Download presentation
Presentation is loading. Please wait.
Published byHorace Harrison Modified over 9 years ago
1
6 Chapter Confidence Intervals © 2012 Pearson Education, Inc.
All rights reserved. 1 of 83
2
Chapter Outline 6.1 Confidence Intervals for the Mean (Large Samples)
6.2 Confidence Intervals for the Mean (Small Samples) 6.3 Confidence Intervals for Population Proportions 6.4 Confidence Intervals for Variance and Standard Deviation © 2012 Pearson Education, Inc. All rights reserved. 2 of 83
3
Confidence Intervals for the Mean (Large Samples)
Section 6.1 Confidence Intervals for the Mean (Large Samples) © 2012 Pearson Education, Inc. All rights reserved. 3 of 83
4
Section 6.1 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. 4 of 83
5
Point Estimate for Population μ
A single value estimate for a population parameter Most unbiased point estimate of the population mean μ is the sample mean Estimate Population Parameter… with Sample Statistic Mean: μ © 2012 Pearson Education, Inc. All rights reserved. 5 of 83
6
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) © 2012 Pearson Education, Inc. All rights reserved. 6 of 83
7
Solution: Point Estimate for Population μ
The sample mean of the data is Your point estimate for the mean number of friends for all users of the website is friends. © 2012 Pearson Education, Inc. All rights reserved. 7 of 83
8
Interval Estimate Interval estimate
An interval, or range of values, used to estimate a population parameter. ( ) Interval estimate Right endpoint Left endpoint 115 120 125 130 135 140 150 145 Point estimate Point estimate 115 120 125 130 135 140 145 150 How confident do we want to be that the interval estimate contains the population mean μ? © 2012 Pearson Education, Inc. All rights reserved. 8 of 83
9
Level of Confidence Level of confidence c
The probability that the interval estimate contains the population parameter. c is the area under the standard normal curve between the critical values. c z z = 0 –zc zc ½(1 – c) ½(1 – c) Use the Standard Normal Table to find the corresponding z-scores. Critical values The remaining area in the tails is 1 – c . © 2012 Pearson Education, Inc. All rights reserved. 9 of 83
10
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. z z = 0 zc c = 0.90 ½(1 – c) = 0.05 ½(1 – c) = 0.05 zc –zc = –1.645 zc = 1.645 The corresponding z-scores are ±1.645. © 2012 Pearson Education, Inc. All rights reserved. 10 of 83
11
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 μ is generally unknown varies from sample to sample © 2012 Pearson Education, Inc. All rights reserved. 11 of 83
12
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. Sometimes called the maximum error of estimate or error tolerance. When n ≥ 30, the sample standard deviation, s, can be used for σ. © 2012 Pearson Education, Inc. All rights reserved. 12 of 83
13
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 53.0. © 2012 Pearson Education, Inc. All rights reserved. 13 of 83
14
Solution: Finding the Margin of Error
First find the critical values z zc z = 0 0.95 0.025 0.025 –zc = –1.96 zc zc = 1.96 95% 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 = 40 ≥ 30.) © 2012 Pearson Education, Inc. All rights reserved. 14 of 83
15
Solution: Finding the Margin of Error
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. © 2012 Pearson Education, Inc. All rights reserved. 15 of 83
16
Confidence Intervals for the Population Mean
A c-confidence interval for the population mean μ The probability that the confidence interval contains μ is c. © 2012 Pearson Education, Inc. All rights reserved. 16 of 83
17
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 In Symbols Find the sample statistics n and . Specify σ, if known. Otherwise, if n ≥ 30, find the sample standard deviation s and use it as an estimate for σ. © 2012 Pearson Education, Inc. All rights reserved. 17 of 83
18
Constructing Confidence Intervals for μ
In Words In Symbols Find the critical value zc that corresponds to the given level of confidence. Find the margin of error E. Find the left and right endpoints and form the confidence interval. Use the Standard Normal Table or technology. Left endpoint: Right endpoint: Interval: © 2012 Pearson Education, Inc. All rights reserved. 18 of 83
19
Example: Constructing a Confidence Interval
Construct a 95% confidence interval for the mean number of friends for all users of the website. Solution: Recall and E ≈ 16.4 Left Endpoint: Right Endpoint: 114.4 < μ < 147.2 © 2012 Pearson Education, Inc. All rights reserved. 19 of 83
20
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 and © 2012 Pearson Education, Inc. All rights reserved. 20 of 83
21
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. © 2012 Pearson Education, Inc. All rights reserved. 21 of 83
22
Solution: Constructing a Confidence Interval σ Known
