This document discusses confidence intervals for population means and proportions. It explains how to construct confidence intervals using the normal distribution for large sample sizes (n ≥ 30) and the t-distribution for small sample sizes. Formulas are provided for calculating margin of error and determining necessary sample size. Guidelines are given for determining whether to use the normal or t-distribution based on sample size and characteristics. Confidence intervals can be constructed for variance and standard deviation using the chi-square distribution.

1. Confidence Intervals Chapter 6

2. § 6.1 Confidence Intervals for the Mean (Large Samples)

3. Point Estimate for Population μ A point estimate is a single value estimate for a population parameter. The most unbiased point estimate of the population mean,  , is the sample mean, Example : A random sample of 32 textbook prices (rounded to the nearest dollar) is taken from a local college bookstore. Find a point estimate for the population mean,  . The point estimate for the population mean of textbooks in the bookstore is $74.22. 101 88 67 45 110 90 68 45 121 98 98 96 95 94 90 87 87 87 86 79 74 67 66 65 65 56 54 45 45 38 34 34

4. Interval Estimate An interval estimate is an interval, or range of values, used to estimate a population parameter. How confident do we want to be that the interval estimate contains the population mean, μ ? Point estimate for textbooks • 74.22 interval estimate

5. Level of Confidence The level of confidence c is the probability that the interval estimate contains the population parameter. c is the area beneath the normal curve between the critical values. The remaining area in the tails is 1 – c . Use the Standard Normal Table to find the corresponding z - scores. z z = 0  z c z c Critical values c (1 – c ) (1 – c )

6. Common Levels of Confidence If the level of confidence is 90%, this means that we are 90% confident that the interval contains the population mean, μ . The corresponding z - scores are ± 1.645.  z c =  1.645 z c = 1.645 z z = 0  z c z c 0.90 0.05 0.05

7. Common Levels of Confidence If the level of confidence is 95%, this means that we are 95% confident that the interval contains the population mean, μ . The corresponding z - scores are ± 1.96.  z c =  1.96 z c = 1.96 z z = 0  z c z c 0.95 0.025 0.025

8. Common Levels of Confidence If the level of confidence is 99%, this means that we are 99% confident that the interval contains the population mean, μ . The corresponding z - scores are ± 2.575.  z c =  2.575 z c = 2.575 z z = 0  z c z c 0.99 0.005 0.005

9. Margin of Error The difference between the point estimate and the actual population parameter value is called the sampling error . Given a level of confidence, the margin of error (sometimes called the maximum error of estimate or error tolerance) E is the greatest possible distance between the point estimate and the value of the parameter it is estimating. When n  30, the sample standard deviation, s , can be used for  . When μ is estimated, the sampling error is the difference μ – . Since μ is usually unknown, the maximum value for the error can be calculated using the level of confidence.

10. Margin of Error Example : A random sample of 32 textbook prices is taken from a local college bookstore. The mean of the sample is = 74.22, and the sample standard deviation is s = 23.44. Use a 95% confidence level and find the margin of error for the mean price of all textbooks in the bookstore. We are 95% confident that the margin of error for the population mean (all the textbooks in the bookstore) is about $8.12. Since n  30, s can be substituted for σ.

11. Confidence Intervals for μ A c - confidence interval for the population mean μ is The probability that the confidence interval contains μ is c . Continued. Construct a 95% confidence interval for the mean price of all textbooks in the bookstore. Example : A random sample of 32 textbook prices is taken from a local college bookstore. The mean of the sample is = 74.22, the sample standard deviation is s = 23.44, and the margin of error is E = 8.12.

12. Confidence Intervals for μ Example continued : Construct a 95% confidence interval for the mean price of all textbooks in the bookstore. s = 23.44 E = 8.12 With 95% confidence we can say that the cost for all textbooks in the bookstore is between $66.10 and $82.34. = 66.1 = 82.34 Left endpoint = ? Right endpoint = ? • • •

14. Confidence Intervals for μ (  Known) Example : A random sample of 25 students had a grade point average with a mean of 2.86. Past studies have shown that the standard deviation is 0.15 and the population is normally distributed. Construct a 90% confidence interval for the population mean grade point average. n = 25  = 0.15 z c = 1.645 2.81 < μ < 2.91 With 90% confidence we can say that the mean grade point average for all students in the population is between 2.81 and 2.91.

15. Sample Size Given a c - confidence level and a maximum error of estimate, 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. Example : You want to estimate the mean price of all the textbooks in the college bookstore. How many books must be included in your sample if you want to be 99% confident that the sample mean is within $5 of the population mean? Continued.

16. Sample Size Example continued : You want to estimate the mean price of all the textbooks in the college bookstore. How many books must be included in your sample if you want to be 99% confident that the sample mean is within $5 of the population mean?   s = 23.44 z c = 2.575 You should include at least 146 books in your sample. (Always round up .)

