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Chapter 6 Confidence Intervals 1 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 2 Section 6.1 Confidence Intervals for the Mean (Large Samples) 3 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 4 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 x Mean: μ 5 Example: Point Estimate for Population μ Market researchers use the number of sentences per advertisement as a measure of readability for magazine advertisements. The following represents a random sample of the number of sentences found in 50 advertisements. Find a point estimate of the population mean, . (Source: Journal of Advertising Research) 9 20 18 16 9 9 11 13 22 16 5 18 6 6 5 12 25 17 23 7 10 9 10 10 5 11 18 18 9 9 17 13 11 7 14 6 11 12 11 6 12 14 11 9 18 12 12 17 11 20 6 Solution: Point Estimate for Population μ The sample mean of the data is x 620 x 12.4 n 50 Your point estimate for the mean length of all magazine advertisements is 12.4 sentences. 7 Interval Estimate Interval estimate • An interval, or range of values, used to estimate a population parameter. Point estimate ( 12.4 • ) Interval estimate How confident do we want to be that the interval estimate contains the population mean μ? 8 Level of Confidence Level of confidence c • The probability that the interval estimate contains the population parameter. c is the area under the c standard normal curve between the critical values. ½(1 – c) ½(1 – c) -zc z=0 Critical values z zc Use the Standard Normal Table to find the corresponding z-scores. The remaining area in the tails is 1 – c . 9 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 zc -zc = -1.645 z=0 zc =zc1.645 z The corresponding z-scores are +1.645. 10 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 x varies from sample to sample 11 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 zcσ x zc σ n When n 30, the sample standard deviation, s, can be used for . • Sometimes called the maximum error of estimate or error tolerance. 12 Example: Finding the Margin of Error Use the magazine advertisement data and a 95% confidence level to find the margin of error for the mean number of sentences in all magazine advertisements. Assume the sample standard deviation is about 5.0. 13 Solution: Finding the Margin of Error • First find the critical values 0.95 0.025 0.025 -zc = -1.96 zc z=0 zczc= 1.96 z 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 ≥ 30.) 14 Solution: Finding the Margin of Error E zc n 1.96 zc s n You don’t know σ, but since n ≥ 30, you can use s in place of σ. 5.0 50 1.4 You are 95% confident that the margin of error for the population mean is about 1.4 sentences. 15 Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ • x E x E where E zc n • The probability that the confidence interval contains μ is c. 16 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. 2. Specify , if known. Otherwise, if n 30, find the sample standard deviation s and use it as an estimate for . In Symbols x x n (x x )2 s n 1 17 Constructing Confidence Intervals for μ In Words 3. Find the critical value zc 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. E zc n Left endpoint: x E Right endpoint: x E Interval: xE xE 18 Example: Constructing a Confidence Interval Construct a 95% confidence interval for the mean number of sentences in all magazine advertisements. Solution: Recall x 12.4 and E = 1.4 Left Endpoint: xE 12.4 1.4 11.0 Right Endpoint: xE 12.4 1.4 13.8 11.0 < μ < 13.8 19 Solution: Constructing a Confidence Interval 11.0 < μ < 13.8 11.0 ( 12.4 13.8 • ) With 95% confidence, you can say that the population mean number of sentences is between 11.0 and 13.8. 20 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. 21 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 zc z z=0 -zc = -1.645 zc zc = 1.645 zc = 1.645 22 Solution: Constructing a Confidence Interval σ Known • Margin of error: E zc n 1.645 • Confidence interval: Left Endpoint: xE 22.9 0.6 22.3 1.5 20 0.6 Right Endpoint: xE 22.9 0.6 23.5 22.3 < μ < 23.5 23 Solution: Constructing a Confidence Interval σ Known 22.3 < μ < 23.5 Point estimate 22.3 ( x E 22.9 23.5 x xE • ) With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years. 24 Using Ti-83 or 84 to find z confidence interval: n 20 1.5 c 0.95 x 22.9 The same as: 22.3 < μ < 23.5 Slide 6- 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. 