1. McGraw-Hill/IrwinCopyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved. Confidence Intervals Chapter 8

2. 8-2 Confidence Intervals 8.1z-Based Confidence Intervals for a Population Mean: σ Known 8.2t-Based Confidence Intervals for a Population Mean: σ Unknown 8.3Sample Size Determination 8.4Confidence Intervals for a Population Proportion 8.5Confidence Intervals for Parameters of Finite Populations (Optional) 8.6A Comparison of Confidence Intervals and Tolerance Intervals (Optional)

3. 8-3 z-Based Confidence Intervals for a Mean: σ Known The starting point is the sampling distribution of the sample mean –Recall that if a population is normally distributed with mean  and standard deviation σ, then the sampling distribution of  is normal with mean   =  and standard deviation –Use a normal curve as a model of the sampling distribution of the sample mean Exactly, because the population is normal Approximately, by the Central Limit Theorem for large samples

4. 8-4 The Empirical Rule 68.26% of all possible sample means are within one standard deviation of the population mean 95.44% of all possible observed values of x are within two standard deviations of the population mean 99.73% of all possible observed values of x are within three standard deviations of the population mean

5. 8-5 Example 8.1: The Car Mileage Case Assume a sample size (n) of 5 Assume the population of all individual car mileages is normally distributed Assume the population standard deviation (σ) is 0.8 The probability that  with be within ±0.7 of µ is 0.9544

6. 8-6 Example 8.1 The Car Mileage Case Continued Assume the sample mean is 31.3 That gives us an interval of [31.3 ± 0.7] = [30.6, 32.0] The probability is 0.9544 that the interval [  ± 2σ] contains the population mean µ

7. 8-7 Example 8.1: The Car Mileage Case #3 That the sample mean is within ±0.7155 of µ is equivalent to…  will be such that the interval [  ± 0.7115] contains µ Then there is a 0.9544 probability that  will be a value so that interval [  ± 0.7115] contains µ –In other words P(  – 0.7155 ≤ µ ≤  + 0.7155) = 0.9544 –The interval [  ± 0.7115] is referred to as the 95.44% confidence interval for µ

8. 8-8 Example 8.1: The Car Mileage Case #4 Three 95.44% Confidence Intervals for  Thee intervals shown Two contain µ One does not

9. 8-9 Example 8.1: The Car Mileage Case #5 According to the 95.44% confidence interval, we know before we sample that of all the possible samples that could be selected … There is 95.44% probability the sample mean is such the interval [  ± 0.7155] will contain the actual (but unknown) population mean µ –In other words, of all possible sample means, 95.44% of all the resulting intervals will contain the population mean µ –Note that there is a 4.56% probability that the interval does not contain µ The sample mean is either too high or too low

10. 8-10 Generalizing In the example, we found the probability that  is contained in an interval of integer multiples of   More usual to specify the (integer) probability and find the corresponding number of   The probability that the confidence interval will not contain the population mean  is denoted by   –In the mileage example,  = 0.0456

11. 8-11 Generalizing Continued The probability that the confidence interval will contain the population mean  is denoted by  –1 –  is referred to as the confidence coefficient –(1 –  )  100% is called the confidence level Usual to use two decimal point probabilities for 1 –  –Here, focus on 1 –  = 0.95 or 0.99

12. 8-12 General Confidence Interval In general, the probability is 1 –  that the population mean  is contained in the interval –The normal point z  /2 gives a right hand tail area under the standard normal curve equal to  /2 –The normal point - z  /2 gives a left hand tail area under the standard normal curve equal to  /2 –The area under the standard normal curve between -z  /2 and z  /2 is 1 – 

13. 8-13 Sampling Distribution Of All Possible Sample Means

14. 8-14 z-Based Confidence Intervals for a Mean with σ Known If a population has standard deviation  (known), and if the population is normal or if sample size is large (n  30), then … … a (1-  )100% confidence interval for  is

15. 8-15 95% Confidence Level For a 95% confidence level, 1 –  = 0.95, so  = 0.05, and  /2 = 0.025 Need the normal point z 0.025 –The area under the standard normal curve between -z 0.025 and z 0.025 is 0.95 –Then the area under the standard normal curve between 0 and z 0.025 is 0.475 –From the standard normal table, the area is 0.475 for z = 1.96 –Then z 0.025 = 1.96

16. 8-16 95% Confidence Interval The 95% confidence interval is

17. 8-17 99% Confidence Interval For 99% confidence, need the normal point z 0.005 –Reading between table entries in the standard normal table, the area is 0.495 for z 0.005 = 2.575 The 99% confidence interval is

18. 8-18 The Effect of a on Confidence Interval Width z  /2 = z 0.025 = 1.96z  /2 = z 0.005 = 2.575

19. 8-19 Example 8.2: Car Mileage Case Given –  = 31.56 –σ = 0.8 –n = 50 Will calculate –95 Percent confidence interval –99 Percent confidence interval

