Section 7.2 Confidence Intervals for Population Proportions


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1 Section 7.2 Confidence Intervals for Population Proportions 2012 Pearson Education, Inc. All rights reserved. 1 of 83
2 Section 7.2 Objectives Find a point estimate for the population proportion Construct a confidence interval for a population proportion Determine the minimum sample size required when estimating a population proportion 2012 Pearson Education, Inc. All rights reserved. 2 of 83
3 Point Estimate for Population p Population Proportion The probability of success in a single trial of a binomial experiment. Denoted by p Point Estimate for p The proportion of successes in a sample. Denoted by x number of successes in sample pˆ = = n sample size read as p hat 2012 Pearson Education, Inc. All rights reserved. 3 of 83
4 Point Estimate for Population p Estimate Population Parameter with Sample Statistic Proportion: p ˆp Point Estimate for q, the population proportion of failures Denoted by Read as q hat ˆ q = 1 ˆ p 2012 Pearson Education, Inc. All rights reserved. 4 of 83
5 Example: Point Estimate for p In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal while at work. Find a point estimate for the population proportion of U.S. adults who say it is acceptable to check personal while at work. (Adapted from Liberty Mutual) Solution: n = 1000 and x = 662 pˆ x = = n = = 66.2% 2012 Pearson Education, Inc. All rights reserved. 5 of 83
6 Confidence Intervals for p A cconfidence interval for a population proportion p ˆ p E < p< pˆ + E where E = Z α 2 pq ˆˆ n The probability that the confidence interval contains p is c Pearson Education, Inc. All rights reserved. 6 of 83
7 Constructing Confidence Intervals for p In Words 1. Identify the sample statistics n and x. 2. Find the point estimate ˆp. 3. Verify that the sampling distribution of pˆ can be approximated by a normal distribution. 4. Find the critical value z c that corresponds to the given level of confidence c. In Symbols npˆ pˆ = x n 5, nqˆ 5 Use the Standard Normal Table or technology Pearson Education, Inc. All rights reserved. 7 of 83
8 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 α 2 pq ˆˆ n Left endpoint: ˆp E Right endpoint: ˆp+ E Interval: pˆ E < p< pˆ + E 2012 Pearson Education, Inc. All rights reserved. 8 of 83
9 Example: Confidence Interval for p In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal 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 while at work. Solution: Recall p ˆ = qˆ = 1 pˆ = = Pearson Education, Inc. All rights reserved. 9 of 83
10 Solution: Confidence Interval for p Verify the sampling distribution of can be approximated by the normal distribution npˆ = = 662 > 5 nqˆ = = 338 > 5 Margin of error: ˆp E pq ˆˆ (0.662) (0.338) = Zα 2 = n Pearson Education, Inc. All rights reserved. 10 of 83
11 Solution: Confidence Interval for p Confidence interval: Left Endpoint: pˆ E < p < Right Endpoint: pˆ = = E 2012 Pearson Education, Inc. All rights reserved. 11 of 83
12 Solution: Confidence Interval for p < p < 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 while at work is between 63.3% and 69.1% Pearson Education, Inc. All rights reserved. 12 of 83
13 Determining Sample Size When p is known, n = Z α 2 E 2 2 pq ˆˆ Where Z α 2 is the value associated with the desired confidence level, and E is the desired margin of error. Round up to the next integer. 13
14 Sample Size for Estimating the Population Proportion When p is unknown, we use n = Z α 2 E
15 Example: 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 p ˆ = 0.5 and q ˆ = p, ˆ 2012 Pearson Education, Inc. All rights reserved. 15 of 83
16 Solution: Sample Size c = 0.95 z α/2 = 1.96 E = 0.03 n 2 [ ] 2 Z ˆˆ α 2 pq 1.96 (0.5)(0.5) = = 2 2 E (0. 03) Round up to the nearest whole number. With no preliminary estimate, the minimum sample size should be at least 1068 voters Pearson Education, Inc. All rights reserved. 16 of 83
17 Example: 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 p ˆ = Solution: Use the preliminary estimate qˆ = 1 pˆ = = ˆ 0.31 p = 2012 Pearson Education, Inc. All rights reserved. 17 of 83