First find the critical values z z = 0 zc c = 0.90 ½(1 – c) = 0.05 ½(1 – c) = 0.05 –zc = –1.645 zc zc = 1.645 zc = 1.645 © 2012 Pearson Education, Inc. All rights reserved. 22 of 83
23
Solution: Constructing a Confidence Interval σ Known
Margin of error: Confidence interval: Left Endpoint: Right Endpoint: 22.3 < μ < 23.5 © 2012 Pearson Education, Inc. All rights reserved. 23 of 83
24
Solution: Constructing a Confidence Interval σ Known
22.3 < μ < 23.5 Point estimate 22.3 22.9 23.5 ( ) • With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years. © 2012 Pearson Education, Inc. All rights reserved. 24 of 83
25
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. © 2012 Pearson Education, Inc. All rights reserved. 25 of 83
26
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. 26 of 83
27
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 If σ is unknown, you can estimate it using s, provided you have a preliminary sample with at least 30 members. © 2012 Pearson Education, Inc. All rights reserved. 27 of 83
28
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 53.0. © 2012 Pearson Education, Inc. All rights reserved. 28 of 83
29
Solution: Sample Size First find the critical values zc = 1.96 z z = 0
0.95 0.025 0.025 –zc = –1.96 zc zc = 1.96 zc = 1.96 © 2012 Pearson Education, Inc. All rights reserved. 29 of 83
30
Solution: Sample Size zc = 1.96 σ ≈ s ≈ 53.0 E = 7
When necessary, round up to obtain a whole number. You should include at least 221 users in your sample. © 2012 Pearson Education, Inc. All rights reserved. 30 of 83
31
Section 6.1 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. 31 of 83
32
Confidence Intervals for the Mean (Small Samples)
Section 6.2 Confidence Intervals for the Mean (Small Samples) © 2012 Pearson Education, Inc. All rights reserved. 32 of 83
33
Section 6.2 Objectives Interpret the t-distribution and use a t-distribution table Construct confidence intervals when n < 30, the population is normally distributed, and σ is unknown © 2012 Pearson Education, Inc. All rights reserved. 33 of 83
34
The t-Distribution 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 t-distribution. Critical values of t are denoted by tc. © 2012 Pearson Education, Inc. All rights reserved. 34 of 83
35
Properties of the t-Distribution
The t-distribution is bell shaped and symmetric about the mean. 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 – Degrees of freedom © 2012 Pearson Education, Inc. All rights reserved. 35 of 83
36
Properties of the t-Distribution
The total area under a t-curve is 1 or 100%. The mean, median, and mode of the t-distribution are equal to zero. 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. = 5 The tails in the t-distribution are “thicker” than those in the standard normal distribution. d.f. = 2 t Standard normal curve © 2012 Pearson Education, Inc. All rights reserved. 36 of 83
37
Example: Critical Values of t
Find the critical value tc for a 95% confidence level when the sample size is 15. Solution: d.f. = n – 1 = 15 – 1 = 14 Table 5: t-Distribution tc = 2.145 © 2012 Pearson Education, Inc. All rights reserved. 37 of 83
38
Solution: Critical Values of t
95% of the area under the t-distribution curve with 14 degrees of freedom lies between t = ±2.145. t –tc = –2.145 tc = 2.145 c = 0.95 © 2012 Pearson Education, Inc. All rights reserved. 38 of 83
39
Confidence Intervals for the Population Mean
A c-confidence interval for the population mean μ The probability that the confidence interval contains μ is c. © 2012 Pearson Education, Inc. All rights reserved. 39 of 83
40
Confidence Intervals and t-Distributions
In Words In Symbols Identify the sample statistics n, , and s. Identify the degrees of freedom, the level of confidence c, and the critical value tc. Find the margin of error E. d.f. = n – 1 © 2012 Pearson Education, Inc. All rights reserved. 40 of 83
41
Confidence Intervals and t-Distributions
In Words In Symbols Find the left and right endpoints and form the confidence interval. Left endpoint: Right endpoint: Interval: © 2012 Pearson Education, Inc. All rights reserved. 41 of 83
42
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 t-distribution (n < 30, σ is unknown, temperatures are approximately normally distributed). © 2012 Pearson Education, Inc. All rights reserved. 42 of 83
43
Solution: Constructing a Confidence Interval
n =16, x = s = c = 0.95 df = n – 1 = 16 – 1 = 15 Critical Value Table 5: t-Distribution tc = 2.131 © 2012 Pearson Education, Inc. All rights reserved. 43 of 83
44
Solution: Constructing a Confidence Interval
Margin of error: Confidence interval: Left Endpoint: Right Endpoint: 156.7 < μ < 167.3 © 2012 Pearson Education, Inc. All rights reserved. 44 of 83
45
Solution: Constructing a Confidence Interval
156.7 < μ < 167.3 Point estimate 156.7 162.0 167.3 ( ) • With 95% confidence, you can say that the population mean temperature of coffee sold is between 156.7ºF and 167.3ºF. © 2012 Pearson Education, Inc. All rights reserved. 45 of 83
46
Normal or t-Distribution?
