STATISTICS INFORMED DECISIONS USING DATA

STATISTICS INFORMED DECISIONS USING DATA
Fifth Edition Chapter 9 Estimating the Value of a Parameter Copyright © 2017, 2013, 2010 Pearson Education, Inc. All Rights Reserved

9.1 Estimating a Population Proportion Learning Objectives
1. Obtain a point estimate for the population proportion 2. Construct and interpret a confidence interval for the population proportion 3. Determine the sample size necessary for estimating a population proportion within a specified margin of error

9. 1 Estimating a Population Proportion 9. 1
9.1 Estimating a Population Proportion Obtain a Point Estimate for the Population Proportion (1 of 3) A point estimate is the value of a statistic that estimates the value of a parameter.

9.1 Estimating a Population Proportion Obtain a Point Estimate for the Population Proportion (2 of 3) Parallel Example 1: Calculating a Point Estimate for the Population Proportion In July of 2008, a Quinnipiac University Poll asked 1783 registered voters nationwide whether they favored or opposed the death penalty for persons convicted of murder were in favor. Obtain a point estimate for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder.

9.1 Estimating a Population Proportion Obtain a Point Estimate for the Population Proportion (3 of 3) Solution Obtain a point estimate for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder.

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (1 of 11) A confidence interval for an unknown parameter consists of an interval of numbers based on a point estimate. The level of confidence represents the expected proportion of intervals that will contain the parameter if a large number of different samples is obtained. The level of confidence is denoted (1 − α) · 100%.

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (2 of 11) For example, a 95% level of confidence (α = 0.05) implies that if 100 different confidence intervals are constructed, each based on a different sample from the same population, we will expect 95 of the intervals to contain the parameter and 5 not to include the parameter.

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (3 of 11) Confidence interval estimates for the population proportion are of the form Point estimate ± margin of error. The margin of error of a confidence interval estimate of a parameter is a measure of how accurate the point estimate is.

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (4 of 11) The margin of error depends on three factors: Level of confidence: As the level of confidence increases, the margin of error also increases. Sample size: As the size of the random sample increases, the margin of error decreases. Standard deviation of the population: The more spread there is in the population, the wider our interval will be for a given level of confidence.

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (5 of 11) NOTE: We also require that each trial be independent; when sampling from finite populations (n ≤ 0.05N).

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (6 of 11) Interpretation of a Confidence Interval A (1 − α) · 100% confidence interval indicates that (1 − α) · 100% of all simple random samples of size n from the population whose parameter is unknown will result in an interval that contains the parameter.

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (7 of 11) Constructing a (1 − α) · 100% Confidence Interval for a Population Proportion Suppose that a simple random sample of size n is taken from a population. A (1 − α) · 100% confidence interval for p is given by the following quantities

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (8 of 11) Margin of Error The margin of error, E, in a (1 − α) · 100% confidence interval for a population proportion is given by

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (9 of 11) Parallel Example 2: Constructing a Confidence Interval for a Population Proportion In July of 2008, a Quinnipiac University Poll asked 1783 registered voters nationwide whether they favored or opposed the death penalty for persons convicted of murder were in favor. Obtain a 90% confidence interval for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder.

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (10 of 11) Solution

9.1 Estimating a Population Proportion Construct and Interpret a Confidence Interval for the Population Proportion (11 of 11) Solution We are 90% confident that the proportion of registered voters who are in favor of the death penalty for those convicted of murder is between 0.61 and 0.65.