26 Confidence Intervals http://www.learner.org/courses/againstallodds/unitpages/unit24.html Slide 6- 27 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 28 Sample Size • Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate the population mean is E zc n 2 2 E 2 zc n z E n zc2 2 zc 2 n 2 ( ) E E 2 • If is unknown, you can estimate it using s provided you have a preliminary sample with at least 30 members. 2 c 2 zc n E 2 29 Example: Sample Size You want to estimate the mean number of sentences in a magazine advertisement. How many magazine advertisements must be included in the sample if you want to be 95% confident that the sample mean is within one sentence of the population mean? Assume the sample standard deviation is about 5.0. 30 Solution: Sample Size • First find the critical values 0.95 0.025 0.025 zc -zc = -1.96 z=0 zczc= 1.96 z zc = 1.96 31 Solution: Sample Size zc = 1.96 s = 5.0 E=1 zc 1.96 5.0 n 96.04 1 E 2 2 When necessary, round up to obtain a whole number. You should include at least 97 magazine advertisements in your sample. 32 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 33 Section 6.2 Confidence Intervals for the Mean (Small Samples) 34 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 35 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. x - t s n • Critical values of t are denoted by tc. 36 In statistics, the t-distribution was first derived in 1876 by Helmert and Lüroth. In the English-language literature it takes its name from William Sealy Gosset's 1908 paper in Biometrika under the pseudonym "Student", published while he worked at the Guinness Brewery in Dublin, Ireland. One version of the origin of the pseudonym is that Gosset's employer forbade members of its staff from publishing scientific papers, so he had to hide his identity. Another version is that Guinness did not want their competition to know that they were using the t-test to test the quality of raw material. The t-test and the associated theory became well-known through the work of Ronald A. Fisher, who called the distribution "Student's distribution". T-distribution’s mathematical formulas: Slide 6- 37 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 x 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 38 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 tdistribution is very close to the standard normal zdistribution. The tails in the tdistribution are “thicker” than those in the standard normal distribution. d.f. = 2 d.f. = 5 Standard normal curve t 0 39 Example: Critical Values of t Find the critical value tc for a 95% confidence when the sample size is 15. Solution: d.f. = n – 1 = 15 – 1 = 14 Table 5: t-Distribution tc = 2.145 40 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 -tc = -2.145 tc = 2.145 41 Using the inverse t with Ti-84: the invT gives the number such that the area to the left of it is given. For example for 95% in the middle we’ll have to enter .95+(.05/2) for the area in the calculator command: invT(.95 +(.05/2), 14) to get: tc = 2.145 and invT(.05/2, 14) to get: -tc = -2.145 c = 0.95 t -tc = -2.145 tc = 2.145 Slide 6- 42 Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ s • x E x E where E tc n • The probability that the confidence interval contains μ is c. 43 Confidence Intervals and t-Distributions 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 tc. 3. Find the margin of error E. In Symbols x (x x )2 x s n 1 n d.f. = n – 1 E tc s n 44 Confidence Intervals and t-Distributions In Words 4. Find the left and right endpoints and form the confidence interval. In Symbols Left endpoint: x E Right endpoint: x E Interval: xE xE s E tc n 45 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 mean temperature. Assume the temperatures are approximately normally distributed. Solution: Use the t-distribution (n < 30, σ is unknown, temperatures are approximately normally distributed.) 