Yes Is n ≥ 30? Use the normal distribution with If σ is unknown, use s instead. No Is the population normally, or approximately normally, distributed? No Cannot use the normal distribution or the t-distribution. Yes Yes Is σ known? Use the normal distribution with No Use the t-distribution with and n – 1 degrees of freedom. © 2012 Pearson Education, Inc. All rights reserved. 46 of 83
47
Example: 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) © 2012 Pearson Education, Inc. All rights reserved. 47 of 83
48
Section 6.2 Summary Interpreted the t-distribution and used a t-distribution table Constructed confidence intervals when n < 30, the population is normally distributed, and σ is unknown © 2012 Pearson Education, Inc. All rights reserved. 48 of 83
49
Confidence Intervals for Population Proportions
Section 6.3 Confidence Intervals for Population Proportions © 2012 Pearson Education, Inc. All rights reserved. 49 of 83
50
Section 6.3 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. 50 of 83
51
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 read as “p hat” © 2012 Pearson Education, Inc. All rights reserved. 51 of 83
52
Point Estimate for Population p
Estimate Population Parameter… with Sample Statistic Proportion: p Point Estimate for q, the population proportion of failures Denoted by Read as “q hat” © 2012 Pearson Education, Inc. All rights reserved. 52 of 83
53
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 = and x = 662 © 2012 Pearson Education, Inc. All rights reserved. 53 of 83
54
Confidence Intervals for p
A c-confidence interval for a population proportion p The probability that the confidence interval contains p is c. © 2012 Pearson Education, Inc. All rights reserved. 54 of 83
55
Constructing Confidence Intervals for p
In Words In Symbols Identify the sample statistics n and x. Find the point estimate Verify that the sampling distribution of can be approximated by a normal distribution. Find the critical value zc that corresponds to the given level of confidence c. Use the Standard Normal Table or technology. © 2012 Pearson Education, Inc. All rights reserved. 55 of 83
56
Constructing Confidence Intervals for p
In Words In Symbols Find the margin of error E. Find the left and right endpoints and form the confidence interval. Left endpoint: Right endpoint: Interval: © 2012 Pearson Education, Inc. All rights reserved. 56 of 83
57
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 © 2012 Pearson Education, Inc. All rights reserved. 57 of 83
58
Solution: Confidence Interval for p
Verify the sampling distribution of can be approximated by the normal distribution Margin of error: © 2012 Pearson Education, Inc. All rights reserved. 58 of 83
59
Solution: Confidence Interval for p
Left Endpoint: Right Endpoint: 0.633 < p < 0.691 © 2012 Pearson Education, Inc. All rights reserved. 59 of 83
60
Solution: Confidence Interval for p
Point estimate 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%. © 2012 Pearson Education, Inc. All rights reserved. 60 of 83
61
Sample Size Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate p is This formula assumes you have an estimate for and . If not, use and © 2012 Pearson Education, Inc. All rights reserved. 61 of 83
62
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 no preliminary estimate is available. Solution: Because you do not have a preliminary estimate for use and © 2012 Pearson Education, Inc. All rights reserved. 62 of 83
63
Solution: Sample Size c = 0.95 zc = 1.96 E = 0.03
Round up to the nearest whole number. With no preliminary estimate, the minimum sample size should be at least 1068 voters. © 2012 Pearson Education, Inc. All rights reserved. 63 of 83
64
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 a preliminary estimate gives Solution: Use the preliminary estimate © 2012 Pearson Education, Inc. All rights reserved. 64 of 83
65
Solution: Sample Size c = 0.95 zc = 1.96 E = 0.03
Round up to the nearest whole number. With a preliminary estimate of , the minimum sample size should be at least 914 voters. Need a larger sample size if no preliminary estimate is available. © 2012 Pearson Education, Inc. All rights reserved. 65 of 83
66
Section 6.3 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. 66 of 83
67
Confidence Intervals for Variance and Standard Deviation