9.1 Estimating a Population Proportion Determine the Sample Size Necessary for Estimating a Population Proportion within a Specified Margin of Error (1 of 6) Sample size needed for a specified margin of error, E, and level of confidence (1 − α):

9.1 Estimating a Population Proportion Determine the Sample Size Necessary for Estimating a Population Proportion within a Specified Margin of Error (2 of 6) Two possible solutions:

9.1 Estimating a Population Proportion Determine the Sample Size Necessary for Estimating a Population Proportion within a Specified Margin of Error (3 of 6) Sample Size Needed for Estimating the Population Proportion p The sample size required to obtain a (1 − α) · 100% confidence interval for p with a margin of error E is given by

9.1 Estimating a Population Proportion Determine the Sample Size Necessary for Estimating a Population Proportion within a Specified Margin of Error (4 of 6) Sample Size Needed for Estimating the Population Proportion p If a prior estimate of p is unavailable, the sample size required is

9.1 Estimating a Population Proportion Determine the Sample Size Necessary for Estimating a Population Proportion within a Specified Margin of Error (5 of 6) Parallel Example 4: Determining Sample Size A sociologist wanted to determine the percentage of residents of America that only speak English at home. What size sample should be obtained if she wishes her estimate to be within 3 percentage points with 90% confidence assuming she uses the estimate obtained from the Census 2000 Supplementary Survey of 82.4%?

9.1 Estimating a Population Proportion Determine the Sample Size Necessary for Estimating a Population Proportion within a Specified Margin of Error (6 of 6) We round this value up to 437. The sociologist must survey 437 randomly selected American residents.

9.2 Estimating a Population Mean Learning Objectives
1. Obtain a point estimate for the population mean 2. State properties of Student’s t-distribution 3. Determine t-values 4. Construct and interpret a confidence interval for a population mean 5. Determine the sample size needed to estimate the population mean within a specified margin of error

9. 2 Estimating a Population Mean 9. 2
9.2 Estimating a Population Mean Obtain a Point Estimate for the Population Mean (1 of 3) A point estimate is the value of a statistic that estimates the value of a parameter.

9.2 Estimating a Population Mean Obtain a Point Estimate for the Population Mean (2 of 3) Parallel Example 1: Computing a Point Estimate Pennies minted after 1982 are made from 97.5% zinc and 2.5% copper. The following data represent the weights (in grams) of 17 randomly selected pennies minted after 1982. Treat the data as a simple random sample. Estimate the population mean weight of pennies minted after 1982.

9.2 Estimating a Population Mean Obtain a Point Estimate for the Population Mean (3 of 3) Solution The sample mean is

9.2 Estimating a Population Mean State Properties of Student’s t-Distribution (1 of 9) Student’s t-Distribution Suppose that a simple random sample of size n is taken from a population. If the population from which the sample is drawn follows a normal distribution, the distribution of

9.2 Estimating a Population Mean State Properties of Student’s t-Distribution (2 of 9) Parallel Example 2: Comparing the Standard Normal Distribution to the t-Distribution Using Simulation Obtain 1,000 simple random samples of size n = 5 from a normal population with μ = 50 and σ = 10. Determine the sample mean and sample standard deviation for each of the samples.

9.2 Estimating a Population Mean State Properties of Student’s t-Distribution (3 of 9) Histogram for z

9.2 Estimating a Population Mean State Properties of Student’s t-Distribution (4 of 9) Histogram for t

9.2 Estimating a Population Mean State Properties of Student’s t-Distribution (5 of 9) CONCLUSION: The histogram for z is symmetric and bell-shaped with the center of the distribution at 0 and virtually all the rectangles between −3 and 3. In other words, z follows a standard normal distribution.

9.2 Estimating a Population Mean State Properties of Student’s t-Distribution (6 of 9) CONCLUSION: The histogram for t is also symmetric and bell-shaped with the center of the distribution at 0, but the distribution of t has longer tails (i.e., t is more dispersed), so it is unlikely that t follows a standard normal distribution. The additional spread in the distribution of t can be attributed to the fact that we use s to find t instead of σ. Because the sample standard deviation is itself a random variable (rather than a constant such as σ), we have more dispersion in the distribution of t.

9.2 Estimating a Population Mean State Properties of Student’s t-Distribution (7 of 9) Properties of the t-Distribution 1. The t-distribution is different for different degrees of freedom. 2. The t-distribution is centered at 0 and is symmetric about 0.

