1. M32 Margin of Error 1  Department of ISM, University of Alabama, 1995-2002 Margin of Error Whose error is it? Why do we care?

2. M32 Margin of Error 2  Department of ISM, University of Alabama, 1995-2002 Lesson Objectives  Learn the meaning of “margin of error,” or “m.o.e.”  Learn how to calculate the m.o.e. for two situations: when the true population std. dev., , is known and when its unknown.  Learn how to use the m.o.e. to construct a confidence interval.

3. M32 Margin of Error 3  Department of ISM, University of Alabama, 1995-2002 “Margin of Error” for estimating the True Mean of a population. m.o.e. at 95% confidence = “The amount that when added and subtracted to the true population mean will define a region that will include the middle 95% of all possible X-bar values.”

4. M32 Margin of Error 4  Department of ISM, University of Alabama, 1995-2002  = true population mean. Find the interval around the mean where 95% of all possible sample means will lie.  X.95 ? ? -1.96 0 +1.96 Z.025 Illustration of “Margin of Error”

5. M32 Margin of Error 5  Department of ISM, University of Alabama, 1995-2002 ? -  Z =  n = ? -   n ? =  +  n Margin of Error

6. M32 Margin of Error 6  Department of ISM, University of Alabama, 1995-2002 1.96  n  x standard error of the mean Margin of Error for 95% confidence: m.o.e. =

7. M32 Margin of Error 7  Department of ISM, University of Alabama, 1995-2002 General form for “margin of error” when  is known: m.o.e. = Z  n  2 Z  2 appropriate percentile from the standard normal distribution, i.e., the Z table. where is the

8. M32 Margin of Error 8  Department of ISM, University of Alabama, 1995-2002 Explanation of symbol: Z ~ N(0,1) / 2/ 2 0 Z / 2Z / 2 Z   / 2 cuts off the top tail at area =  /2

9. M32 Margin of Error 9  Department of ISM, University of Alabama, 1995-2002 Amount of confidence Area in each tail Table value.95.90.80.98.0250 ? ?.0100 1.96 ? ? 2.33 1 -  / 2 / 2 Z  / 2 Examples

10. M32 Margin of Error 10  Department of ISM, University of Alabama, 1995-2002 (1-  )100% Confidence Interval: Point estimate  m.o.e. (1-  )100% is the amount of confidence desired..95 confidence  =.05 risk

11. M32 Margin of Error 11  Department of ISM, University of Alabama, 1995-2002 m.o.e. = Z  n  2 Estimate with 98% confidence the mean gallons of water used per shower for Dallas Cowboys after a game if the true standard deviation is known to be 10 gallons. The sample mean for 16 showers is 30.00 gal. Example 1

12. M32 Margin of Error 12  Department of ISM, University of Alabama, 1995-2002 Point estimate  m.o.e. Example 1, continued... 98% confidence interval:

13. M32 Margin of Error 13  Department of ISM, University of Alabama, 1995-2002 I am 95% confident that the true mean gallons of water used per shower for Dallas Cowboys after a game will fall within the interval 24.175 to 35.825 gallons. Example 1, Statement in the L.O.P. A statement in L.O.P. must contain four parts: 1. amount of confidence. 1. amount of confidence. 2. the parameter being estimated in L.O.P. 2. the parameter being estimated in L.O.P. 3. the population to which we generalize in L.O.P. 3. the population to which we generalize in L.O.P. 4. the calculated interval. 4. the calculated interval.

14. M32 Margin of Error 14  Department of ISM, University of Alabama, 1995-2002 Replace  with. replace z with. What do we do if the true population standard deviation  is unknown?

15. M32 Margin of Error 15  Department of ISM, University of Alabama, 1995-2002 General form for “margin of error” when  is UN-known: m.o.e. = t s n  2,n–1 s x “estimated” standard error of the mean t  2,n–1 where is the appropriate percentile from the t-distribution.

16. M32 Margin of Error 16  Department of ISM, University of Alabama, 1995-2002 Explanation of symbol: t-distribution / 2/ 2 0 t  / 2, n–1 t   / 2, n-1 cuts off the top tail at area =  /2 Use the t-table to find the value.

