Statistics and Sampling distributions

Transcription

1 Statistics and Sampling distributions a statistic is a numerical summary of sample data. It is a rv. The distribution of a statistic is called its sampling distribution. The rv s X 1, X 2,, X n are said to form a random sample of size n if the X i s are independent and each X i has the same probability distribution (identically distributed).

2 Derive sampling distribution Example: P(X = 1) = 1 3, P(X = 0) = 2 3. Let X = 1 2 (X 1 + X 2 ). P( X = 0) = P(X 1 = 0, X 2 = 0) = 4/9. P( X = 1 2 ) = P(X 1 = 0, X 2 = 1) + P(X 1 = 1, X 2 = 0) = 4 9. P( X = 1) = P(X 1 = 1, X 2 = 1) = 1 9. If X = 1 3 (X 1 + X 2 + X 3 ), then 1 x p( x) x 0 8 p( x) Note E( X ) = µ X = 1 3 = E(X ).

3 Simulation Specify 1. The statistic of interest. 2. The population distribution. 3. The sample size n. 4. The number of replication k. For each sample, compute the statistic and the histogram of the k values gives the approximate distribution of the statistic.

4 the distribution of the sample mean proposition: let X 1, X 2,, X n be a random sample from a distribution with mean µ and standard deviation σ, then 1. E( X ) = µ X = µ, 2. V ( X ) = σ 2 X = σ2 n and σ X = σ n example: A soft-drink vending machine is set so that the amount of drink dispensed is a rv with a mean of 200 milliliters and a standard deviation of 15 milliliters. Let X be the mean amount dispensed in a random sample of size 36. Then E( X ) = µ X = 200, σ X = = 2.5.

5 Normal population distribution proposition: Let X 1, X 2,, X n, iid N(µ, σ 2 ), then X N(µ, σ2 n ). example: The time that it takes a randomly selected rat to find its way through a maze is a normal rv with µ = 1.5 min and σ = 0.35 min. Let X 1,, X 5 be the time in the maze of 5 randomly selected rats. Then the sample mean X is also normal with µ X = 1.5, σ X = = and the probability the sample mean is at most 2 min is P( X 2) = P(Z ) = P(Z 3.19) =

6 the Central Limit Theorem the Central Limit Theorem (CLT): Let X 1, X 2,, X n be a random sample from a population with mean µ and variance σ 2. Then as n, X µ σ/ n N(0, 1). Roughly speaking, when n is large ( 30), X approximately N(µ, σ2 n ). Example: Rockwell hardness of pins is known to have a mean value of 50 and a standard deviation of 1.2. What is the probability that the sample mean hardness for a random sample of 36 pins is at least 51? P( X 51) = P(Z / = P(Z 5)

7 exercise The tip percentage at a restaurant has a mean value of 18% and a standard deviation of 6%. What is the approximate probability that the sample mean tip percentage for a random sample of 40 bills is between 16% and 19%? Can we compute the probability if the sample size is 15 rather than 40?

8 Let X be the sample mean for a random sample of 40 bills, then by the CLT, X approximately has a normal distribution with µ X = 18 percent and σ X = 6 40 = percent. So P(16 X 19) = P( Z ) = P( 2.11 Z 1.05) = =

9 The Law of Large Numbers X µ as n. example: Flip a coin n times. Let X i = 1 for a head and X i = 0 for a tail. Then X = fraction of heads. X 0.5 as n.

10 The distribution of a linear combination Given a collection of n rvs and n constants a 1,, a n, the rv Y = a 1 X a n X n is called a linear combination of the X i s. proposition: 1. E(a 1 X a n X n ) = a 1 E(X 1 ) + a 2 E(X 2 ) + a n E(X n ) 2. V (a 1 X a n X n ) = a1 2V (X 1) + a2 2V (X 2) + anv 2 (X n ) if X i s are independent. In particular, E(X 1 X 2 ) = E(X 1 ) E(X 2 ) and V (X 1 X 2 ) = V (X 1 ) + V (X 2 ) if X 1 and X 2 are independent.