18 Solution: Sample Size c = 0.95 z α/2 = 1.96 E = 0.03 n Z ˆ ˆ α 2 pq = = 2 E 2 [ ] (0.31)(0.69) (0.03) Round up to the nearest whole number. With a preliminary estimate of p ˆ = 0.31, the minimum sample size should be at least 914 voters. Need a larger sample size if no preliminary estimate is available Pearson Education, Inc. All rights reserved. 18 of 83
19 Section 7.3 Confidence Intervals for the Mean (σ known) 2012 Pearson Education, Inc. All rights reserved. 19 of 83
20 Section 7.3 Objectives Find a point estimate and a margin of error Construct and interpret confidence intervals for the population mean Determine the minimum sample size required when estimating μ 2012 Pearson Education, Inc. All rights reserved. 20 of 83
21 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 x Estimate Population with Sample Parameter Statistic Mean: μ x 2012 Pearson Education, Inc. All rights reserved. 21 of 83
22 Example: 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) Pearson Education, Inc. All rights reserved. 22 of 83
23 Solution: Point Estimate for Population μ The sample mean of the data is Σx 5232 x = = = n 40 Your point estimate for the mean number of friends for all users of the website is friends Pearson Education, Inc. All rights reserved. 23 of 83
24 Interval Estimate Interval estimate An interval, or range of values, used to estimate a population parameter. Left endpoint Point estimate x =130.8 ( ) Interval estimate Right endpoint How confident do we want to be that the interval estimate contains the population mean μ? 2012 Pearson Education, Inc. All rights reserved. 24 of 83
25 Level of Confidence Level of confidence (CLevel) The probability that the interval estimate contains the population parameter. α/2 /2 z z = 0 Critical values The remaining area in the tails is 1 c. c is the area under the Clevel standard normal curve between the critical values. z c Use the Standard Normal Table to find the corresponding zscores Pearson Education, Inc. All rights reserved. 25 of 83 z
26 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. CLevel = 0.90 α /2= 0.05 α/2= 0.05 z α/2 z = c z = 0 Zz α/2 c = The corresponding zscores are ± z 2012 Pearson Education, Inc. All rights reserved. 26 of 83
27 Sampling error Sampling Error The difference between the point estimate and the actual population parameter value. For μ: the sampling error is the difference x μ μ is generally unknown varies from sample to sample x 2012 Pearson Education, Inc. All rights reserved. 27 of 83
28 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 α /2 σ n When n 30, the sample standard deviation, s, can be used for σ. Sometimes called the maximum error of estimate or error tolerance Pearson Education, Inc. All rights reserved. 28 of 83
29 Example: 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 Pearson Education, Inc. All rights reserved. 29 of 83
30 Solution: Finding the Margin of Error First find the critical values z α/2 = z 1.96 c z = 0 z α/2 = % of the area under the standard normal curve falls within 1.96 standard deviations of the mean. (You can approximate the distribution of the sample means with a normal curve by the Central Limit Theorem, because n = ) 2012 Pearson Education, Inc. All rights reserved. 30 of 83 z c z
31 Solution: Finding the Margin of Error σ E = Z Z n α 2 α 2 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 Pearson Education, Inc. All rights reserved. 31 of 83
32 Confidence Intervals for the Population Mean A cconfidence interval for the population mean μ x E < µ < x + E where E = z c σ n The probability that the confidence interval contains μ is c Pearson Education, Inc. All rights reserved. 32 of 83
33 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. x In Symbols Σx x = 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 = Σ( x x) n Pearson Education, Inc. All rights reserved. 33 of 83
34 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. σ E = Zα 2 n Left endpoint: x E Right endpoint: x + E Interval: x E < µ < x + E 2012 Pearson Education, Inc. All rights reserved. 34 of 83