17. M32 Margin of Error 17  Department of ISM, University of Alabama, 1995-2002 t-dist.Z Bell shaped, symmetric Mean Std. dev. As “n–1” increases, “t n–1 ” approaches “Z” Degrees of freedom = 0 > 1 Yes = 0 = 1 “n-1” “”“” Comparison of “Z” and “t”

18. 0 t Table gives right-tail area. (e.g., for a right-tail area of 0.025 and d.f. = 15, the t value is 2.131.) 0.1 0.05 0.025 0.01 0.005 d.f. = 13.0786.31412.70631.821 63.656 21.8862.9204.3036.9659.925 31.6382.3533.1824.5415.841 41.5332.1322.7763.7474.604 51.4762.0152.5713.3654.032 61.4401.9432.4473.1433.707 71.4151.8952.3652.9983.499 81.3971.8602.3062.8963.355 91.3831.8332.2622.8213.250 101.3721.8122.2282.7643.169 111.3631.7962.2012.7183.106 121.3561.7822.1792.6813.055 131.3501.7712.1602.6503.012 141.3451.7612.1452.6242.977 151.3411.7532.1312.6022.947 161.3371.7462.1202.5832.921 171.3331.7402.1102.5672.898 181.3301.7342.1012.5522.878 191.3281.7292.0932.5392.861 201.3251.7252.0862.5282.845 211.3231.7212.0802.5182.831 221.3211.7172.0742.5082.819 231.3191.7142.0692.5002.807 241.3181.7112.0642.4922.797 251.3161.7082.0602.4852.787 261.3151.7062.0562.4792.779 271.3141.7032.0522.4732.771 281.3131.7012.0482.4672.763 291.3111.6992.0452.4622.756 301.3101.6972.0422.4572.750 311.3091.6962.0402.4532.744 321.3091.6942.0372.4492.738 331.3081.6922.0352.4452.733 341.3071.6912.0322.4412.728 351.3061.6902.0302.4382.724 0.1 0.05 0.025 0.01 0.005 361.3061.6882.0282.4342.719 371.3051.6872.0262.4312.715 381.3041.6862.0242.4292.712 391.3041.6852.0232.4262.708 401.3031.6842.0212.4232.704 411.3031.6832.0202.4212.701 421.3021.6822.0182.4182.698 431.3021.6812.0172.4162.695 441.3011.6802.0152.4142.692 451.3011.6792.0142.4122.690 971.2901.6611.9852.3652.627 981.2901.6611.9842.3652.627 991.2901.6601.9842.3652.626 1001.2901.6601.9842.3642.626 1.2821.6451.9602.3262.576 Want 95% CI, n = 20,  /2 =.025 d.f. =19 t.025, 19 = 2.093

19. 0 t 0.1 0.05 0.025 0.01 0.005 d.f. = 13.0786.31412.70631.821 63.656 21.8862.9204.3036.9659.925 31.6382.3533.1824.5415.841 41.5332.1322.7763.7474.604 51.4762.0152.5713.3654.032 61.4401.9432.4473.1433.707 71.4151.8952.3652.9983.499 81.3971.8602.3062.8963.355 91.3831.8332.2622.8213.250 101.3721.8122.2282.7643.169 111.3631.7962.2012.7183.106 121.3561.7822.1792.6813.055 131.3501.7712.1602.6503.012 141.3451.7612.1452.6242.977 151.3411.7532.1312.6022.947 161.3371.7462.1202.5832.921 171.3331.7402.1102.5672.898 181.3301.7342.1012.5522.878 191.3281.7292.0932.5392.861 201.3251.7252.0862.5282.845 211.3231.7212.0802.5182.831 221.3211.7172.0742.5082.819 231.3191.7142.0692.5002.807 241.3181.7112.0642.4922.797 251.3161.7082.0602.4852.787 261.3151.7062.0562.4792.779 271.3141.7032.0522.4732.771 281.3131.7012.0482.4672.763 291.3111.6992.0452.4622.756 301.3101.6972.0422.4572.750 311.3091.6962.0402.4532.744 321.3091.6942.0372.4492.738 331.3081.6922.0352.4452.733 341.3071.6912.0322.4412.728 351.3061.6902.0302.4382.724 0.1 0.05 0.025 0.01 0.005 361.3061.6882.0282.4342.719 371.3051.6872.0262.4312.715 381.3041.6862.0242.4292.712 391.3041.6852.0232.4262.708 401.3031.6842.0212.4232.704 411.3031.6832.0202.4212.701 421.3021.6822.0182.4182.698 431.3021.6812.0172.4162.695 441.3011.6802.0152.4142.692 451.3011.6792.0142.4122.690 971.2901.6611.9852.3652.627 981.2901.6611.9842.3652.627 991.2901.6601.9842.3652.626 1001.2901.6601.9842.3642.626 1.2821.6451.9602.3262.576 Want 98% CI, n = 33,  /2 = d.f. = t, = Table gives right-tail area. (e.g., for a right-tail area of 0.025 and d.f. = 15, the t value is 2.131.)

