1. Confidence Intervals for the Mean (Large Samples) Larson/Farber 4th ed 1 Section 6.1

2. Section 6.1 Objectives Larson/Farber 4th ed 2 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

3. Point Estimate for Population μ Larson/Farber 4th ed 3 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

4. Example: Point Estimate for Population μ Larson/Farber 4th ed 4 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

5. Solution: Point Estimate for Population μ Larson/Farber 4th ed 5 The sample mean of the data is Your point estimate for the mean length of all magazine advertisements is 12.4 sentences.

6. Interval Estimate Larson/Farber 4th ed 6 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

7. Level of Confidence Larson/Farber 4th ed 7 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.

8. zczc Level of Confidence Larson/Farber 4th ed 8 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

9. Sampling Error Larson/Farber 4th ed 9 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

10. Margin of Error Larson/Farber 4th ed 10 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 .

11. Example: Finding the Margin of Error Larson/Farber 4th ed 11 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.

12. zczc Solution: Finding the Margin of Error Larson/Farber 4th ed 12 First find the critical values z zczc z = 0 0.95 0.025 -z c = -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.) z c = 1.96

13. Solution: Finding the Margin of Error Larson/Farber 4th ed 13 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.

14. Confidence Intervals for the Population Mean Larson/Farber 4th ed 14 A c-confidence interval for the population mean μ The probability that the confidence interval contains μ is c.

15. Constructing Confidence Intervals for μ Larson/Farber 4th ed 15 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 .

16. Constructing Confidence Intervals for μ Larson/Farber 4th ed 16 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

17. Example: Constructing a Confidence Interval Larson/Farber 4th ed 17 Construct a 95% confidence interval for the mean number of sentences in all magazine advertisements. Solution: Recall and E = 1.4 11.0 < μ < 13.8 Left Endpoint:Right Endpoint:

18. ( ) Solution: Constructing a Confidence Interval Larson/Farber 4th ed 18 11.0 < μ < 13.8 12.4 11.013.8 With 95% confidence, you can say that the population mean number of sentences is between 11.0 and 13.8.

19. Example: Constructing a Confidence Interval σ Known Larson/Farber 4th ed 19 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.

20. zczc Solution: Constructing a Confidence Interval σ Known Larson/Farber 4th ed 20 First find the critical values z z = 0zczc c = 0.90 ½(1 – c) = 0.05 -z c = -1.645 z c = 1.645

21. Solution: Constructing a Confidence Interval σ Known Larson/Farber 4th ed 21 Margin of error: Confidence interval: Left Endpoint:Right Endpoint: 22.3 < μ < 23.5

22. Solution: Constructing a Confidence Interval σ Known Larson/Farber 4th ed 22 22.3 < μ < 23.5 ( ) 22.9 22.323.5 With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years. Point estimate

23. Interpreting the Results Larson/Farber 4th ed 23 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.

24. Interpreting the Results Larson/Farber 4th ed 24 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

25. Sample Size Larson/Farber 4th ed 25 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.

26. Example: Sample Size Larson/Farber 4th ed 26 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.

27. zczc Solution: Sample Size Larson/Farber 4th ed 27 First find the critical values z c = 1.96 z z = 0zczc 0.95 0.025 -z c = -1.96 z c = 1.96

28. Solution: Sample Size Larson/Farber 4th ed 28 z c = 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.

29. Section 6.1 Summary Larson/Farber 4th ed 29 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