Def 1 A population consists of the totality of the observations with which we are concerned.

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1 Chapter 6 Sampling Distributions 6.1 Random Sampling Def 1 A population consists of the totality of the observations with which we are concerned. Remark 1. The size of a populations may be finite or infinite. For example, Finite population the blood types for students in a scholl. Infinite population the observations of atmospheric pressure every day from past to future.. The observations from a population have a common distribution. We will use X i to denote the value of the ith observation. Def A sample is a subset of population. Def 3 A random sample of size n is a collection of n indenpendent random variables with a common distribtion. Remark The probability distribtion of a random) sample, say, x 1, x,..., x n, of size n is described as the following jpdf fx 1, x,..., x n ) = fx 1 )fx ) fx n ) 6. Some Important Statistics Def 4 Any function T X 1, X,..., X n ) of the observations X 1, X,..., X n, is called a statistic. Central Tendency in a sample Let X 1, X,..., X n represent a random sample of size n. 1

2 1. sample mean X = 1 X i n. sample median Y n+1)/ if n is odd X = Y n/ + Y n/)+1 if n is even where Y i s denote the order statistics corresponding to X 1,..., X n. 3. mode the value in the sample that occurs most often or with the greatest frequency. The mode may not exist, and when it does it is not necessary unique. Example 1 The lengths of time, in minutes, that 10 patients waited in a doctor s office before receiving treatment were recorded as follows: Find 1. the mean;. the median; 3. the mode. 5, 11, 9, 5, 10, 15, 6, 10, 5, 10. sol) 1. x = ) = 8.6 mimutes; 10. x = 9.5 minutes; 3. MoX) = 5 and 10 minutes. Facts 1. MoX) X X or X X MoX).. 3. X i X X i a for any a. X i X) X i a) for any a. 4. X MoX) 3X X).

3 Variability in the Sample 1. range Let X n) = maxx 1,..., X n ) and X 1) = minx 1,..., X n ). The range of a sample is defined to be X n) X 1).. sample variance S = = = 1 n 1 1 n 1 1 n 1 X i X) n X i 1 n { n } Xi nx ) X i 3. sample deviation S. Example The IQ s of a random sample of five members of a sorority are 108, 11, 17, 118, and 113. Find 1. the range;. the sample variance; 3. the sample deviation. sol) 1. Range = = x i = = 67, x i = = 578 S = 1 ) = S =

4 6.3 Sampling Distributions Sampling Distributions of Means 1. E[X] = µ, V ar[x] = σ /n.. From central limit theorem, Z = lim n X µ σ/ n N0, 1) 3. Suppose there are two populations with means and variances being µ 1, σ 1) and µ, σ ), respectively. If two random sample of size n 1 and n are drawn at random from two populations. Then As n 1 and n, E[X 1 ± X ] = µ 1 ± µ V ar[x 1 ± X ] = σ 1 n 1 + σ n Z = X 1 ± X ) µ 1 ± µ ) N0, 1) σ1 n 1 + σ n 4. The above fact trivially holds if both populations are normally distributed. Example 3 If all possible sample size of 16 are drawn from a normal population with mean equal to 50 and standard deviation equal to 5, what is the probability that a sample mean X will fall in the interval from µ X 1.9σ X to µ X 0.4σ X? sol) We want to find P µ X 1.9σ X < X < µ X 0.4σ X ). Clearly, X N50, 5 /16) N50, 1.565). P µ X 1.9σ X < X < µ X 0.4σ X ) = P X < µ X 0.4σ X ) P X < µ X 1.9σ X ) = Φ 0.4) Φ 1.9) = =