20. 0 t 0.1 0.05 0.025 0.01 0.005 d.f. = 13.0786.31412.70631.821 63.656 21.8862.9204.3036.9659.925 31.6382.3533.1824.5415.841 41.5332.1322.7763.7474.604 51.4762.0152.5713.3654.032 61.4401.9432.4473.1433.707 71.4151.8952.3652.9983.499 81.3971.8602.3062.8963.355 91.3831.8332.2622.8213.250 101.3721.8122.2282.7643.169 111.3631.7962.2012.7183.106 121.3561.7822.1792.6813.055 131.3501.7712.1602.6503.012 141.3451.7612.1452.6242.977 151.3411.7532.1312.6022.947 161.3371.7462.1202.5832.921 171.3331.7402.1102.5672.898 181.3301.7342.1012.5522.878 191.3281.7292.0932.5392.861 201.3251.7252.0862.5282.845 211.3231.7212.0802.5182.831 221.3211.7172.0742.5082.819 231.3191.7142.0692.5002.807 241.3181.7112.0642.4922.797 251.3161.7082.0602.4852.787 261.3151.7062.0562.4792.779 271.3141.7032.0522.4732.771 281.3131.7012.0482.4672.763 291.3111.6992.0452.4622.756 301.3101.6972.0422.4572.750 311.3091.6962.0402.4532.744 321.3091.6942.0372.4492.738 331.3081.6922.0352.4452.733 341.3071.6912.0322.4412.728 351.3061.6902.0302.4382.724 0.1 0.05 0.025 0.01 0.005 361.3061.6882.0282.4342.719 371.3051.6872.0262.4312.715 381.3041.6862.0242.4292.712 391.3041.6852.0232.4262.708 401.3031.6842.0212.4232.704 411.3031.6832.0202.4212.701 421.3021.6822.0182.4182.698 431.3021.6812.0172.4162.695 441.3011.6802.0152.4142.692 451.3011.6792.0142.4122.690 971.2901.6611.9852.3652.627 981.2901.6611.9842.3652.627 991.2901.6601.9842.3652.626 1001.2901.6601.9842.3642.626 1.2821.6451.9602.3262.576 Want 90% CI, n = 600,  /2 = d.f. = Same as Normal Table gives right-tail area. (e.g., for a right-tail area of 0.025 and d.f. = 15, the t value is 2.131.) t, =

21. M32 Margin of Error 21  Department of ISM, University of Alabama, 1995-2002 Estimate with 98% confidence the mean gallons of water used per shower for Dallas Cowboys after a game. The sample mean of 16 showers is 30.00 gallons and the standard deviation is 10.4 gallons. Example 2 m.o.e. = s n t  2, n-1

22. M32 Margin of Error 22  Department of ISM, University of Alabama, 1995-2002 Point estimate  m.o.e. Example 2, continued... 98% confidence interval:

23. M32 Margin of Error 23  Department of ISM, University of Alabama, 1995-2002 I am 95% confident that the true mean gallons of water used per shower for Dallas Cowboys after a game will fall within the interval 23.235 to 36.765 gallons. Example 2, Statement in the L.O.P. A statement in L.O.P. must contain four parts: 1. amount of confidence. 1. amount of confidence. 2. the parameter being estimated in L.O.P. 2. the parameter being estimated in L.O.P. 3. the population to which we generalize in L.O.P. 3. the population to which we generalize in L.O.P. 4. the calculated interval. 4. the calculated interval.