35 Example: Constructing a Confidence Interval Construct a 95% confidence interval for the mean number of friends for all users of the website. Solution: Recall x =130.8 and E 16.4 Left Endpoint: x E = x < μ < Right Endpoint: E = Pearson Education, Inc. All rights reserved. 35 of 83
36 Solution: Constructing a Confidence Interval < μ < With 95% confidence, you can say that the population mean number of friends is between and Pearson Education, Inc. All rights reserved. 36 of 83
37 Example: 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 Pearson Education, Inc. All rights reserved. 37 of 83
38 Solution: Constructing a Confidence Interval σ Known First find the critical values α = 0.90 α/2 = 0.05 α/2 = 0.05 z c = z c z = 0 z c = z c z z c = Pearson Education, Inc. All rights reserved. 38 of 83
39 Solution: Constructing a Confidence Interval σ Known Margin of error: σ E 1.5 = Zα 2 = n 20 Confidence interval: Left Endpoint: x = E < μ < 23.5 Right Endpoint: x + E = Pearson Education, Inc. All rights reserved. 39 of 83
40 Solution: Constructing a Confidence Interval σ Known 22.3 < μ < 23.5 Point estimate x ( ) E x + x E With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years Pearson Education, Inc. All rights reserved. 40 of 83
41 Interpreting the Results μ is a fixed 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, approximately 90% of these intervals will contain μ Pearson Education, Inc. All rights reserved. 41 of 83
42 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 2012 Pearson Education, Inc. All rights reserved. 42 of 83
43 Sample Size Given a cconfidence level and a margin of error E, the minimum sample size n needed to estimate the population mean µ is n z c σ = E If σ is unknown, you can estimate it using s, provided you have a preliminary sample with at least 30 members Pearson Education, Inc. All rights reserved. 43 of 83
44 Example: 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 Pearson Education, Inc. All rights reserved. 44 of 83
45 Solution: Sample Size First find the critical values z c = z 1.96 c z = 0 z c = 1.96 z c z z c = Pearson Education, Inc. All rights reserved. 45 of 83
46 Solution: Sample Size z c = 1.96 σ s 53.0 E = 7 n z c σ = E When necessary, round up to obtain a whole number. You should include at least 221 users in your sample Pearson Education, Inc. All rights reserved. 46 of 83
47 Section 7.3 Summary Found a point estimate and a margin of error Constructed and interpreted confidence intervals for the population mean Determined the minimum sample size required when estimating μ 2012 Pearson Education, Inc. All rights reserved. 47 of 83
48 Section 7.4 Confidence Intervals for the Mean (σ unknown) 2012 Pearson Education, Inc. All rights reserved. 48 of 83
49 Section 7.4 Objectives Interpret the tdistribution and use a tdistribution table Construct confidence intervals when n < 30, the population is normally distributed, and σ is unknown 2012 Pearson Education, Inc. All rights reserved. 49 of 83
50 The tdistribution When the population standard deviation is unknown, the sample size is less than 30, and the random variable x is approximately normally distributed, it follows a tdistribution. t = x µ s n Critical values of t are denoted by t c Pearson Education, Inc. All rights reserved. 50 of 83
51 Properties of the tdistribution 1. The tdistribution is bell shaped and symmetric about the mean. 2. The tdistribution 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 x is calculated. When you use a tdistribution 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 2012 Pearson Education, Inc. All rights reserved. 51 of 83
52 Properties of the tdistribution 3. The total area under a tcurve is 1 or 100%. 4. The mean, median, and mode of the tdistribution are equal to zero. 5. As the degrees of freedom increase, the tdistribution 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 Pearson Education, Inc. All rights reserved. 52 of 83
53 Example: 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: tdistribution t c = Pearson Education, Inc. All rights reserved. 53 of 83