11 A shipping company handles containers in three different sizes: 27, 125 and 512 ft 3. Let X 1, X 2, X 3 be the number of type 1, 2, 3 containers shipped during a given week respectively. We have µ 1 = 200, µ 2 = 250, µ 3 = 100 and σ 1 = 10, σ 2 = 12, σ 3 = 8. 1). Suppose X 1, X 2, X 3 are independent, calculate the expected value and variance of the total volume (V = 27X X X 3 ) shipped. E(V ) = 27µ µ µ 3 = = 87850ft 3. V (V ) = 27 2 V (X 1 ) V (X 2 ) V (X 3 ) = = 19, 100, 116.

12 Normal rvs proposition: If X 1, X 2,, X n are independent, normally distributed rv s, then any linear combination of the X i s also has a normal distribution. example: For males the expected pulse rate is 70 per min and the standard deviation is 10 per min. For women the expected pulse rate is 77 and the standard deviation is 12. Let X be the sample average pulse rate for a random sample of 40 men and Ȳ be the sample average for a random sample of 36 women. 1. Then approximately X N(70, ) and Ȳ N(77, 36 ). 2. Approximately X Ȳ N with E( X Ȳ ) = = 7 and V ( X Ȳ ) = V ( X ) + V (Ȳ ) = = 6.5. P( X 1 ( 7) Ȳ 1) = P(Z 6.5 ) = P(Z 3.14) =

13 Distributions based on a normal sample propositions 1. Z 2 χ If X 1 χ 2 ν 1, X 2 χ 2 ν 2, and they are independent, then X 1 + X 2 χ 2 ν 1 +ν n i=1 Z i 2 χ 2 n for independent Z i s. 4. For a random sample from a normal distribution, (n 1)S 2 χ 2 σ 2 n If Z N(0, 1), X χ 2 ν, and Z and X are independent, then T = Z is said to have a t distribution with ν degrees of X /ν freedom. Theorem: If X 1, X 2,, X n is random sample from N(µ, σ 2 ), then T = X µ S/ n t n 1.

14 F distribution F distribution: If X 1 χ 2 ν 1, X 2 χ 2 ν 2 and X 1 and X 2 are independent, then F = X 1/ν 1 X 2 /ν 2 F ν1,ν 2. If we have a random sample of size m from N(µ 1, σ1 2 ) and an independent sample of size n from N(µ 2, σ2 2 ), then F = (m 1)S 2 1 /σ2 1 m 1 (n 1)S 2 2 /σ2 2 n 1 = S2 1 /σ2 1 S 2 2 /σ2 2 F m 1,n 1 F distribution can be used to make inference about the ratio of two population variances.

15 exercise The expected time spent on standing and walking a day for lean people is 500 minutes with a standard deviation 102 minutes, while the expected time for mildly obese people is 370 minutes with a standard deviation 68 minutes. Let X, Ȳ be the sample average time of a sample of 36 lean people and 49 obese people respectively. 1). What is the approximate distribution of X, Ȳ? Write down E( X ), E(Ȳ ), V ( X ), V (Ȳ ). 2). What is the approximate distribution of X Ȳ? Find E( X Ȳ ), V ( X Ȳ ). 3). Find P( X Ȳ ).

16 1). Since the sample sizes are large enough, by the Central Limit Theorem, X and Ȳ approximately follow normal distributions with E( X ) = 500, V ( X ) = ) = 289 and E(Ȳ ) = 370, V (Ȳ ) = ) = ). Since X Ȳ is a linear combination of two independent normal random variables, it has a normal distribution with E( X Ȳ ) = = 130, V ( X Ȳ ) = = ). P( X Ȳ ) = P( X Ȳ 0) = P(Z ) = P(Z 6.64) 1.00.