5 Example 4 Given the discrete uniform population fx) = { 1 3 x =, 4, 6 0 otherwise find the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4. sol) We want to find P 4.1 < X < 4.4). µ X = ) = 4 3 σ X = ) 4 = 8 3 n = 54 > 30, X N4, /9) ). Hence, P 4.1 < X < 4.4) = P X < 4.4) P X 4.1) ) ) = Φ Φ /9 /9 = Φ1.8) Φ0.45) = = Example 5 A random sample of size 5 is taken from a normal population having a mean of 80 and a standard deviation of 5. A second random sample of size 36 is taken from a different normal population having a mean of 75 and a standard deviation of 3. Find the probability that the sample mean computed from the 5 measurements will exceed the sample mean computed from the 36 measurements by at least 3.4 but less than 5.9. n 1 = 5, µ 1 = 80, σ 1 = 5. n = 36, µ = 75, σ = 3. We want to find P 3.4 X 1 X < 5.9). Clearly, X 1 N80, 5 /5), X 75, 3 /36) = X 1 X ) N5, 5/4). 5

6 P 3.4 X 1 X < 5.9) = P X 1 X < 5.9) P X 1 X < 3.4) = Φ Φ /4 5/4 = Φ0.8050) Φ ) = = Remark Let Z N0, 1). The critical value z α is defined to be Several often used z α are We can find z α from the probability table. Chi-square distribution P Z z α ) = α z =.58 z 0.01 =.33 z 0.05 = 1.96 z 0.05 = Def 5 A random variable X is said to possess a chi-square distribution with ν degrees of freedom, denoted by X χ ν, if it has the pdf Remark 1. χ ν Γn/, 1/). fx) =. X N0, 1) = X χ 1. { ν/ Γν/) xν/) 1 e x/ x 0 0 x 0 3. Let X i N0, 1), i = 1,..., n be n independent random variables. Then, Xi χ n. 4. Let X i Nµ, σ ), i = 1,..., n be n independent random variables. Then, ) Xi µ χ σ n. 6

7 5. Let χ χ ν. The chi-square critical value χ α,ν is defined such that P χ χ α,ν) = α We can find the critical value from the table. Example 6 Given a sample of size 16 from a normal population N50, 10 ), 1. find b such that P X 50 b) ;. find c such that P X 50 c) ; n ) 3. find d such that P X i 50) d. sol) 1.. P X 50 b) ) X 50 = P 10/ 16 b 10/ 16 = P Z b.5 ) = P Z z 0.05 ) = b.5 = z 0.05 = b = z = = P X 50 c) ) X 50 = P 10/ 16 c 10/ 16 = P Z c ).5 = P Z c ).5 = P Z z 0.05 ) = c.5 = z 0.05 = c = z = = 4.9

8 3. = P P X i 50) d ) ) Xi 50 d ) = P χ d ) 100 = P χ χ 0.05,16 ) = d 100 = χ 0.05,16 = d = 100 χ 0.05,16 = = 796. Sampling Distributions of S Let E[X i ] = µ and V ar[x i ] = σ. Then, 1. [ ] 1 E[X] = E X i = 1 E[X i ] = µ n n V ar[x] = 1 [ n ] n V ar X i = 1 n nσ = σ n. X i a) = X i X) = X i X) = X i X) + nx a) Xi nx = Xi 1 n ) X i n X i µ) nx µ) [ n ] [ n ] E X i X) = E X i µ) ne [ X µ) ] = n 1)σ [ ] E[S 1 ] = E X i X) = σ n 1 5. X and S are independent. 8

9 6. ) Xi µ = σ n 1)S σ + ) X µ σ/ n 7. Therefore, if X i Nµ, σ ), then Xi µ σ X µ σ/ n σ ) χ n; ) χ 1; n 1)S χ n 1; E [ n 1)S σ ] = n 1; V ar [ n 1)S σ ] = n 1). Example 7 A manufacturer of car batteries guarantees that his batteries will last, on the average, µ = 3 years with a standard deviation of σ = 1 year. Assume that the battery lifetime follows a normal distribution. If a sample of size 5 is taken to estimate the mean lifetime and variance. Find 1. P X µ b) ;. Find c such that P S c) ; 3. Find d such that P 1 5 ) 5 X i µ) d. sol) 1. X N3, 1/5). P X 3 b) X 3 = P 1/5 b 1/5 = P Z b 1/5 = P Z z 0.05 ) = 1/5 b = z 0.05 = b = z /5 = ) =