17. § 6.2 Confidence Intervals for the Mean (Small Samples)

20. Critical Values of t Example : Find the critical value t c for a 95% confidence when the sample size is 5. Appendix B: Table 5: t -Distribution Continued. d.f. = n – 1 = 5 – 1 = 4 c = 0.95 t c = 2.776 3.365 2.571 2.015 1.476 .727 5 3.747 2.776 2.132 1.533 .741 4 4.541 3.182 2.353 1.638 .765 3 6.965 4.303 2.920 1.886 .816 2 31.821 12.706 6.314 3.078 1.000 1 0.02 0.05 0.10 0.20 0.50 Two tails,  d.f. 0.01 0.025 0.05 0.10 0.25 One tail,  0.98 0.95 0.90 0.80 0.50 Level of confidence, c

21. Critical Values of t Example continued : Find the critical value t c for a 95% confidence when the sample size is 5. 95% of the area under the t -distribution curve with 4 degrees of freedom lies between t = ±2.776. t  t c =  2.776 t c = 2.776 c = 0.95

23. Constructing a Confidence Interval Example : In a random sample of 20 customers at a local fast food restaurant, the mean waiting time to order is 95 seconds, and the standard deviation is 21 seconds. Assume the wait times are normally distributed and construct a 90% confidence interval for the mean wait time of all customers. We are 90% confident that the mean wait time for all customers is between 86.9 and 103.1 seconds. s = 21 t c = 1.729 n = 20 d.f. = 19 86.9 < μ < 103.1

24. Normal or t -Distribution? Is n  30? Is the population normally, or approximately normally, distributed? You cannot use the normal distribution or the t -distribution. Is  known? No Yes No Use the normal distribution with If  is unknown, use s instead. Yes No Use the normal distribution with Yes Use the t - distribution with and n – 1 degrees of freedom.

25. Normal or t -Distribution? Example : Determine whether to use the normal distribution, the t - distribution, or neither. a.) n = 50, the distribution is skewed, s = 2.5 The normal distribution would be used because the sample size is 50. b.) n = 25, the distribution is skewed, s = 52.9 Neither distribution would be used because n < 30 and the distribution is skewed. c.) n = 25, the distribution is normal,  = 4.12 The normal distribution would be used because although n < 30, the population standard deviation is known.

26. § 6.3 Confidence Intervals for Population Proportions

27. Point Estimate for Population p The probability of success in a single trial of a binomial experiment is p . This probability is a population proportion . The point estimate for p , the population proportion of successes, is given by the proportion of successes in a sample and is denoted by where x is the number of successes in the sample and n is the number in the sample. The point estimate for the proportion of failures is = 1 – The symbols and are read as “ p hat” and “ q hat.”

28. Point Estimate for Population p Example : In a survey of 1250 US adults, 450 of them said that their favorite sport to watch is baseball. Find a point estimate for the population proportion of US adults who say their favorite sport to watch is baseball. The point estimate for the proportion of US adults who say baseball is their favorite sport to watch is 0.36, or 36%. n = 1250 x = 450

29. Confidence Intervals for p A c - confidence interval for the population proportion p is where The probability that the confidence interval contains p is c . Example : Construct a 90% confidence interval for the proportion of US adults who say baseball is their favorite sport to watch. Continued. n = 1250 x = 450

30. Confidence Intervals for p Example continued : With 90% confidence we can say that the proportion of all US adults who say baseball is their favorite sport to watch is between 33.8% and 38.2%. n = 1250 x = 450 Left endpoint = ? Right endpoint = ? • • •

32. 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 Example : You wish to find out, with 95% confidence and within 2% of the true population, the proportion of US adults who say that baseball is their favorite sport to watch. Continued.

33. Sample Size You should sample at least 2213 adults to be 95% confident. (Always round up .) Example continued : You wish to find out, with 95% confidence and within 2% of the true population, the proportion of US adults who say that baseball is their favorite sport to watch. n = 1250 x = 450

34. § 6.4 Confidence Intervals for Variance and Standard Deviation

35. The Chi - Square Distribution The point estimate for  2 is s 2 , and the point estimate for  is s . s 2 is the most unbiased estimate for  2 . 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.

37. Critical Values for X 2 There are two critical values for each level of confidence. The value χ 2 R represents the right - tail critical value and χ 2 L represents the left - tail critical value. The area between the left and right critical values is c . X 2 X 2 R Area to the right of X 2 R X 2 X 2 L Area to the right of X 2 L X 2 X 2 R X 2 L c

38. Critical Values for X 2 Example : Find the critical values χ 2 R and χ 2 L for an 80% confidence when the sample size is 18. Continued. Because the sample size is 18, there are d.f. = n – 1 = 18 – 1 = 17 degrees of freedom, Use the Chi - square distribution table to find the critical values. Area to the right of χ 2 R = Area to the right of χ 2 L =

39. Critical Values for X 2 Example continued : Appendix B: Table 6: χ 2 - Distribution χ 2 R = 24.769 χ 2 L = 10.085 6.251 4.605 2.706 0.10 0.216 0.051 0.001 0.975 7.815 0.584 0.352 0.115 0.072 3 5.991 0.211 0.103 0.020 0.010 2 3.841 0.016 0.004 - - 1 0.05 0.90 0.95 0.99 0.995 freedom  Degrees of 25.989 24.769 23.542 8.231 7.564 6.908 28.869 10.865 9.390 7.015 6.265 18 27.587 10.085 8.672 6.408 5.697 17 26.296 9.312 7.962 5.812 5.142 16

40. Confidence Intervals for  2 and  A c - confidence interval for a population variance and standard deviation is as follows. The probability that the confidence intervals contain  2 or  is c . Confidence Interval for  2 : Confidence Interval for  :

43. Constructing a Confidence Interval Example : You randomly select and weigh 41 samples of 16-ounce bags of potato chips. The sample standard deviation is 0.05 ounce. Assuming the weights are normally distributed, construct a 90% confidence interval for the population standard deviation. d.f. = n – 1 = 41 – 1 = 40 degrees of freedom, The critical values are χ 2 R = 55.758 and χ 2 L = 26.509. Continued. Area to the right of χ 2 R = Area to the right of χ 2 L =

44. Constructing a Confidence Interval Example continued : χ 2 R = 55.758 χ 2 L = 26.509 With 90% confidence we can say that the population standard deviation is between 0.04 and 0.06 ounces. Left endpoint = ? Right endpoint = ? • •