Section 6.4 Confidence Intervals for Variance and Standard Deviation © 2012 Pearson Education, Inc. All rights reserved. 67 of 83
68
Section 6.4 Objectives Interpret the chi-square distribution and use a chi-square distribution table Use the chi-square distribution to construct a confidence interval for the variance and standard deviation © 2012 Pearson Education, Inc. All rights reserved. 68 of 83
69
The Chi-Square Distribution
The point estimate for σ2 is s2 The point estimate for σ is s s2 is the most unbiased estimate for σ2 Estimate Population Parameter… with Sample Statistic Variance: σ2 s2 Standard deviation: σ s © 2012 Pearson Education, Inc. All rights reserved. 69 of 83
70
The Chi-Square Distribution
You can use the chi-square 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 forms a chi-square distribution for samples of any size n > 1. © 2012 Pearson Education, Inc. All rights reserved. 70 of 83
71
Properties of The Chi-Square Distribution
All chi-square values χ2 are greater than or equal to zero. The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for σ2, use the χ2-distribution with degrees of freedom equal to one less than the sample size. d.f. = n – Degrees of freedom The area under each curve of the chi-square distribution equals one. © 2012 Pearson Education, Inc. All rights reserved. 71 of 83
72
Properties of The Chi-Square Distribution
Chi-square distributions are positively skewed. Chi-square Distributions © 2012 Pearson Education, Inc. All rights reserved. 72 of 83
73
Critical Values for χ2 There are two critical values for each level of confidence. The value χ2R represents the right-tail critical value The value χ2L represents the left-tail critical value. χ2 c The area between the left and right critical values is c. © 2012 Pearson Education, Inc. All rights reserved. 73 of 83
74
Example: Finding Critical Values for χ2
Find the critical values and 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 chi-square curve to the right of the critical value. Area to the right of χ2R = Area to the right of χ2L = © 2012 Pearson Education, Inc. All rights reserved. 74 of 83
75
Solution: Finding Critical Values for χ2
Table 6: χ2-Distribution 7.564 30.191 95% of the area under the curve lies between and © 2012 Pearson Education, Inc. All rights reserved. 75 of 83
76
Confidence Intervals for σ2 and σ
Confidence Interval for σ2: Confidence Interval for σ: The probability that the confidence intervals contain σ2 or σ is c. © 2012 Pearson Education, Inc. All rights reserved. 76 of 83
77
Confidence Intervals for σ2 and σ
In Words In Symbols Verify that the population has a normal distribution. Identify the sample statistic n and the degrees of freedom. Find the point estimate s2. Find the critical values χ2R and χ2L that correspond to the given level of confidence c. d.f. = n – 1 Use Table 6 in Appendix B. © 2012 Pearson Education, Inc. All rights reserved. 77 of 83
78
Confidence Intervals for 2 and
In Words In Symbols Find the left and right endpoints and form the confidence interval for the population variance. Find the confidence interval for the population standard deviation by taking the square root of each endpoint. © 2012 Pearson Education, Inc. All rights reserved. 78 of 83
79
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. © 2012 Pearson Education, Inc. All rights reserved. 79 of 83
80
Solution: Constructing a Confidence Interval
Area to the right of χ2R = Area to the right of χ2L = The critical values are χ2R = and χ2L = © 2012 Pearson Education, Inc. All rights reserved. 80 of 83
81
Solution: Constructing a Confidence Interval
Confidence Interval for σ2: Left endpoint: Right endpoint: 0.80 < σ2 < 3.18 With 99% confidence, you can say that the population variance is between 0.80 and 3.18. © 2012 Pearson Education, Inc. All rights reserved. 81 of 83
82
Solution: Constructing a Confidence Interval
Confidence Interval for σ : 0.89 < σ < 1.78 With 99% confidence, you can say that the population standard deviation is between 0.89 and 1.78 milligrams. © 2012 Pearson Education, Inc. All rights reserved. 82 of 83
83
Section 6.4 Summary Interpreted the chi-square distribution and used a chi-square distribution table Used the chi-square distribution to construct a confidence interval for the variance and standard deviation © 2012 Pearson Education, Inc. All rights reserved. 83 of 83
Similar presentations
© 2024 SlidePlayer.com Inc.
All rights reserved.