10 . 4S 1 χ 4. P S c) 4S = P 1 4c ) 1 = P χ χ 0.05,4) = = P χ 4c ) = 4c = χ 0.05,4 = c = χ 0.05,4/4 = 9.488/4 = ) Xi µ χ σ 5. ) 1 5 P X i µ) d ) Xi µ = P d ) 5 σ σ = P χ 5d ) σ = P χ χ 0.05,5) = 5d σ = χ 0.05,5 = d = σ χ 0.05,5 = /5 =.14 t-distribution Let X Nµ, σ ). Then X Nµ, σ /n) or, eqivalently, Z = X µ σ/ n N0, 1) In most cases, the value of σ is not available. Thus, we will use S to estimate σ. The t-distribution deals with the distribution about the statistic T defined by T = X µ S/ n Def 6 Let Z N0, 1) and W χ ν be two independent random variables. The random variable T = Z W/ν is said to possess a t-distribution with ν degrees of freedom and is denoted by T t v. 10

11 Facts Let X t ν. Then: 1. f T t) = Γ ) ν+1 ) ν+1)/ Γ ) ν 1 + t, < t < πν ν. E[T ] = 0 V ar[t ] = ν ν, ν > 3. As ν, t ν N0, 1). 4. Let X 1,..., X n be a random sample from a normal population Nµ, σ ). Then X and S are independent. Z = X µ σ/ n Hence, 5. The value of t α,ν, defined by N0, 1), W = n 1)S σ T = can be found from the probability table. 6. t 1 α,ν = t α,ν. χ n 1. Z W/n 1) = X µ S/ n t n 1 P T t α,ν ) = α, Example 8 The gas consumption liters/hr) of automobiles manufactured by a company is normally distributed but with mean µ and variance σ being unknown. Now, a random sample of size 16 is taken to estimate µ by X. Find c in terms of sample variance s such that P X µ < c) = sol) P X µ < c) = 0.95 = 1 P X µ c) = 0.95 = P X µ c) ) X µ = P S/ 16 c S/ 16 11

12 = ) c P T S/ 16 = P T t 0.05,15 ) = c s/ 16 = t 0.05,15 = c = t 0.05,15 s/ 16 =.131/4)s = 0.538s F -distribution Def 7 Let W 1 χ ν 1 and W χ ν be two independent random variables. The the random variable F = W 1/ν 1 W /ν is said to possess and an F distribution with ν 1 and ν degrees of freedom and is denoted by F F ν1,ν. Facts 1. f F x) = Γ ) ν 1 +ν ν ν 1 / 1 ν ν / Γ ) ) ν 1 Γ ν x ν 1/) 1 ν + ν 1 x) ν 1+ν )/ x > 0 0 x 0. Let X 1,..., X m be a sample of size m arising from a normal population Nµ 1, σ 1), and let Y 1,..., Y n be a sample of size n arising from a normal population Nµ, σ ). Suppose that the two population are independent. Then pf) Let F = S 1/σ 1 S /σ F m 1,n 1 W 1 = m 1)S 1 ; σ1 W = n 1)S. σ Clearly, W 1 χ m 1 and W χ n 1. Hence, W 1 /m 1) W /n 1) = S 1/σ 1 S /σ F m 1,n 1 1

13 3. In the above, if σ 1 = σ, we have S1 S F m 1,n 1 4. The critical value f α,ν1,ν is defined such that 5. f 1 α,ν,ν 1 = 1 f α,ν1,ν. pf) P F f α,ν1,ν ) = α. 1 α = 1 P F f α,ν1,ν ) ) W1 /ν 1 = 1 P f α,ν1,ν W /ν W /ν = 1 P 1 ) W 1 /ν 1 f α,ν1,ν W /ν = P 1 ) W 1 /ν 1 f α,ν1,ν According to the defintion regarding to the F critical value, we have f 1 α,ν,ν 1 = 1 f α,ν1,ν 6. The F critical value can be found from the probability table. Example 9 f 0.95,11,8 =? sol) From the probability table, f 0.05,8,11 =.95. Hence, f 0.95,11,8 = 1 = 1 f 0.05,8,11.95 = 0.34 Example 10 Two samples of size 5 are taken from sample two independent normal populations. Assume that the two populations have the same variance. 1. Find b such that P S 1/S b).. Find c such taht P S 1/S c). sol) Clearly, S 1/S F 4,4. 13

