156-159 Chapter 4.1 Summary of Necessary Theoretical Results

# 156-159 Chapter 4.1 Summary of Necessary Theoretical...

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Unformatted text preview: CHAPTER APPLICATIONS OF STATISTICAL INFERENCE @ SUMMARY or In order tocarry out many more statistical inferences other than the simple ea . NECESSARY ﬁdence intervals that we have considered. we need some additional theereti results which we give in this section- Same of these will be proved in Chapter it so we state a few of them as theorems and We give an example of each. THEORETICAL RESULTS . Remark If the instructer en EhDOE-B-ﬁ Sec—tint: 6.2 mt the moment-generaliuaf, funetinn ("1.31.3 aflinear functions can he studied at this time. Theorem 4.1-1 Let X].X:.. . ..Xn be n independent chi-square random variables with I f[,f1,. Mr“ degrees of freedom, respectively. Then the random variable. Y = X; +3} +- -- +Jlt',1 has aehiequare distribution with n + r; + - -- + 5,. degrees of freedom. EXAMPLE Lei M, £2.13 be three independent chi~square random variables with r: = 2, r2 5, r3. = 4 degrees of freedom. Then the random variable Y = XI + X2 + X3 :_ 211 .8 :U. 5 b1 IV. xi iﬂﬂdHY-clﬁiﬁ} '3 fromTtt e Theorem 4.1-2 [f X1.X2..-.,Xn are independent nerrnel random variables with means tithtuyu wit“ and variances 0?.n3...._aﬁ. reapeetiveiy. then the random variable 1" = L, 0.x] has a normal distribution with mean in: :1 EL, mp4 and variance 0.. 2 L1 tea}. EXAMPLE 4.1-2 Theorem 4.1-3 Section 4.1 Summary of Necessary Theoretical Results 157 Let X1.X3,X3 be three independent normal random variables with means in = 4, n2 2 —5, M3 = 2 and variances of = 4, 022 = 16, 032 = 9, respectively. Then the random variable Y = X1 —X2 is normal with My 2 (1)(4)+ (—1)(—5) : 9 and 03 = (1)2(4) + (—1)2(16) = 20. Moreover. the random variable Z = X1+ X2 + X3 is normal with mean n2 2 4 + (—5) + 2 = 1 and variance a§=4+16+9=29x. ___ Let if and 5'2 be the mean and the variance of a random sample of size n EXAMPLE 4.1-3 EXAMPLE 4.1-4 from a distribution which is Min. 02}. Then ? and S2 are independent random variables with distributions such that f is Nmmﬂmi ~ 2 w- is fin—l). 52 Let X1.X2.....X,, be a random sample from a distribution which is N(,u,02). We know from Section 3.4 that (X,- -— 1;.)2/02 is x2(1),i = 1,2,...,n. From Theorem 4.1-1. it must be true that 2,1431 — ii)2/cr2 is x202) as we are adding 21 independent chi-square variables together. From Theorem 4.1-3, note that if we replace it by its estimator Y we obtain ()1 — 1)Sz/<r2 = 27:1(Xi — 70/02, which is gratin-4). That is, we have lost one degree of freedom by replacing a parameter in a chisquare variable by its estimator. Later, we see that this is generalized: pr parameters in a x2(r) random variable, p < r, are replaced by “reasonable” estimators, the resulting chi—square variable has r —p degrees of freedom. __ ___ _ Before we close this section on results needed to make a number of statistical inferences, we consider twa important distributions. We do not try to develop them here, but only give the deﬁnitions of these two random variables and refer the reader to Tables VI and VII in which a few probabilities concerning these T and F random variables are given. Also see Examples 52-12 and 52-13 for the p.d.f.s and some graphs for the T and F distributions. Let where Z has a N(0.1) distribution, U has a X2(r) distribution, and Z and U are independent. We say that T has a Student’s t distribution with r degrees of freedom. For illustration, if r = 10, P[T 5 2.764] = 0.99 from Table VI in the Appendix. In general. from Table VI, we can ﬁnd la(r) so that P[T > ta(r)] = a. (Continuation of Example 4.1-3) With Theorem 4.1-3, Y has a N(,u,(72/I1) distribu- tion so that _ X — it ONE N(0, 1). 15B Chaptem Applleaﬁons of Statistlcal Inference Of course (11 — 1)52/o2 is X2(n—1) and 7 and 52 are independent. Henoe' 7-H T: a/ﬁ ZY-u (II—1)52 S/ﬁ T ("-1) has a Student’s 1 distribution with n — 1 degree of freedom. Let _ Him [7— . Uzt’rz’ where U, is x2(r,-). t' = 1,2.and U1 and U2 are independent. Then we say that Fhasa Fisher's F distribution (some call it Snedecors F } with r1 and r2 degrees of freedom. denoted by F(r;,r2). For illustration,ifr1 = 5 and r; = 10, then P(F g 4.24) = 0.935 and P(F > 5.64) = 0.01 from Table VII in the Appendix. In general from that : table, we can ﬁnd Fa so that P(F > Fa) = or for selected values ofot. EXAMPLE Let X1,X2,...,X,, and Y1,Y2,...,Ym be independent random samples from NU”, 02.) and N(uy,a,2,), respective] . We know, from Theorem 4.1-3, that (n — l) AJo? is X201 — 1) and (m —- “Swing is x2(m —- l) and the two are indepen- dent since the X s and Ys are independent. Thus (:1 —- US} _ an“ 1) _ stars ’ W? ‘ Sta)? 