9.2 Estimating a Population Mean State Properties of Student’s t-Distribution (8 of 9) Properties of the t-Distribution 5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution, because we are using s as an estimate of σ, thereby introducing further variability into the t- statistic. 6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because, as the sample size n increases, the values of s get closer to the values of σ, by the Law of Large Numbers.

9.2 Estimating a Population Mean State Properties of Student’s t-Distribution (9 of 9)

9.2 Estimating a Population Mean 9.2.3 Determine t-Values (1 of 3)
The notation tα is the t-value such that the area under the standard normal curve to the right is α. The figure illustrates the notation.

Parallel Example 3: Finding t-values Find the t-value such that the area under the t-distribution to the right of the t-value is 0.2 assuming 10 degrees of freedom. That is, find t0.20 with 10 degrees of freedom.

Solution The figure to the right shows the graph of the t-distribution with 10 degrees of freedom.

9.2 Estimating a Population Mean Construct and Interpret a Confidence Interval for the Population Mean (1 of 11) Constructing a (1 − α)100% Confidence Interval for μ Provided sample data come from a simple random sample or randomized experiment, sample size is small relative to the population size (n ≤ 0.05N), and the data come from a population that is normally distributed, or the sample size is large.

9.2 Estimating a Population Mean Construct and Interpret a Confidence Interval for the Population Mean (2 of 11) Constructing a (1 − α)100% Confidence Interval for μ A (1 − α) · 100% confidence interval for μ is given by

9.2 Estimating a Population Mean Construct and Interpret a Confidence Interval for the Population Mean (3 of 11) Parallel Example: Using Simulation to Demonstrate the Idea of a Confidence Interval We will use Minitab to simulate obtaining 30 simple random samples of size n = 8 from a population that is normally distributed with μ = 50 and σ = 10. Construct a 95% confidence interval for each sample. How many of the samples result in intervals that contain μ = 50?

9.2 Estimating a Population Mean Construct and Interpret a Confidence Interval for the Population Mean (4 of 11) SAMPLE MEAN 95.0% CI C1 47.07 (40.14, 54.00) C2 49.33 (42.40, 56.26) C3 50.62 (43.69, 57.54) C4 47.91 (40.98, 54.84) C5 44.31 (37.38, 51.24) C6 51.50 (44.57, 58.43) C7 52.47 (45.54, 59.40) C8 59.62 (52.69, 66.54) C9 43.49 (36.56, 50.42) C10 55.45 (48.52, 62.38)

9.2 Estimating a Population Mean Construct and Interpret a Confidence Interval for the Population Mean (7 of 11) Note that 28 out of 30, or 93%, of the confidence intervals contain the population mean μ = 50. In general, for a 95% confidence interval, any sample mean that lies within 1.96 standard errors of the population mean will result in a confidence interval that contains μ.

9.2 Estimating a Population Mean Construct and Interpret a Confidence Interval for the Population Mean (8 of 11) Parallel Example 4: Constructing a Confidence Interval about a Population Mean Construct a 99% confidence interval about the population mean weight (in grams) of pennies minted after Assume μ = 0.02 grams.

9.2 Estimating a Population Mean Construct and Interpret a Confidence Interval for the Population Mean (9 of 11)

9.2 Estimating a Population Mean Construct and Interpret a Confidence Interval for the Population Mean (10 of 11)

9.2 Estimating a Population Mean Construct and Interpret a Confidence Interval for the Population Mean (11 of 11) We are 99% confident that the mean weight of pennies minted after 1982 is between and grams.

9.2 Estimating a Population Mean Determine the Sample Size Needed to Estimate the Population Mean within a Specified Margin of Error (1 of 3) Determining the Sample Size n The sample size required to estimate the population mean, µ, with a level of confidence (1 − α)·100% with a specified margin of error, E, is given by

9.2 Estimating a Population Mean Determine the Sample Size Needed to Estimate the Population Mean within a Specified Margin of Error (2 of 3) Parallel Example 7: Determining the Sample Size Back to the pennies. How large a sample would be required to estimate the mean weight of a penny manufactured after within grams with 99% confidence? Assume s = 0.02.

