1. Section 6.1 Confidence Intervals for the Mean (Large Samples) © 2012 Pearson Education, Inc. All rights reserved. 1 of 83

2. 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. 2 of 83

3. 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 Estimate Population Parameter… with Sample Statistic Mean: μ © 2012 Pearson Education, Inc. All rights reserved. 3 of 83

4. 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, . (Source: Facebook) 140 105 130 97 80 165 232 110 214 201 122 98 65 88 154 133 121 82 130 211 153 114 58 77 51 247 236 109 126 132 125 149 122 74 59 218 192 90 117 105 © 2012 Pearson Education, Inc. All rights reserved. 4 of 83

5. 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 130.8 friends. © 2012 Pearson Education, Inc. All rights reserved. 5 of 83

6. What is a point estimate?? In your own words….What is a point estimate?

7. Try it Yourself 1 1621141318710863 249135172196127100 146214805571130 95156201227137125 14517974215137124

9. Interval Estimate Interval estimate An interval, or range of values, used to estimate a population parameter. Point estimate 12.4 How confident do we want to be that the interval estimate contains the population mean μ? ( ) Interval estimate © 2012 Pearson Education, Inc. All rights reserved. 9 of 83

10. Level of Confidence Level of confidence c The probability that the interval estimate contains the population parameter. z z = 0-zc-zc zczc Critical values ½(1 – c) c is the area under the standard normal curve between the critical values. The remaining area in the tails is 1 – c. c Use the Standard Normal Table to find the corresponding z-scores. © 2012 Pearson Education, Inc. All rights reserved. 10 of 83

11. zc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. z z = 0zczc The corresponding z-scores are +1.645. c = 0.90 ½(1 – c) = 0.05 -z c = -1.645 z c = 1.645 © 2012 Pearson Education, Inc. All rights reserved. 11 of 83

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 © 2012 Pearson Education, Inc. All rights reserved. 12 of 83

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 . © 2012 Pearson Education, Inc. All rights reserved. 13 of 83

14. 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. 14 of 83

15. zczc Solution: Finding the Margin of Error First find the critical values z zczc z = 0 -zc =-zc = 95% of the area under the standard normal curve falls within ________ 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.) zc =zc = © 2012 Pearson Education, Inc. All rights reserved. 15 of 83

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 16.4 friends. © 2012 Pearson Education, Inc. All rights reserved. 16 of 83

17. p. 306 Try it yourself 2

18. 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. 18 of 83

19. Constructing Confidence Intervals for μ Finding a Confidence Interval for a Population Mean (n  30 or σ known with a normally distributed population) In WordsIn Symbols 1.Find the sample statistics n and. 2.Specify , if known. Otherwise, if n  30, find the sample standard deviation s and use it as an estimate for . © 2012 Pearson Education, Inc. All rights reserved. 19 of 83

20. Constructing Confidence Intervals for μ 3.Find the critical value z c 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. Use the Standard Normal Table. Left endpoint: Right endpoint: Interval: In WordsIn Symbols © 2012 Pearson Education, Inc. All rights reserved. 20 of 83

21. 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: © 2012 Pearson Education, Inc. All rights reserved. 21 of 83

22. Solution: Constructing a Confidence Interval With 95% confidence, you can say that the population mean number of friends is between ____ and_____. © 2012 Pearson Education, Inc. All rights reserved. 22 of 83

23. p. 308 Try it Yourself 3

24. 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. 24 of 83

25. zczc Solution: Constructing a Confidence Interval σ Known First find the critical values z z = 0zczc c = ½(1 – c) = -z c = zc =zc = z c = 1.645 © 2012 Pearson Education, Inc. All rights reserved. 25 of 83

26. Margin of error: Confidence interval: Solution: Constructing a Confidence Interval σ Known Left Endpoint:Right Endpoint: © 2012 Pearson Education, Inc. All rights reserved. 26 of 83

27. Solution: Constructing a Confidence Interval σ Known ( ) With 90% confidence, you can say that the mean age of all the students is between _____ and ____ years. Point estimate © 2012 Pearson Education, Inc. All rights reserved. 27 of 83