03011 — 1) has an F distribution with n — 1 and m — 1 degrees of freedom. _P_— EXERCISES 4.1 4.1-1 Let X; and X3 be independent chi—square variables with r1 = 4 and r2 = 2 degrees of freedom, respectively. (a) Determine P(2.204 < X1 +X2 < 16.81). (b) Find P(12.59 < X1+X2). 4.1-2 Let X1 and X2 be independent random variables. Let Y 2 X1 + X2 be x204) and let X1 be X2(3). (3) Guess the distribution of X2. Note that we prove this in Section 6.2. ([1) Determine H1053 < X; < 24.72). 4.1-3 Let the independent random variables X1 and X2 be N041 = 3. 012 = 9) and N(u2 = 6, (722 = 16), respectively. Determine P(—10 < Y < 5), where Y : X; ~ X2. 4.1-4 Three random steps in series are needed to complete a certain procedure. The means and the standard deviations of the respective steps are m = 6 hours, in = 4 hours. it; = 5 hours and a] = 2 hours, 02 = 2 hours, 03 == 3 hours. Assuming , independence and normal distributions. compute the probability that the procedure will be completed in less than 20 hours. Section 4.1 Summary of Necessary Theoretical Results 159 4.1-5 Let T and .52 be the mean and the variance of a random sample of size n = 16 from the normal distribution N01, 02). (Ed Find at from Table V! such that Y— [1 P —d < < d = 0.95. < S/x/16 ) (bl Rewrite the inequalities in part (a) so that P[u(3(',\$) < u < 1415)] : 0.95. Find u()_(, S) and MY, S) so that once i? and s are computed the interval from u(,Y,s) to u(E,s) provides a 95% conﬁdence interval for u. 4.1-6 (continuation of Example 4.15) In Example 4.1-5 let a}? = 6,2, = (:2, which is unknown. (a) We know that Y is N0”, oz/n) and 7 is N(uy, og/m); so argue that Y — 7 is N[llX — ,Lly, 02(1/11 + 1/m)]. (b) Argue that [(n — US; + (m — 1)S%,]/cr2 is x2(n+m —2). (c) Show that 7 — 7 — (MX — MY) t/02(1/n +1/m) [(n _ ms; + (m — 1)SZYl/a2 —m— T: has a Student’s 1 distribution with n + m — 2 degrees of freedom and eliminate a in that T. 4.1-7 (continuation of Exercise 4.1-6.) In Exercise 4.1-6. let n = 8 and m = 10. (a) Find d from Table VI so that P(—d < T < d) = 0.90. (b) Rewrite the inequalities in this probability statement in part (a) so that nX — ny is “trapped” in the middle. (c) If} = 7.3. s. = 12.7 and y = 6.4, s). = 10.3, using the result of part (b) compute a 90% conﬁdence interval for uX - /.l.y. Assuming that (7} might not equal 0,2,, use the F of Example 4.1—5 to ﬁnd C and d from Table VII so that P(c < F < d): 0.95 and P(F < d) = 0.975. HINT: Note that P(c < F) = P(1/c > l/F) = 0.975 and that l/F has an F(m—1,n—1) distribution. Rewrite P(c < F < d) = 0.95 so that P[u(52 5%,) < (ﬁg/oi, < as? ,Si)] 2 0.95. Use the values in part (c) to ﬁnd a 95% conﬁdence interval for air/ale. 350 Appendix c Answers to Odd-Numbered Exercises 3.7-5 0.6915 using normal approximation, 0.7030 using binomial. 3.7—? 0.3085. 3.7-9 0.6247 using normal approximation, 0.6148 115ng Poisson. 3.7-” 0.5548; (b) 0.3823. 3.7-]3 0.6813 using normal approximation. 0.6788 using binomial. 3.715 0.3802; (b) 0.7571. 3.7-17 0.4734 using normal approximation, 0.4749 using Poisson approximation with A = 50, 0.4769 using b(5000, 0.01). 3.7-19 0.6455 using normal approximation, 0.6449 using Poisson. 3.7-2! (a) 0.8289 using normal approximation, 0.8294 using Poisson. (b) 0.0261 using tables in book, 0.0218 using Maple. CHAPTER 4 4.1—3 (3) 0.890; (b) 0.05. 4.1—3 0.8644. 4.1-5 (3) d=2.131‘, 11(15): 3 — 2.1313/4, no.5) = 3 + 2.1315/4. (a) d = 1.746; _ _ /7S§+9S§. /1 1_ X—Yi1.746 T\§+1—0, (c) [—8.5517. 10.3517]; c = 1/482, d = 4.20; S2 2 4.20 2 P( Y2 <a—g< 25Y>=0.95; 4.82SX 0X SK (0 [0.1365, 2.7626]. (3) s = 6.144; (b) [440610.142] or [41079521]. [ ;;1<X.-— 102 ZL1(X1— 102] X37201) , Mag/201) 0.4987; (b) [0,1835]. [1947,2233]. )7 = 3.580; 5 = 0.512; [0, 3.877]. f = 25.475, S = 2.4935; (b) [24.059, [—115.480, 129.105]; (C) no. (9) (n —1)s§/d+(m—1)s§(a’ 1) + 11+ m — 2 2 ~ y :i: ta/72(n+m 2)\/ nm' ...
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## This note was uploaded on 12/28/2010 for the course ECON 41 taught by Professor Guggenberger during the Fall '07 term at UCLA.

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156-159 Chapter 4.1 Summary of Necessary Theoretical...

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