20. 8-20 Example 8.2: 95 Percent Confidence Interval

21. 8-21 Example 8.2: 99 Percent Confidence Interval

22. 8-22 Notes on the Example The 99% confidence interval is slightly wider than the 95% confidence interval –The higher the confidence level, the wider the interval Reasoning from the intervals: –The target mean should be at least 31 mpg –Both confidence intervals exceed this target –According to the 95% confidence interval, we can be 95% confident that the mileage is between 31.33 and 31.78 mpg We can be 95% confident that, on average, the mean exceeds the target by at least 0.33 and at most 0.78 mpg

23. 8-23 t-Based Confidence Intervals for a Mean:  Unknown If  is unknown (which is usually the case), we can construct a confidence interval for  based on the sampling distribution of If the population is normal, then for any sample size n, this sampling distribution is called the t distribution

24. 8-24 The t Distribution The curve of the t distribution is similar to that of the standard normal curve –Symmetrical and bell-shaped –The t distribution is more spread out than the standard normal distribution –The spread of the t is given by the number of degrees of freedom Denoted by df For a sample of size n, there are one fewer degrees of freedom, that is, df = n – 1

25. 8-25 Degrees of Freedom and the t-Distribution As the number of degrees of freedom increases, the spread of the t distribution decreases and the t curve approaches the standard normal curve

26. 8-26 The t Distribution and Degrees of Freedom As the sample size n increases, the degrees of freedom also increases As the degrees of freedom increase, the spread of the t curve decreases As the degrees of freedom increases indefinitely, the t curve approaches the standard normal curve –If n ≥ 30, so df = n – 1 ≥ 29, the t curve is very similar to the standard normal curve

27. 8-27 t and Right Hand Tail Areas Use a t point denoted by t  –t  is the point on the horizontal axis under the t curve that gives a right hand tail equal to a –So the value of t  in a particular situation depends on the right hand tail area a and the number of degrees of freedom df = n – 1  = 1 – a, where 1 – a is the specified confidence coefficient

28. 8-28 t and Right Hand Tail Areas

29. 8-29 Using the t Distribution Table Rows correspond to the different values of df Columns correspond to different values of a See Table 8.3, Tables A.4 and A.20 in Appendix A and the table on the inside cover –Table 8.3 and A.4 gives t points for df 1 to 30, then for df = 40, 60, 120 and ∞ On the row for ∞, the t points are the z points –Table A.20 gives t points for df from 1 to 100 For df greater than 100, t points can be approximated by the corresponding z points on the bottom row for df = ∞ –Always look at the accompanying figure for guidance on how to use the table

30. 8-30 Using the t Distribution Example: Find t  for a sample of size n=15 and right hand tail area of 0.025 –For n = 15, df = 14 –  = 0.025 Note that a = 0.025 corresponds to a confidence level of 0.95 –In Table 8.3, along row labeled 14 and under column labeled 0.025, read a table entry of 2.145 –So t  = 2.145

31. 8-31 Using the t Distribution Continued

32. 8-32 t-Based Confidence Intervals for a Mean:  Unknown If the sampled population is normally distributed with mean , then a (1­  )100% confidence interval for  is t  /2 is the t point giving a right-hand tail area of  /2 under the t curve having n­1 degrees of freedom

33. 8-33 Example 8.4 Debt-to-Equity Ratios Estimate the mean debt-to-equity ratio of the loan portfolio of a bank Select a random sample of 15 commercial loan accounts –  = 1.34 – s = 0.192 – n = 15 Want a 95% confidence interval for the ratio Assume all ratios are normally distributed but σ unknown

34. 8-34 Example 8.4 Debt-to-Equity Ratios Continued Have to use the t distribution At 95% confidence, 1 –  = 0.95 so  = 0.05 and  /2 = 0.025 For n = 15, df = 15 – 1 = 14 Use the t table to find t  /2 for df = 14, t  /2 = t 0.025 = 2.145 The 95% confidence interval:

35. 8-35 Sample Size Determination (z) If  is known, then a sample of size so that  is within B units of , with 100(1-  )% confidence

36. 8-36 Sample Size Determination (t) If σ is unknown and is estimated from s, then a sample of size so that  is within B units of , with 100(1-  )% confidence. The number of degrees of freedom for the t  /2 point is the size of the preliminary sample minus 1

37. 8-37 Example 8.6: The Car Mileage Case What sample size is needed to make the margin of error for a 95 percent confidence interval equal to 0.3? Assume we do not know σ Take prior example as preliminary sample –s = 0.7583 –z  /2 = z 0.025 = 1.96 –t  /2 = t 0.025 = 2.776 based on n-1 = 4 d.f.

38. 8-38 Example 8.6: The Car Mileage Case Continued

39. 8-39 Example 8.7: The Car Mileage Case Want to see that the sample of 50 mileages produced a 95 percent confidence interval with a margin or error of ±0.3 The margin of error is 0.227, which is smaller than the 0.3 desired