54 Solution: Critical Values of t 95% of the area under the tdistribution curve with 14 degrees of freedom lies between t = ± c = 0.95 t c = t c = t 2012 Pearson Education, Inc. All rights reserved. 54 of 83
55 Confidence Intervals for the Population Mean A cconfidence interval for the population mean μ s x E < µ < x + E where E = tc n The probability that the confidence interval contains μ is c Pearson Education, Inc. All rights reserved. 55 of 83
56 Confidence Intervals and tdistributions In Words 1. Identify the sample statistics n, x, and s. 2. Identify the degrees of freedom, the level of confidence c, and the critical value t c. In Symbols Σx x = s = n d.f. = n 1 ( x x) n Find the margin of error E. E = t c s n 2012 Pearson Education, Inc. All rights reserved. 56 of 83
57 Confidence Intervals and tdistributions In Words 4. Find the left and right endpoints and form the confidence interval. In Symbols Left endpoint: x Right endpoint: x + Interval: x E < µ < x + E E E 2012 Pearson Education, Inc. All rights reserved. 57 of 83
58 Example: 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 approximately normally distributed. Solution: Use the tdistribution (n < 30, σ is unknown, temperatures are approximately normally distributed) Pearson Education, Inc. All rights reserved. 58 of 83
59 Solution: Constructing a Confidence Interval n =16, x = s = 10.0 c = 0.95 df = n 1 = 16 1 = 15 Critical Value Table 5: tdistribution t c = Pearson Education, Inc. All rights reserved. 59 of 83
60 Solution: Constructing a Confidence Interval Margin of error: s 10 E = t c = n 16 Confidence interval: Left Endpoint: x E = < μ < Right Endpoint: x + E = Pearson Education, Inc. All rights reserved. 60 of 83
61 Solution: Constructing a Confidence Interval < μ < Point estimate x ( ) E + x x E With 95% confidence, you can say that the population mean temperature of coffee sold is between 156.7ºF and 167.3ºF Pearson Education, Inc. All rights reserved. 61 of 83
62 Normal or tdistribution? Is n 30? No Is the population normally, or approximately 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 tdistribution. Is σ known? No Use the tdistribution with s E = t c and n 1 degrees of freedom. n Yes Use the normal distribution with σ E = z c. n 2012 Pearson Education, Inc. All rights reserved. 62 of 83
63 Example: Normal or tdistribution? 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 tdistribution, 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) 2012 Pearson Education, Inc. All rights reserved. 63 of 83
64 Section 7.3 Summary Interpreted the tdistribution and used a tdistribution table Constructed confidence intervals when n < 30, the population is normally distributed, and σ is unknown 2012 Pearson Education, Inc. All rights reserved. 64 of 83
65 Section 7.3 Summary Found a point estimate for the population proportion Constructed a confidence interval for a population proportion Determined the minimum sample size required when estimating a population proportion 2012 Pearson Education, Inc. All rights reserved. 65 of 83
66 Section 7.5 Confidence Intervals for Variance and Standard Deviation 2012 Pearson Education, Inc. All rights reserved. 66 of 83
67 Section 7.5 Objectives Interpret the chisquare distribution and use a chisquare distribution table Use the chisquare distribution to construct a confidence interval for the variance and standard deviation 2012 Pearson Education, Inc. All rights reserved. 67 of 83
68 The ChiSquare Distribution The point estimate for σ 2 is s 2 The point estimate for σ is s s 2 is the most unbiased estimate for σ 2 Estimate Population Parameter with Sample Statistic Variance: σ 2 s 2 Standard deviation: σ s 2012 Pearson Education, Inc. All rights reserved. 68 of 83
69 The ChiSquare Distribution You can use the chisquare distribution to construct a confidence interval for the variance and standard deviation. If the random variable x has a normal distribution, then the distribution of 2 2 ( n 1) s χ = 2 σ forms a chisquare distribution for samples of any size n > Pearson Education, Inc. All rights reserved. 69 of 83