14 1. P S1/S b) = P F b) = P F f 0.05,4,4 ) = b = f 0.05,4,4 = Exercises P S 1/S c) = 1 P S 1/S > c) = P F c) = 0.95 = P F f 0.95,4,4 ) = 0.95 = b = f 0.95,4,4 = 1 = 1 f 0.05,4, = A finite population contains six numbers 1,,3,4,5,6. a) Find the mean µ and variance σ of the numbers in the population. b) Randomly taking a sample of size without replacement from the population. Find the distribution of the sample mean X. c) Show that the sample mean in b) satisfies E[X] = µ, and V ar[x] = N n σ N 1 n.. Let X 1, X, X 3, X 4 be a sample from a normal population N0, 1). What value of c can make the statistic T = cx 1 + X ) X3 + X4 a t-distribution? Determine the number of degrees of freedom for the t-distribution. 3. Let Y 1, Y be a random sample from a normal population N0, 1), and let X 1, X be a random sample from another independent normal population N1, 1). Find a) the distribution of X + Y ; Y 1 + Y b) the distribution of X X 1 ) + Y Y 1 ) ; c) the distribution of X X 1 ) + Y Y 1 ) ; d) the distribution of X + X 1 ). X X 1 ) 14

15 4. Let S 1 and S be the sample variances from two independent normal population Nµ 1, 10) and Nµ, 15), respectively. Suppose n 1 = n = 10. a) Find b such that P S 1/S b) = b) Find c such that P S 1/S c). 5. Find a) χ 0.05,0 b) χ 0.95,15 c) χ 0.05,10 d) t 0.05,10 e) t 0.95,0 f) f 0.05,10,11 g) f 0.99,6,8. 6. Let X 1 and X be sample means from two independent normal populations with a common variance σ. Define S p, called pooled variance, by Show that a) E[S p ] = σ ; b) n 1 + n )S p σ χ n 1 +n. S p = n 1 1)S 1 + n 1)S n 1 + n 7. Let X 1,..., X n is a sample from a Bernoulli population. Let a) Determine the distribution of Y. Y = X i b) If n = 100 and p. Use the other two possible distributions to approximate P Y = 3). c) Let ˆp = X = Y/n. For n > 40, find c such that P ˆp 0.05 < c) = The lifetime of a system is Y = X 1 + X + X 3 + X 4, wherex 1, X, X 3, X 4 are the lifetimes of its subsystems. Suppose that each subsystem is independent, and the lifetime is exponentially distributed with MTBF mean time between failure) being 3 hours. Find the probability that the system can survive at least 18 hours. 9. Let X 1,..., X 100 be a sample from a normal population N0, 1). a) Find the pdf of sample variance b) Find E[S]. S = 1 99 c) Find P X 1 + X X 3 X 4 ). 100 X i X). 10. Let Y 1 < Y < Y 3 < Y 4 be the order statistics of a random sample of size n = 4 from a continuous symmetric distribution with mean µ and variance σ. Find the probability of Y 3 < µ. 15

16 11. Suppose X 1,..., X n is a random sample from Nµ, σ ) population, where µ, σ are both unknown. Let X n = 1 X i, Sn = 1 n n X i X n ). If X n+1 is an additional observation, find the constant k so that kx n X n+1 )/S n has a t-distribution. 1. Let X i N0, 1), i = 1,,..., 6 be mutually independent. Let Y = X 1 X + X 3 ) + 3X 4 X 5 3X 6 ). Find the constant k so that ky has a chi-square distribution. 16