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Confidence Intervals

Chapter 6. Confidence Intervals. Confidence Intervals for the Mean (Large Samples). § 6.1.   74.22. Point Estimate for Population μ.

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Confidence Intervals

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  1. Chapter 6 Confidence Intervals

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

  3.  74.22 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 thesample 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.

  4. Point estimate for textbooks • 74.22 interval estimate 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?

  5. c (1 – c) (1 – c) z zc z = 0 zc Critical values 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. Use the Standard Normal Table to find the corresponding z-scores. The remaining area in the tails is 1 – c .

  6. 0.90 0.05 0.05 z z = 0 zc zc 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. zc= 1.645 zc=  1.645 The corresponding z-scores are ± 1.645.

  7. 0.95 0.025 0.025 z z = 0 zc zc 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. zc= 1.96 zc=  1.96 The corresponding z-scores are ± 1.96.

  8. 0.99 0.005 0.005 z z = 0 zc zc 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. zc= 2.575 zc=  2.575 The corresponding z-scores are ± 2.575.

  9. When n 30, the sample standard deviation, s, can be used for . Margin of Error The difference between the point estimate and the actual population parameter value is called the sampling error. When μ is estimated, the sampling error is the difference μ– . Since μis usually unknown, the maximum value for the error can be calculated using the level of confidence. 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.

  10. Since n 30, s can be substituted for σ. 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 iss = 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.

  11. Confidence Intervals for μ A c-confidence interval for the population mean μ is   E < μ< +E. The probability that the confidence interval contains μis c. 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 iss = 23.44, and the margin of error is E = 8.12. Construct a 95% confidence interval for the mean price of all textbooks in the bookstore. Continued.

  12. Left endpoint = ? Right endpoint = ? • • •  = 74.22 Confidence Intervals for μ Example continued: Construct a 95% confidence interval for the mean price of all textbooks in the bookstore. = 74.22 s = 23.44 E= 8.12   E = 74.22 – 8.12  + E = 74.22 + 8.12 = 66.1 = 82.34 With 95% confidence we can say that the cost for all textbooks in the bookstore is between $66.10 and $82.34.

  13. Finding 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 . • 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. Left endpoint: ERight endpoint: +E Interval: E < μ< +E

  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 = 2.86  =0.15 zc = 1.645 2.81 < μ< 2.91  ± E = 2.86 ± 0.05 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? = 74.22   s = 23.44 zc = 2.575 (Always round up.) You should include at least 146 books in your sample.

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