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Chapter 6

Chapter 6. Confidence Intervals. 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

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Chapter 6

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  1. Chapter 6 Confidence Intervals Larson/Farber 4th ed

  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 (Not part of our syllabus) Larson/Farber 4th ed

  3. Section 6.1 Confidence Intervals for the Mean (Large Samples) Larson/Farber 4th ed

  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 Larson/Farber 4th ed

  5. 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 Larson/Farber 4th ed

  6. 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 Larson/Farber 4th ed

  7. Solution: Point Estimate for Population μ The sample mean of the data is Your point estimate for the mean length of all magazine advertisements is 12.4 sentences. Larson/Farber 4th ed

  8. Point estimate 12.4 • Interval Estimate Interval estimate • An interval, or range of values, used to estimate a population parameter. ( ) Interval estimate How confident do we want to be that the interval estimate contains the population mean μ? Larson/Farber 4th ed

  9. c ½(1 – c) ½(1 – c) z -zc z = 0 zc Critical values 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. Use the Standard Normal Table to find the corresponding z-scores. The remaining area in the tails is 1 – c . Larson/Farber 4th ed

  10. c = 0.90 ½(1 – c) = 0.05 ½(1 – c) = 0.05 z z = 0 zc 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. zc zc= 1.645 -zc= -1.645 The corresponding z-scores are +1.645. Larson/Farber 4th ed

  11. c = 0.99 ½(1 – c) = 0.005 ½(1 – c) = 0.005 z z = 0 zc Level of Confidence Find the z-score corresponding to level of confidence of 99%. You can use calculator to find the z-score by using Invnorm function we use to use in section 5.3 Invnorm(0.005) = 2.575. We use always + z-score. So the z-score for 99% of confidence level is 2.575 -zc Larson/Farber 4th ed

  12. 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 Larson/Farber 4th ed

  13. 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 . Larson/Farber 4th ed

  14. 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. Larson/Farber 4th ed

  15. 0.95 0.025 0.025 z z = 0 zc Solution: Finding the Margin of Error • First find the critical values zc zc= 1.96 -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 ≥ 30.) Larson/Farber 4th ed

  16. 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 1.4 sentences. Larson/Farber 4th ed

  17. Page 311 Exercise 12: Find the sampling error. The diagram shows the value of So the sampling error Interpretation: The population mean was under estimated by 1.56 points by the sample mean (the point estimate). Exercise 16: Find the margin of error. c = 0.975, s = 4.6 and n = 100 The margin error E = Larson/Farber 4th ed

  18. Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ • The probability that the confidence interval contains μis c. Larson/Farber 4th ed

  19. Constructing Confidence Intervals for μ Finding a Confidence Interval for a Population Mean (n 30 or σknown with a normally distributed population) In Words In Symbols • Find the sample statistics n and. • Specify , if known. Otherwise, if n  30, find the sample standard deviation s and use it as an estimate for . Larson/Farber 4th ed

  20. Constructing Confidence Intervals for μ In Words In Symbols Use the Standard Normal Table. • 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. Left endpoint: Right endpoint:Interval: Larson/Farber 4th ed

  21. Example: Constructing a Confidence Interval Construct a 95% confidence interval for the mean number of sentences in all magazine advertisements. Solution: Recall and E = 1.4 Left Endpoint: Right Endpoint: 11.0 < μ < 13.8 Larson/Farber 4th ed

  22. Solution: Constructing a Confidence Interval 11.0 < μ < 13.8 11.0 13.8 12.4 • ( ) With 95% confidence, you can say that the population mean number of sentences is between 11.0 and 13.8. Larson/Farber 4th ed

  23. Example: Constructing a Confidence Interval σKnown 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. Larson/Farber 4th ed

  24. c = 0.90 ½(1 – c) = 0.05 ½(1 – c) = 0.05 z z = 0 zc Solution: Constructing a Confidence Interval σKnown • First find the critical values zc= 1.645 zc -zc= -1.645 zc= 1.645 Larson/Farber 4th ed