9.2 Estimating a Population Mean Determine the Sample Size Needed to Estimate the Population Mean within a Specified Margin of Error (3 of 3) Rounding up, we find n = 107.

9.3 Estimating a Population Standard Deviation Learning Objectives
1. Find critical values for the chi-square distribution 2. Construct and interpret confidence intervals for the population variance and standard deviation

9. 3 Estimating a Population Standard Deviation 9. 3
9.3 Estimating a Population Standard Deviation Find Critical Values for the Chi-Square Distribution (1 of 5) Chi-Square Distribution If a simple random sample of size n is obtained from a normally distributed population with mean μ and standard deviation σ, then

9.3 Estimating a Population Standard Deviation Find Critical Values for the Chi-Square Distribution (2 of 5) Characteristics of the Chi-Square Distribution It is not symmetric. The shape of the chi-square distribution depends on the degrees of freedom, just like the Student’s t-distribution. As the number of degrees of freedom increases, the chi- square distribution becomes more nearly symmetric. The values of χ2 are nonnegative (greater than or equal to 0).

9.3 Estimating a Population Standard Deviation Find Critical Values for the Chi-Square Distribution (3 of 5)

9.3 Estimating a Population Standard Deviation Find Critical Values for the Chi-Square Distribution (4 of 5) Parallel Example 1: Finding Critical Values for the Chi- Square Distribution Find the chi-square values that separate the middle 95% of the distribution from the 2.5% in each tail. Assume 18 degrees of freedom.

9.3 Estimating a Population Standard Deviation Find Critical Values for the Chi-Square Distribution (5 of 5) Solution Find the chi-square values that separate the middle 95% of the distribution from the 2.5% in each tail. Assume 18 degrees of freedom.

9.3 Estimating a Population Standard Deviation Construct and Interpret Confidence Intervals for the Population Variance and Standard Deviation (1 of 4) A (1 − ) · 100% Confidence Interval for σ2 If a simple random sample of size n is taken from a normal population with mean μ and standard deviation σ, then a (1 − α) · 100% confidence interval for χ2 is given by

9.3 Estimating a Population Standard Deviation Construct and Interpret Confidence Intervals for the Population Variance and Standard Deviation (2 of 4) A (1 − ) · 100% Confidence Interval for σ To find a (1 − ) · 100% confidence interval about σ, take the square root of the lower bound and upper bound.

9.3 Estimating a Population Standard Deviation Construct and Interpret Confidence Intervals for the Population Variance and Standard Deviation (3 of 4) Parallel Example 2: Constructing a Confidence Interval for a Population Variance and Standard Deviation One way to measure the risk of a stock is through the standard deviation rate of return of the stock. The following data represent the weekly rate of return (in percent) of Microsoft for 15 randomly selected weeks. Compute the 90% confidence interval for the risk of Microsoft stock. − − − −3.90 − −1.61 −3.31 Source: Yahoo!Finance

9.3 Estimating a Population Standard Deviation Construct and Interpret Confidence Intervals for the Population Variance and Standard Deviation (4 of 4) Solution A normal probability plot and boxplot indicate the data is approximately normal with no outliers. s = ; s2 = χ20.95 = and χ20.05 = for 15 − 1 = 14 degrees of freedom

9.4 Putting It Together: Which Procedure Do I Use? Learning Objective
1. Determine the appropriate confidence interval to construct

9. 4 Putting It Together: Which Procedure Do I Use. 9. 4
9.4 Putting It Together: Which Procedure Do I Use? Determine the Appropriate Confidence Interval to Construct

9.5 Estimating with Bootstrapping Learning Objectives
1. Estimate a parameter using the bootstrap method

9. 5 Estimating with Bootstrapping 9. 5
9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (1 of 12) Bootstrapping is a computer-intensive approach to statistical inference whereby parameters are estimated by treating a set of sample data as a population. A computer is used to resample with replacement n observations from the sample data. This process is repeated many (say, 1000) times. For each resample, the statistic (such as the sample mean) is obtained.

