 5: t-distribution t c = 2.131
Solution: Constructing a Confidence Interval Margin of error: s 10 E t c 2.131 5.3 n 16 Confidence interval: Left Endpoint: E 162 5.3 156.7 156.7 < μ < 167.3 Right Endpoint: E 162 5.3 167.3
Solution: Constructing a Confidence Interval 156.7 < μ < 167.3 Point estimate 156.7 162.0 167.3 ( ) E E With 95% confidence, you can say that the population mean temperature of coffee sold is between 156.7ºF and 167.3ºF.
Normal or t-distribution? Is n 30? No Is the population normally, or approimately normally, distributed? Yes Yes No Use the normal distribution with σ E z c n If σ is unknown, use s instead. Cannot use the normal distribution or the t-distribution. Is σ known? No Use the t-distribution with s E t c and n 1 degrees of freedom. n Yes Use the normal distribution with σ E z c. n
Eample: Normal or t-distribution? You randomly select 25 newly constructed houses. The sample mean construction cost is $181,000 and the population standard deviation is $28,000. Assuming construction costs are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 95% confidence interval for the population mean construction cost? Solution: Use the normal distribution (the population is normally distributed and the population standard deviation is known)
Section 8.3 Confidence Intervals for Population Proportions
Point Estimate for Population p Population Proportion The probability of success in a single trial of a binomial eperiment. Denoted by p Point Estimate for p The proportion of successes in a sample. Denoted by number of successes in sample pˆ n sample size read as p hat
Point Estimate for Population p Estimate Population Parameter Proportion: p with Sample Statistic ˆp Point Estimate for q, the population proportion of failures Denoted by Read as q hat ˆ q 1 ˆ p
Eample: Point Estimate for p In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal e-mail while at work. Find a point estimate for the population proportion of U.S. adults who say it is acceptable to check personal e-mail while at work. (Adapted from Liberty Mutual) Solution: n = 1000 and = 662 pˆ 662 n 1000 0.662 66.2%
Confidence Intervals for p A c-confidence interval for a population proportion p pˆ E p pˆ E where E zc pq ˆˆ n The probability that the confidence interval contains p is c.
Constructing Confidence Intervals for p In Words 1. Identify the sample statistics n and. 2. Find the point estimate 3. Verify that the sampling distribution of pˆ can be approimated by a normal distribution. 4. Find the critical value z c that corresponds to the given level of confidence c. ˆp. In Symbols npˆ pˆ n 5, nqˆ5 Use the Standard Normal Table or technology.
Constructing Confidence Intervals for p In Words 5. Find the margin of error E. 6. Find the left and right endpoints and form the confidence interval. In Symbols E z c pq ˆˆ n Left endpoint: ˆp E Right endpoint: ˆp E Interval: pˆ E p pˆ E
Eample: Confidence Interval for p In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal e-mail while at work. Construct a 95% confidence interval for the population proportion of U.S. adults who say that it is acceptable to check personal e-mail while at work. Solution: Recall pˆ 0.662 qˆ 1 pˆ 10.662 0.338
Solution: Confidence Interval for p Verify the sampling distribution of can be approimated by the normal distribution npˆ 1000 0.662 662 5 nqˆ 1000 0.338 338 5 Margin of error: ˆp E z c pq ˆˆ (0.662) (0. 338) 1.96 0.029 n 1000 2012 Pearson Education, Inc. All rights reserved. 82 of 83
Solution: Confidence Interval for p Confidence interval: Left Endpoint: pˆ E 0.633 < p < 0.691 Right Endpoint: pˆ 0.662 0.029 0.662 0.029 0.633 0.691 E
Solution: Confidence Interval for p 0.633 < p < 0.691 Point estimate ˆp ˆp E ˆp E With 95% confidence, you can say that the population proportion of U.S. adults who say that it is acceptable to check personal e-mail while at work is between 63.3% and 69.1%.
Sample Size Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate p is zc n pq ˆˆ E This formula assumes you have an estimate for and qˆ. If not, use pˆ 0.5 and qˆ 0.5. 2 ˆp
Eample: Sample Size You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if 1. no preliminary estimate is available. Solution: Because you do not have a preliminary estimate for use pˆ 0.5 and qˆ 0. 5. p, ˆ
Solution: Sample Size c = 0.95 z c = 1.96 E = 0.03 zc 2 2 1.96 n pq ˆ ˆ (0.5)(0.5) 1067.11 E 0. 03 Round up to the nearest whole number. With no preliminary estimate, the minimum sample size should be at least 1068 voters.
Eample: Sample Size You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if 2. a preliminary estimate gives pˆ 0.31. Solution: Use the preliminary estimate qˆ 1 pˆ 10.31 0. 69 ˆ 0.31 p
Solution: Sample Size c = 0.95 z c = 1.96 E = 0.03 z c 2 2 1.96 n pq ˆˆ (0.31)(0.69) 913.02 E 0. 03 Round up to the nearest whole number. With a preliminary estimate of pˆ 0.31, the minimum sample size should be at least 914 voters. Need a larger sample size if no preliminary estimate is available.