46 Solution: Constructing a Confidence Interval • n =16, x = 162.0 s = 10.0 c = 0.95 • df = n – 1 = 16 – 1 = 15 • Critical Value Table 5: t-Distribution tc = 2.131 47 Solution: Constructing a Confidence Interval • Margin of error: s 10 E tc 2.131 5.3 n 16 • Confidence interval: Left Endpoint: xE 162 5.3 156.7 Right Endpoint: xE 162 5.3 167.3 156.7 < μ < 167.3 48 Solution: Constructing a Confidence Interval • 156.7 < μ < 167.3 Point estimate 156.7 ( x E 162.0 •x 167.3 ) xE With 95% confidence, you can say that the mean temperature of coffee sold is between 156.7ºF and 167.3ºF. 49 n =16, x = 162.0 s = 10.0 c = 0.95 df = n – 1 = 16 – 1 = 15 The same as: 156.7 < μ < 167.3 Slide 6- 50 Normal or t-Distribution? Is n 30? Yes No Is the population normally, or approximately normally, distributed? Use the normal distribution with σ E zc n If is unknown, use s instead. No Cannot use the normal distribution or the t-distribution. Yes Use the normal distribution with E z σ Yes Is known? No c n Use the t-distribution with E tc s n and n – 1 degrees of freedom. 51 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) 52 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 53 Section 6.3 Confidence Intervals for Population Proportions 54 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 55 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 number in sample read as “p hat” 56 Point Estimate for Population p Estimate Population with Sample Parameter… Statistic Proportion: p p̂ Point Estimate for q, the proportion of failures • Denoted by qˆ 1 pˆ • Read as “q hat” 57 Example: Point Estimate for p In a survey of 1219 U.S. adults, 354 said that their favorite sport to watch is football. Find a point estimate for the population proportion of U.S. adults who say their favorite sport to watch is football. (Adapted from The Harris Poll) Solution: n = 1219 and x = 354 x 354 pˆ 0.290402 29.0% n 1219 58 Confidence Intervals for p Think of proportion as: x pˆ where x is binomial. n So the mean of p̂ is the mean of x divided by n, and std is std of x divided by n: Remember : the mean for the pˆ np p Binomial distribution is np n npq pq and variance is npq. 2 pˆ 2 n n Also, when we divide the data points by pq a number the new mean gets divided by pˆ n that number, and the variance gets divided 59 by the square of the number! Constructing Confidence Intervals for p In Words In Symbols 1. Identify the sample statistics n and x. 2. Find the point estimate p̂. 3. Verify that the sampling distribution of p̂ can be approximated by the normal distribution. 4. Find the critical value zc that corresponds to the given level of confidence c. pˆ x n npˆ 5, nqˆ 5 Use the Standard Normal Table 60 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 zc pq ˆˆ n Left endpoint: p̂ E Right endpoint: p̂ E Interval: pˆ E p pˆ E 61 Example: Confidence Interval for p In a survey of 1219 U.S. adults, 354 said that their favorite sport to watch is football. Construct a 95% confidence interval for the proportion of adults in the United States who say that their favorite sport to watch is football. Solution: Recall pˆ 0.290402 qˆ 1 pˆ 1 0.290402 0.709598 62 Solution: Confidence Interval for p • Verify the sampling distribution of p̂ can be approximated by the normal distribution npˆ 1219 0.290402 354 5 nqˆ 1219 0.709598 865 5 • Margin of error: E zc pq (0.290402) (0.709598) ˆˆ 1.96 0.025 n 1219 63 Solution: Confidence Interval for p • Confidence interval: Left Endpoint: pˆ E 0.29 0.025 0.265 Right Endpoint: pˆ E 0.29 0.025 0.315 0.265 < p < 0.315 64 Solution: Confidence Interval for p • 0.265 < p < 0.315 Point estimate 0.265 ( p̂ E 0.29 0.315 p̂ p̂ E • ) With 95% confidence, you can say that the proportion of adults who say football is their favorite sport is between 26.5% and 31.5%. 65 Confidence Interval for p using Ti-83 or 84: The same as: 0.265 < p < 0.315 Slide 6- 66 Sample Size • Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate p is 2 zc ˆ ˆ n pq E • This formula assumes you have an estimate for p̂ and qˆ . • If not, use pˆ 0.5 and qˆ 0.5. 67 Sample Size: E zc E 2 pq ˆˆ n 2 zc pq ˆˆ n 2 zc pq ˆˆ n 2 E Slide 6- 68 If there is no preliminary estimate for the population proportion, use : pˆ 0.5 and qˆ 0.5 The product of pq is the largest when p and q are both 0.5. y pq p q 1 y p(1 p) y p p Max at p=0.5, q=0.5 2 Slide 6- 69 Example: Sample Size You are running a political campaign and wish to estimate, with 95% confidence, the proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population. Find the minimum sample size needed if 1. no preliminary estimate is available. Solution: Because you do not have a preliminary estimate for p̂ use pˆ 0.5 and qˆ 0.5. 