70 Properties of The ChiSquare Distribution 1. All chisquare values χ 2 are greater than or equal to zero. 2. The chisquare distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for σ 2, use the χ 2 distribution with degrees of freedom equal to one less than the sample size. d.f. = n 1 Degrees of freedom 3. The area under each curve of the chisquare distribution equals one Pearson Education, Inc. All rights reserved. 70 of 83
71 Properties of The ChiSquare Distribution 4. Chisquare distributions are positively skewed. Chisquare Distributions 2012 Pearson Education, Inc. All rights reserved. 71 of 83
72 Critical Values for χ 2 There are two critical values for each level of confidence. The value χ 2 R represents the righttail critical value The value χ 2 L represents the lefttail critical value. 1 c 2 c 1 c 2 The area between the left and right critical values is c. 2 χ L 2 χ R χ Pearson Education, Inc. All rights reserved. 72 of 83
73 Example: Finding Critical Values for χ Find the critical values χ R and χ L for a 95% confidence interval when the sample size is 18. Solution: d.f. = n 1 = 18 1 = 17 d.f. Each area in the table represents the region under the chisquare curve to the right of the critical value. Area to the right of χ 2 R = Area to the right of χ 2 L = 1 c = = c = = Pearson Education, Inc. All rights reserved. 73 of 83
74 Solution: Finding Critical Values for χ 2 Table 6: χ 2 Distribution 2 χ L = χ R = % of the area under the curve lies between and Pearson Education, Inc. All rights reserved. 74 of 83
75 Confidence Intervals for σ 2 and σ Confidence Interval for σ 2 : ( n 1) s ( n 1) s χ < σ < 2 R χl Confidence Interval for σ: 2 2 ( n 1) s ( n 1) s < σ < χ 2 2 R χl The probability that the confidence intervals contain σ 2 or σ is c Pearson Education, Inc. All rights reserved. 75 of 83
76 Confidence Intervals for σ 2 and σ In Words 1. Verify that the population has a normal distribution. 2. Identify the sample statistic n and the degrees of freedom. 3. Find the point estimate s Find the critical values χ 2 R and χ 2 L that correspond to the given level of confidence c. In Symbols s d.f. = n ( x x) = n 1 Use Table 6 in Appendix B Pearson Education, Inc. All rights reserved. 76 of 83
77 Confidence Intervals for σ 2 and σ In Words 5. Find the left and right endpoints and form the confidence interval for the population variance. 6. Find the confidence interval for the population standard deviation by taking the square root of each endpoint. In Symbols < σ < 2 R χl ( n 1) s ( n 1) s χ 2 2 ( n 1) s ( n 1) s < σ < χ 2 2 R χl 2012 Pearson Education, Inc. All rights reserved. 77 of 83
78 Example: Constructing a Confidence Interval You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.20 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation. Solution: d.f. = n 1 = 30 1 = 29 d.f Pearson Education, Inc. All rights reserved. 78 of 83
79 Solution: Constructing a Confidence Interval Area to the right of χ 2 R = 1 c = = Area to the right of χ 2 L = 1+ c = = The critical values are χ 2 R = and χ 2 L = Pearson Education, Inc. All rights reserved. 79 of 83
80 Solution: Constructing a Confidence Interval Confidence Interval for σ 2 : Left endpoint: ( n 1) s χ 2 R 2 = (30 1)(1.20) Right endpoint: ( n 1) s χ 2 L 2 = (30 1)(1.20) < σ 2 < 3.18 With 99% confidence, you can say that the population variance is between 0.80 and Pearson Education, Inc. All rights reserved. 80 of 83
81 Solution: Constructing a Confidence Interval Confidence Interval for σ : 2 2 ( n 1) s ( n 1) s < σ < χ 2 2 R χl 2 2 (30 1)(1.20) (30 1)(1.20) < σ < < σ < 1.78 With 99% confidence, you can say that the population standard deviation is between 0.89 and 1.78 milligrams Pearson Education, Inc. All rights reserved. 81 of 83
82 Section 7.5 Summary Interpreted the chisquare distribution and used a chisquare distribution table Used the chisquare distribution to construct a confidence interval for the variance and standard deviation 2012 Pearson Education, Inc. All rights reserved. 82 of 83
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