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (2 of 12) To use a bootstrap, two basic requirements must be satisfied: The “center” of the bootstrap distribution must be close to the “center” of the original sample data. For example, the mean of all bootstrap means must be close to the mean of the original data. The distribution of the bootstrap sample statistic must be symmetric.

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (3 of 12) The Bootstrap Algorithm Step 1 Select B independent bootstrap samples of size n with replacement. Note that n is the number of observations in the original sample. So, if the original sample has 10 observations, each bootstrap sample will have 10 observations. Sampling with replacement means that once an observation is selected to be in the sample, its value is “put back into the hat” so it may be sampled again. The number of resamples B required is typically

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (4 of 12) The Bootstrap Algorithm Step 2 Determine the value of the statistic of interest, such as the mean, for each of the B samples. Step 3 Use the distribution of the B statistics to make a judgment about the value of the parameter. For example, find the 2.5th and 97.5th percentiles to determine the lower and upper bound of a 95% confidence interval.

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (5 of 12) Example 1: Using the Bootstrap Method to Construct a 95% Confidence Interval The website fueleconomy.gov allows drivers to report the miles per gallon of their vehicle. The data in Table 7 shows the reported miles per gallon of 2011 Ford Focus automobiles for 16 different owners. Treat the sample as a simple random sample of all Ford Focus automobiles. Construct a 95% confidence interval for the mean miles per gallon of a 2011 Ford Focus using a bootstrap sample. Interpret the interval.

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (6 of 12) Example 1: Using the Bootstrap Method to Construct a 95% Confidence Interval Table 7 35.7 37.2 34.1 38.9 32.0 41.3 32.5 37.1 37.3 38.8 38.2 39.6 32.2 40.9 37.0 36.0 Source:

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (7 of 12) Solution We will use Minitab to obtain the 95% confidence interval using a bootstrap sample. The steps for constructing confidence intervals using a bootstrap sample using Minitab and StatCrunch are given in the Technology Step-by-Step in the text.

Observations From Each Sample
9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (8 of 12) Solution Step 1 Obtain 1000 independent bootstrap samples of size n = 16 with replacement. Table 8 shows the first few bootstrap samples. Notice that sampling with replacement means that some observations from the original sample may be selected more than once. Note: Your results will differ. Table 8 blank Bootstrap Sample Observations From Each Sample 1 2 3

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (9 of 12) Solution Step 2 Determine the sample mean for each of the bootstrap samples. Step 3 Identify the 2.5th (the 25th observation with the data written in ascending order) and the 97.5th percentile (the 975th observation with the data written in ascending order). The lower bound is the 2.5th percentile, mpg, and the upper bound is the 97.5th percentile, mpg.

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (10 of 12) Solution Figure 27 shows a histogram of the 1000 sample means with the lower and upper bounds labeled. This histogram represents an approximation of the sampling distribution of the sample mean based on the bootstrap resamples. Notice that the distribution is symmetric and centered around the mean of the original data in

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (11 of 12) Solution

9.5 Estimating with Bootstrapping Estimate a Parameter Using the Bootstrap Method (12 of 12) Solution Interpretation We are 95% confident the mean miles per gallon of a 2011 Ford Focus is between and miles per gallon.

STATISTICS INFORMED DECISIONS USING DATA

Similar presentations

Presentation on theme: "STATISTICS INFORMED DECISIONS USING DATA"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

STATISTICS INFORMED DECISIONS USING DATA

Similar presentations

Presentation on theme: "STATISTICS INFORMED DECISIONS USING DATA"— Presentation transcript:

Similar presentations

About project

Feedback