70 Solution: Sample Size • c = 0.95 zc = 1.96 2 E = 0.03 2 zc 1.96 ˆ ˆ (0.5)(0.5) n pq 1067.11 0.03 E Round up to the nearest whole number. With no preliminary estimate, the minimum sample size should be at least 1068 voters. 71 Example: Sample Size You are running a political campaign and wish to estimate, with 95% confidence, the proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population. Find the minimum sample size needed if 2. a preliminary estimate gives pˆ 0.31 . Solution: Use the preliminary estimate pˆ 0.31 qˆ 1 pˆ 1 0.31 0.69 72 Solution: Sample Size • c = 0.95 zc = 1.96 2 E = 0.03 2 zc 1.96 ˆ ˆ (0.31)(0.69) n pq 913.02 0.03 E 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. 73 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 74 Section 6.4 Confidence Intervals for Variance and Standard Deviation 75 Section 6.4 Objectives • Interpret the chi-square distribution and use a chisquare distribution table • Use the chi-square distribution to construct a confidence interval for the variance and standard deviation 76 The Chi-Square Distribution • The point estimate for 2 is s2 • The point estimate for is s • s2 is the most unbiased estimate for 2 Estimate Population with Sample Parameter… Statistic Variance: σ2 s2 Standard deviation: σ s 77 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 2 (n 1)s 2 σ2 forms a chi-square distribution for samples of any size n > 1. 78 The Chi-Square Distribution Slide 6- 79 Properties of The Chi-Square Distribution 1. All chi-square values χ2 are greater than or equal to zero. 2. The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for 2, use the χ2-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 chi-square distribution equals one. 80 Properties of The Chi-Square Distribution 4. Chi-square distributions are positively skewed. chi-square distributions 81 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. 1 c 2 c The area between the left and right critical values is c. 1 c 2 L2 R2 χ2 82 Example: Finding Critical Values for χ2 Find the critical values R2 and L2 for a 90% confidence interval when the sample size is 20. Solution: • d.f. = n – 1 = 20 – 1 = 19 d.f. • Each area in the table represents the region under the chi-square curve to the right of the critical value. 1 c 1 0.90 0.05 2 2 • Area to the right of χ2R = • Area to the right of 1 c 2 χ L= 2 1 0.90 0.95 2 83 Solution: Finding Critical Values for χ2 Table 6: χ2-Distribution R2 30.144 L2 10.117 90% of the area under the curve lies between 10.117 and 30.144 84 Slide 6- 85 Confidence Intervals for 2 and Confidence Interval for 2: 2 2 ( n 1) s ( n 1) s 2 • σ 2 R L2 Confidence Interval for : • (n 1)s 2 R2 σ Note : R2 L2 (n 1)s 2 R2 (n 1)s 2 L2 (n 1)s 2 L2 • The probability that the confidence intervals contain σ2 or σ is c. 86 Confidence Intervals for 2 and In Words In Symbols 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 s2. 4. Find the critical value χ2R and χ2L that correspond to the given level of confidence c. d.f. = n – 1 2 ( x x ) s2 n 1 Use Table 6 in Appendix B 87 Confidence Intervals for 2 and 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 (n 1)s 2 R2 (n 1)s 2 R2 σ2 σ (n 1)s 2 L2 (n 1)s 2 L2 88 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. 89 Solution: Constructing a Confidence Interval 1 c 1 0.99 0.005 R= 2 2 • Area to the right of χ2 • Area to the right of 1 c 2 χ L= 2 1 0.99 0.995 2 • The critical values are χ2R = 52.336 and χ2L = 13.121 90 Solution: Constructing a Confidence Interval Confidence Interval for 2: Left endpoint: (n 1)s 2 R2 (30 1)(1.20)2 0.80 52.336 2 2 ( n 1) s (30 1)(1.20) Right endpoint: 3.18 2 13.121 L 0.80 < σ2 < 3.18 With 99% confidence you can say that the population variance is between 0.80 and 3.18 milligrams. 91 Solution: Constructing a Confidence Interval Confidence Interval for : (n 1)s 2 2 R σ (n 1)s 2 2 L (30 1)(1.20)2 (30 1)(1.20)2 52.336 13.121 0.89 < σ < 1.78 With 99% confidence you can say that the population standard deviation is between 0.89 and1.78 milligrams. 