Christopher Dougherty EC220 - Introduction to econometrics (review chapter) Slideshow: confidence intervals Original citation: Dougherty, C. (2012) EC220 - Introduction to econometrics (review chapter). [Teaching Resource] © 2012 The Author This version available at: Available in LSE Learning Resources Online: May 2012 This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License. This license allows the user to remix, tweak, and build upon the work even for commercial purposes, as long as the user credits the author and licenses their new creations under the identical terms

00  sd  0 –sd  0 –1.96sd  0 +sd 2.5% CONFIDENCE INTERVALS probability density function of X conditional on  =  0 being true null hypothesis H 0 :  =  0 In the sequence on hypothesis testing, we started with a given hypothesis, for example H 0 :  =  0, and considered whether an estimate X derived from a sample would or would not lead to its rejection. 1

00  sd  0 –sd  0 –1.96sd  0 +sd 2.5% CONFIDENCE INTERVALS probability density function of X conditional on  =  0 being true null hypothesis H 0 :  =  0 1 Using a 5% significance test, this estimate would not lead to the rejection of H 0.

00  sd  0 –sd  0 –1.96sd  0 +sd 2.5% CONFIDENCE INTERVALS probability density function of X conditional on  =  0 being true null hypothesis H 0 :  =  0 1 This estimate would cause H 0 to be rejected.

00  sd  0 –sd  0 –1.96sd  0 +sd 2.5% CONFIDENCE INTERVALS probability density function of X conditional on  =  0 being true null hypothesis H 0 :  =  0 1 So would this one.

00  sd  0 –sd  0 –1.96sd  0 +sd 2.5% CONFIDENCE INTERVALS probability density function of X conditional on  =  0 being true null hypothesis H 0 :  =  0 1 We ended by deriving the range of estimates that are compatible with H 0 and called it the acceptance region. acceptance region for X

6 Now we will do exactly the opposite. Given a sample estimate, we will find the set of hypotheses that would not be contradicted by it, using a two-tailed 5% significance test. CONFIDENCE INTERVALS X

00 null hypothesis H 0 :  =  0 1 X Suppose that someone came up with the hypothesis H 0 :  =  0.

00  sd  0 –sd  0 –1.96sd CONFIDENCE INTERVALS probability density function of X conditional on  =  0 being true null hypothesis H 0 :  =  0 1 To see if it is compatible with our sample estimate X, we need to draw the probability distribution of X conditional on H 0 :  =  0 being true. X

00  sd  0 –sd  0 –1.96sd CONFIDENCE INTERVALS probability density function of X conditional on  =  0 being true null hypothesis H 0 :  =  0 1 X To do this, we need to know the standard deviation of the distribution. For the time being we will assume that we do know it, although in practice we have to estimate it.

00  sd  0 –sd  0 –1.96sd CONFIDENCE INTERVALS probability density function of X conditional on  =  0 being true null hypothesis H 0 :  =  0 1 X Having drawn the distribution, it is clear this null hypothesis is not contradicted by our sample estimate.

CONFIDENCE INTERVALS null hypothesis H 0 :  =  1 1 X Here is another null hypothesis to consider. 11

12 Having drawn the distribution of X conditional on this hypothesis being true, we can see that this hypothesis is also compatible with our estimate. CONFIDENCE INTERVALS 11  sd  1 +sd  1 –1.96sd probability density function of X conditional on  =  1 being true null hypothesis H 0 :  =  1 X

13 Here is another null hypothesis. CONFIDENCE INTERVALS 22 X null hypothesis H 0 :  =  2

14 It is incompatible with our estimate because the estimate would lead to its rejection. CONFIDENCE INTERVALS  2 –sd  2 –1.96sd  2 +sd 22 null hypothesis H 0 :  =  2 probability density function of X conditional on  =  2 being true X

15 The highest hypothetical value of  not contradicted by our sample estimate is shown in the diagram. CONFIDENCE INTERVALS  max  max +1.96sd  max –sd  max +sd null hypothesis H 0 :  =  max probability density function of X conditional on  =  max being true X

15 CONFIDENCE INTERVALS  max  max +1.96sd  max –sd  max +sd null hypothesis H 0 :  =  max probability density function of X conditional on  =  max being true X When the probability distribution of X conditional on it is drawn, it implies X lies on the edge of the left 2.5% tail. We will call this maximum value  max.

15 CONFIDENCE INTERVALS  max  max +1.96sd  max –sd  max +sd null hypothesis H 0 :  =  max probability density function of X conditional on  =  max being true X If the hypothetical value of  were any higher, it would be rejected because X would lie inside the left tail of the probability distribution.

15 CONFIDENCE INTERVALS  max  max +1.96sd  max –sd  max +sd null hypothesis H 0 :  =  max probability density function of X conditional on  =  max being true X Since X lies on the edge of the left 2.5% tail, it must be 1.96 standard deviations less than  max. reject any  >  max X =  max – 1.96 sd  max = X sd reject any  > X sd

15 CONFIDENCE INTERVALS  max  max +1.96sd  max –sd  max +sd null hypothesis H 0 :  =  max probability density function of X conditional on  =  max being true X reject any  >  max X =  max – 1.96 sd  max = X sd reject any  > X sd Hence, knowing X and the standard deviation, we can calculate  max.