  25. Solution: Constructing a Confidence Interval σKnown • Margin of error: • Confidence interval: Left Endpoint: Right Endpoint: 22.3 < μ < 23.5 Larson/Farber 4th ed

  26. Solution: Constructing a Confidence Interval σKnown 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. Larson/Farber 4th ed

  27. 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. Larson/Farber 4th ed

  28. 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 Larson/Farber 4th ed

  29. Exercise: Construct a Confidence Interval To construct a confidence interval we need to know sample mean and the margin of error. So the confidence interval is = 13.5 – 0.39 <  < 13.5 + 0.39 = 13.11 <  < 13.89 Interpretation: If large number of samples are collected and a confidence interval is built for each sample then 99% of them will contain the mean between 13.11 and 13.89 Larson/Farber 4th ed

  30. Page 312 Exercise 40: Repair costs of Refrigerators Construct a 99% C.I for the mean repair cost of refrigerators. To calculate the confidence interval we need to know So the confidence interval is 150 – 5.15 <  < 150 + 5.15 144.85 <  < 155.15 So if large number of samples of size 60 are collected from the population and a confidence interval is built for each one of them then 99% of them will have the mean repair of cost of refrigerators between $144.85 and $155.15 Larson/Farber 4th ed

  31. 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 • If  is unknown, you can estimate it using s provided you have a preliminary sample with at least 30 members. Larson/Farber 4th ed

  32. 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. Larson/Farber 4th ed

  33. 0.95 0.025 0.025 z z = 0 zc Solution: Sample Size • First find the critical values zc= 1.96 zc -zc= -1.96 zc= 1.96 Larson/Farber 4th ed

  34. Solution: Sample Size zc = 1.96   s = 5.0 E = 1 When necessary, round up to obtain a whole number. You should include at least 97 magazine advertisements in your sample. Larson/Farber 4th ed

  35. 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 Note: Try to think about the questions in Exercise 49 on page 319. Larson/Farber 4th ed

  36. Section 6.2 Confidence Intervals for the Mean (Small Samples) Larson/Farber 4th ed

  37. 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 Larson/Farber 4th ed

  38. 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. Larson/Farber 4th ed

  39. 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 – 1 Degrees of freedom Larson/Farber 4th ed

  40. d.f. = 2 d.f. = 5 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. The tails in the t-distribution are “thicker” than those in the standard normal distribution. t 0 Standard normal curve Larson/Farber 4th ed

  41. 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 Larson/Farber 4th ed

  42. c = 0.95 t -tc = -2.145 tc = 2.145 Solution: Critical Values of t 95% of the area under the t-distribution curve with 14 degrees of freedom lies between t = +2.145. Larson/Farber 4th ed

  43. Critical values of t-distribution using Graphing Calculator • Critical values for 95% confidence level with 14 degrees of freedom. • Since the area in the middle of curve is 0.95 the area in each of the two far tails of the curve is (1 - 0.95)  2 = 0.025 • Now press 2nd key and the VARS key and use item invT • Critical value tc = invT(0.025, 14) = -2.145 and + 2.145 • Critical values for 90% confidence level with n = 20. • Since n = 20 the degrees of freedom is 19. • c = 0.90, so the area in each tail is (1 – 0.90)  2 = 0.05 • Critical value tc = invT(0.05, 19) = -1.729 and + 1.729 Larson/Farber 4th ed

  44. Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ • The probability that the confidence interval contains μis c. Larson/Farber 4th ed

  45. 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 Larson/Farber 4th ed

  46. Confidence Intervals and t-Distributions In Words In Symbols Left endpoint: Right endpoint: Interval: Find the left and right endpoints and form the confidence interval. Larson/Farber 4th ed

  47. 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.) Larson/Farber 4th ed

  48. 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 Larson/Farber 4th ed

  49. Solution: Constructing a Confidence Interval • Margin of error: • Confidence interval: Left Endpoint: Right Endpoint: 156.7 < μ < 167.3 Larson/Farber 4th ed

  50. 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 mean temperature of coffee sold is between 156.7ºF and 167.3ºF. Larson/Farber 4th ed

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