92 Slide 6- 93 Section 6.4 Summary • Interpreted the chi-square distribution and used a chisquare distribution table • Used the chi-square distribution to construct a confidence interval for the variance and standard deviation 94 Control Charts http://www.learner.org/courses/againstallodds/unitpages/unit23.html Slide 6- 95 Chapter 6: Confidence Intervals Elementary Statistics: Picturing the World Fifth Edition by Larson and Farber Slide 4- 96 A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a point estimate for the mean price of all textbooks. A. $42 B. $12.30 C. $114.50 D. $2.73 Slide 6- 97 A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a point estimate for the mean price of all textbooks. A. $42 B. $12.30 x C. $114.50 D. $2.73 Slide 6- 98 Find the critical value zc necessary to form a 98% confidence interval. A. 2.33 B. 2.05 C. 0.5040 D. 0.8365 Slide 6- 99 Find the critical value zc necessary to form a 98% confidence interval. A. 2.33 B. 2.05 C. 0.5040 D. 0.8365 Slide 6- 100 A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a 98% confidence interval for the mean price of all textbooks. A. (111.38, 117.62) B. (110.08, 118.92) C. (110.78, 118.22) D. (109.61, 119.39) Slide 6- 101 A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a 98% confidence interval for the mean price of all textbooks. A. (111.38, 117.62) B. (110.08, 118.92) C. (110.78, 118.22) D. (109.61, 119.39) Slide 6- 102 Determine the minimum sample size needed to construct a 95% confidence interval for the mean age of employees at a company. The estimate must be accurate to within 0.5 year. Assume the standard deviation is 4.8 years. A. 18 B. 19 C. 354 D. 355 Slide 6- 103 Determine the minimum sample size needed to construct a 95% confidence interval for the mean age of employees at a company. The estimate must be accurate to within 0.5 year. Assume the standard deviation is 4.8 years. A. 18 B. 19 zc (1.96)(4.8) n 354.04 0.5 E 2 2 C. 354 D. 355 Slide 6- 104 Find the critical value tc necessary to form a 95% confidence interval with a sample size of 15. A. 1.960 B. 2.145 C. 2.131 D. 2.120 Slide 6- 105 Find the critical value tc necessary to form a 95% confidence interval with a sample size of 15. A. 1.960 B. 2.145 C. 2.131 D. 2.120 Slide 6- 106 A random sample of 15 DVD players has a mean price of $64.30 and a standard deviation of $5.60. Find a 95% confidence interval for the mean price of all DVD players. A. (61.20, 67.40) B. (61.47, 67.13) C. (61.22, 67.38) D. (61.10, 67.51) Slide 6- 107 A random sample of 15 DVD players has a mean price of $64.30 and a standard deviation of $5.60. Find a 95% confidence interval for the mean price of all DVD players. A. (61.20, 67.40) B. (61.47, 67.13) C. (61.22, 67.38) D. (61.10, 67.51) Slide 6- 108 In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a point estimate for the proportion of all Internet users who have high-speed Internet access at home. A. 1.28 B. 0.78 C. 0.22 D. 195 Slide 6- 109 In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a point estimate for the proportion of all Internet users who have high-speed Internet access at home. A. 1.28 B. 0.78 C. 0.22 195 pˆ p 0.78 250 D. 195 Slide 6- 110 In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a 90% confidence interval for the proportion of all Internet users who have high-speed Internet access at home. A. (1.19, 1.38) B. (0.728, 0.832) C. (0.731, 0.829) D. (0.737, 0.823) Slide 6- 111 In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a 90% confidence interval for the proportion of all Internet users who have high-speed Internet access at home. A. (1.19, 1.38) B. (0.728, 0.832) C. (0.731, 0.829) D. (0.737, 0.823) Slide 6- 112 You want to estimate, with 95% confidence, the proportion of households with pets. Your estimate must be accurate within 3% of the population proportion. No preliminary estimate is available. Find the minimum sample size needed. A. 1141 B. 3267 C. 1068 D. 1067 Slide 6- 113 You want to estimate, with 95% confidence, the proportion of households with pets. Your estimate must be accurate within 3% of the population proportion. No preliminary estimate is available. Find the minimum sample size needed. A. 1141 B. 3267 C. 1068 D. 1067 Let pˆ 0.5 and qˆ 0.5 2 2 zc 1.96 ˆ ˆ (0.5)(0.5) n pq 1067.11 0.03 E Slide 6- 114 Find the critical values and for a 98% confidence interval when the sample size is 25. 2 R A. R2 2.492 L2 2.492 2 B. R 2.326 L2 2.326 2 L 2 2 C. R 44.314 L 11.524 2 2 42 . 980 D. R L 10.856 Slide 6- 115 Find the critical values and for a 98% confidence interval when the sample size is 25. 2 R 2 L R2 has (1-0.98)/2 area to the right of it. L2 has 0.98 + (1-0.98)/2 to the right of it. Remember in both cases you are looking for areas to their right! Slide 6- 116

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