20 In the same way, we can identify the lowest hypothetical value of  2 that is not contradicted by our estimate. CONFIDENCE INTERVALS b2b2 22  2 -sd  sd  2 +sd min probability density function of b 2 conditional on  2 =  2 being true min null hypothesis H 0 :  2 =  2 min

CONFIDENCE INTERVALS b2b2 22  2 -sd  sd  2 +sd min probability density function of b 2 conditional on  2 =  2 being true min null hypothesis H 0 :  2 =  2 min It implies that b 2 lies on the edge of the right 2.5% tail of the associated conditional probability distribution. We will call it  2 min. 21

CONFIDENCE INTERVALS b2b2 22  2 -sd  sd  2 +sd min probability density function of b 2 conditional on  2 =  2 being true min null hypothesis H 0 :  2 =  2 min Any lower hypothetical value would be incompatible with our estimate. 22

CONFIDENCE INTERVALS b2b2 22  2 -sd  sd  2 +sd min probability density function of b 2 conditional on  2 =  2 being true min null hypothesis H 0 :  2 =  2 min Since b 2 lies on the edge of the right 2.5% tail, b 2 is equal to  2 min plus 1.96 standard deviations. Hence  2 min is equal to b 2 minus 1.96 standard deviations. 23 reject any  2 <  2 b 2 =  sd  2 = b 2 – 1.96 sd reject any  2 < b 2 – 1.96 sd min

probability density function of b 2 (1) conditional on  2 =  2 being true (2) conditional on  2 =  2 being true CONFIDENCE INTERVALS b2b2 22  2 -sd  sd  2 +sd min max The diagram shows the limiting values of the hypothetical values of  2, together with their associated probability distributions for b 22  sd  2 - sd  2 + sd max (1)(2)

CONFIDENCE INTERVALS b2b2 22  2 -sd  sd  2 +sd min 22  sd  2 - sd  2 + sd max (1)(2) Any hypothesis lying in the interval from  2 min to  2 max would be compatible with the sample estimate (not be rejected by it). We call this interval the 95% confidence interval. 25 reject any  2 >  2 = b sd reject any  2 <  2 = b 2 – 1.96 sd 95% confidence interval: b 2 – 1.96 sd <  2 < b sd max min

CONFIDENCE INTERVALS b2b2 22  2 -sd  sd  2 +sd min 22  sd  2 - sd  2 + sd max (1)(2) max min The name arises from a different application of the interval. It can be shown that it will include the true value of the coefficient with 95% probability, provided that the model is correctly specified. 26 reject any  2 >  2 = b sd reject any  2 <  2 = b 2 – 1.96 sd 95% confidence interval: b 2 – 1.96 sd <  2 < b sd

CONFIDENCE INTERVALS In exactly the same way, using a 1% significance test to identify hypotheses compatible with our sample estimate, we can construct a 99% confidence interval. 27 Standard deviation known 95% confidence interval b 2 – 1.96 sd <  2 < b sd 99% confidence interval b 2 – 2.58 sd <  2 < b sd Standard deviation estimated by standard error 95% confidence interval b 2 – t crit (5%) se <  2 < b 2 + t crit (5%) se 99% confidence interval b 2 – t crit (1%) se <  2 < b 2 + t crit (1%) se

Standard deviation known 95% confidence interval b 2 – 1.96 sd <  2 < b sd 99% confidence interval b 2 – 2.58 sd <  2 < b sd Standard deviation estimated by standard error 95% confidence interval b 2 – t crit (5%) se <  2 < b 2 + t crit (5%) se 99% confidence interval b 2 – t crit (1%) se <  2 < b 2 + t crit (1%) se CONFIDENCE INTERVALS  2 min and  2 max will now be 2.58 standard deviations to the left and to the right of b 2, respectively. 28

CONFIDENCE INTERVALS Until now we have assumed that we know the standard deviation of the distribution. In practice we have to estimate it. 29 Standard deviation known 95% confidence interval b 2 – 1.96 sd <  2 < b sd 99% confidence interval b 2 – 2.58 sd <  2 < b sd Standard deviation estimated by standard error 95% confidence interval b 2 – t crit (5%) se <  2 < b 2 + t crit (5%) se 99% confidence interval b 2 – t crit (1%) se <  2 < b 2 + t crit (1%) se

CONFIDENCE INTERVALS As a consequence, the t distribution has to be used instead of the normal distribution when locating  2 min and  2 max. 30 Standard deviation known 95% confidence interval b 2 – 1.96 sd <  2 < b sd 99% confidence interval b 2 – 2.58 sd <  2 < b sd Standard deviation estimated by standard error 95% confidence interval b 2 – t crit (5%) se <  2 < b 2 + t crit (5%) se 99% confidence interval b 2 – t crit (1%) se <  2 < b 2 + t crit (1%) se

CONFIDENCE INTERVALS This implies that the standard error should be multiplied by the critical value of t, given the significance level and number of degrees of freedom, when determining the limits of the interval. Standard deviation known 95% confidence interval b 2 – 1.96 sd <  2 < b sd 99% confidence interval b 2 – 2.58 sd <  2 < b sd Standard deviation estimated by standard error 95% confidence interval b 2 – t crit (5%) se <  2 < b 2 + t crit (5%) se 99% confidence interval b 2 – t crit (1%) se <  2 < b 2 + t crit (1%) se 31

Copyright Christopher Dougherty These slideshows may be downloaded by anyone, anywhere for personal use. Subject to respect for copyright and, where appropriate, attribution, they may be used as a resource for teaching an econometrics course. There is no need to refer to the author. The content of this slideshow comes from Section R.12 of C. Dougherty, Introduction to Econometrics, fourth edition 2011, Oxford University Press. Additional (free) resources for both students and instructors may be downloaded from the OUP Online Resource Centre Individuals studying econometrics on their own and who feel that they might benefit from participation in a formal course should consider the London School of Economics summer school course EC212 Introduction to Econometrics or the University of London International Programmes distance learning course 20 Elements of Econometrics