24. M32 Margin of Error 24  Department of ISM, University of Alabama, 1995-2002 To get a smaller Margin of Error:   = 1.96  n Margin of Error for 95% confidence:

25. M32 Margin of Error 25  Department of ISM, University of Alabama, 1995-2002 Confidence vs. Probability  BEFORE a sample is collected, there is a 95% probability that the future to be computed sample mean, will fall within m.o.e. units of .  AFTER the sample is collected, the computed sample mean either fell within m.o.e. units of  or it did not. After the event, it does not make sense to talk about probability. Analogy: Suppose you own 95 tickets in a 100-ticket lottery. The drawing was held one hour ago, but you don’t know the result. P(win) = 0 or 1, but you are very CONFIDENT that you have won the lottery.

26. M32 Margin of Error 26  Department of ISM, University of Alabama, 1995-2002 What sample size is needed to estimate the mean mpg of Toyota Camrys with an m.o.e. of 0.2 mpg at 90%confidence if the pop. std. dev. is 0.88 mpg? m.o.e. = Z  n  2

27. M32 Margin of Error 27  Department of ISM, University of Alabama, 1995-2002 Interpretations of the Confidence Interval for  If we took many, many samples of size n, and calculated a confidence interval for each, then I would expect that 95% of all these many intervals would contain the true mean, and 5% would not.

28. M32 Margin of Error 28  Department of ISM, University of Alabama, 1995-2002 Example 3: A car rental agency wants to estimate the average mileage driven by its customers. A sample of 225 customer receipts, selected at random, yields an average of 325 miles. Assuming that the population standard deviation is 120 miles, construct a 95% confidence interval for the population average mileage.

29. M32 Margin of Error 29  Department of ISM, University of Alabama, 1995-2002

30. M32 Margin of Error 30  Department of ISM, University of Alabama, 1995-2002 Example 4: An insurance company collected data to estimate the mean value of personal property held by apartment renters in Tuscaloosa. In a random sample of 45 renters, the average value of personal property was $14,280 and the standard deviation was $6,540. Example 4:

31. M32 Margin of Error 31  Department of ISM, University of Alabama, 1995-2002 a. At 95% confidence, find the margin of error for estimating the true population mean? Example 4:

32. M32 Margin of Error 32  Department of ISM, University of Alabama, 1995-2002 b. Construct a 95% confidence interval for the true mean value of personal property owned by renters in Tuscaloosa. Example 4:

33. M32 Margin of Error 33  Department of ISM, University of Alabama, 1995-2002 c. Three years ago the true mean value of personal property owned by renters in Tuscaloosa was $13,050. Based on your confidence interval, is there evidence that the true mean has changed? Example 4:

34. M32 Margin of Error 34  Department of ISM, University of Alabama, 1995-2002 d. Assuming that the true mean is unchanged from three years ago, find the probability that a future sample of 45 renters would result in a mean that is more extreme than the sample the insurance company just took. P(X > 14,280 | true mean = 13,050) What distribution must be used? Example 4:

35. M32 Margin of Error 35  Department of ISM, University of Alabama, 1995-2002

36. M32 Margin of Error 36  Department of ISM, University of Alabama, 1995-2002 e. Based on this probability, would you conclude that the true mean has changed from three years ago? or, equivalently Is there sufficient evidence to conclude that the true mean has changed? or, equivalently Was the sample mean of $14,280 too close to $13,050 to call it unusual, or was this value a rare event? Had the sample mean been $15,000, would your conclusion be different? Example 4: