Interval Estimation (Confidence Interval = C.I.): An interval estimate of some population parameter is an interval of the form (, ),

Transcription

1 Cofidece Iterval

2 Estimatio Problems Suppose we have a populatio with some ukow parameter(s). Example: Normal(,) ad are parameters. We eed to draw coclusios (make ifereces) about the ukow parameters. We select samples, compute some statistics, ad make ifereces about the ukow parameters based o the samplig distributios of the statistics. Statistical Iferece (1) Estimatio of the parameters Poit Estimatio Iterval Estimatio (Cofidece Iterval) () Tests of hypotheses about the parameters

3 Classical Methods of Estimatio: Poit Estimatio: A poit estimate of some populatio parameter is a sigle valueˆ of a statistic ˆ. For example, the value of the statistic computed from a sample of size is a poit estimate of the populatio mea. x Iterval Estimatio (Cofidece Iterval = C.I.): A iterval estimate of some populatio parameter is a iterval of the form (, ), i.e, ˆL <<. This iterval cotais the true value of "with probability 1", that is P( ˆL << ˆU )=1 ˆU ˆ L ˆ U

4 Example of Poit Estimatio

5 Iterval Estimatio (Cofidece Iterval) of the Mea (): If i1 i / is the sample mea of a radom sample of size from a populatio (distributio) with mea ad kow variace, the a (1)100% cofidece iterval for ca be calculated as follows depedig o whether the populatio variace is kow or ot.

6 ), ( where is the -value leavig a area of / to the right; i.e., P(> )=/, or equivaletly, P(< )=1/. Note: We are (1)100% cofidet that ), ( (i) First Case: is kow: The value is called the -score ad the test is called the -test

7 Example The average zic cocetratio recorded from a sample of zic measuremets i 36 differet locatios is foud to be.6 gram/milliliter. Fid a 95% ad 99% cofidece iterval (C.I.) for the mea zic cocetratio i the river. Assume that the populatio stadard deviatio is 0.3. Solutio: = the mea zic cocetratio i the river. Populatio Sample =?? =36 =0.3 =.6 First, a poit estimate for is =.6. (a) We wat to fid 95% C.I. for. =?? 95% = (1)100% = (1) =0.05 / = 0.05

8 = 0.05 = 1.96 A 95% C.I. for is (1.96).6 (1.96) < < < <.698 (.50,.698) We are 95% cofidet that (.50,.698).

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10 (b) Similarly, we ca fid that a 99% C.I. for is.471 < <.79 (.471,.79) We are 99% cofidet that (.471,.79) Notice that a 99% C.I. is wider that a 95% C.I. This is a tradeoff betwee accuracy ad precisio Theorem If is used as a estimate of, we ca the be (1)100% cofidet that the error (i estimatio) will ot exceed

11 Example: I previous example, we are 95% cofidet that the sample mea differs from the true mea by a amout less tha.6 (1.96) Note: Let e be the maximum amout of the error, that is e, the: e e e Theorem : If is used as a estimate of, we ca the be (1)100% cofidet that the error (i estimatio) will ot exceed a specified amout e whe the sample size is e

12 Example How large a sample is required i previous example if we wat to be 95% cofidet that our estimate of is off by less tha 0.05? Solutio: We have = 0.3, e=0.05. The by Theorem, e 0.05 Therefore, we ca be 95% cofidet that a radom sample of size =139 will provide a estimate differig from by a amout less tha e=0.05.

13 t-distributio: Recall that, if 1,,, is a radom sample of size from a ormal distributio with mea ad variace, i.e. N(,), the / ~ N(0,1) We ca apply this result oly whe is kow ad umber of samples is 30 or more! If is ukow, we replace the populatio variace with the ( ) sample variace 1 i i S to have the followig statistic 1 T S /

14 Result: If 1,,, is a radom sample of size from a ormal distributio with mea ad variace, i.e. N(,), the the statistic T S / has a t-distributio with =1degrees of freedom (df), ad we write T~ t(). Note: t-distributio is a cotiuous distributio. The shape of t-distributio is similar to the shape of the stadard ormal distributio.

15 t = The t-value above which we fid a area equal to, that is P(T> t ) = Sice the curve of the pdf of T~ t() is symmetric about 0, we have t 1 = t Values of t are tabulated i Tables.

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17 of: Example: Fid the t-value with =14 (df) that leaves a area (a) (b) 0.95 to the left to the right. Solutio: = 14 (df); T~ t(14) (a) The t-value that leaves a area of 0.95 to the left is t 0.05 = 1.761

18 (b) The t-value that leaves a area of 0.95 to the right is t 0.95 = t = t 0.05 = 1.761

19 Example: For = 10 degrees of freedom (df), fid t 0.10 ad t Solutio: t 0.10 = 1.37 t 0.85 = t = t 0.15 = (t 0.15 = 1.093)

20 If ad are the sample mea ad the sample stadard deviatio of a radom sample of size from a ormal populatio (distributio) with ukow variace, the a (1)100% cofidece iterval for is : i i / 1 i i S 1 1) /( ) ( Result: ), ( S t S t S t S t S t Iterval Estimatio (Cofidece Iterval) of the Mea (): (ii) Secod Case: is ukow: Recall: 1) ~ t( S/ T

21 t where is the t-value with =1 degrees of freedom leavig t a area of / to the right; i.e., P(T> )=/, or equivaletly, P(T< )=1/. t Example The cotets of 7 similar cotaiers of sulfuric acid are 9.8, 10., 10.4, 9.8, 10.0, 10., ad 9.6 liters. Fid a 95% C.I. for the mea of all such cotaiers, assumig a approximate ormal distributio. Solutio:.=7 i / 10.0 S ( ) /( 1) i1 i1 First, a poit estimate for is i / i1 i

22 Now, we eed to fid a cofidece iterval for. =?? 95%=(1)100% 0. 95=(1) =0.05 /=0.05 = t 0.05 =.447 (with =1=6 degrees of freedom) t A 95% C.I. for is t S S t t (.447) 7 S < < < < 10.6 ( 9.74, 10.6) We are 95% cofidet that ( 9.74, 10.6) (.447) 7

23 Estimatio of the Mea (): Recall: E ( ) Var( ) ~ N, / T S/ ~ N(0,1) ~ t( 1) ( is kow) We use the samplig distributio of ifereces about. ( is ukow) to make