Chapter 3 Selected Basic Concepts in Statistics n Expected Value, Variance, Standard Deviation n Numerical summaries of selected statistics n Sampling.

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Presentation transcript:

Chapter 3 Selected Basic Concepts in Statistics n Expected Value, Variance, Standard Deviation n Numerical summaries of selected statistics n Sampling distributions

Expected Value Weighted average Not the value of y you “expect”; a long-run average

E(y) Example 1 Toss a fair die once. Let y be the number of dots on upper face. y p(y)1/6

E(y) Example 2: Green Mountain Lottery Choose 3 digits between 0 and 9. Repeats allowed, order of digits counts. If your 3-digit number is selected, you win $500. Let y be your winnings (assume ticket cost $0) y$0$500 p(y)

US Roulette Wheel and Table n The roulette wheel has alternating black and red slots numbered 1 through 36. n There are also 2 green slots numbered 0 and 00. n A bet on any one of the 38 numbers (1-36, 0, or 00) pays odds of 35:1; that is... n If you bet $1 on the winning number, you receive $36, so your winnings are $35 American Roulette (The European version has only one 0.)

US Roulette Wheel: Expected Value of a $1 bet on a single number n Let y be your winnings resulting from a $1 bet on a single number; y has 2 possible values y-135 p(y)37/381/38 n E(y)= -1(37/38)+35(1/38)= -.05 n So on average the house wins 5 cents on every such bet. A “fair” game would have E(y)=0. n The roulette wheels are spinning 24/7, winning big $$ for the house, resulting in …

Variance and Standard Deviation n Measure spread around the middle, where the middle is measured by 

Variance Example Toss a fair die once. Let y be the number of dots on upper face. y p(y)1/6 Recall  = 3.5

V(y) Example 2: Green Mountain Lottery y$0$500 p(y) Recall  =.50

Estimators for ,  2,  n s 2 “average” squared deviation from the middle n Automate these calculations n Examples

Linear Transformations of Random Variables and Sample Statistics n Random variable y with E(y) and V(y) n Lin trans y*=a+by, what is E(y*) and V(y*) in terms of original E(y) and V(y)? n Data y 1, y 2, …, y n with mean y and standard deviation s n Lin trans y* = a + by; new data y 1 *, y 2 *, …, y n *; what is y* and s* in terms of y and s

n E(y*)=E(a+by) = a + bE(y) n V(y*)=V(a+by) = b 2 V(y) n SD(y*)=SD(a+by) =|b|SD(y) n y* = a + by n s* 2 = b 2 s 2 n s* =  b  s Linear Transformations Rules for E(y*), V(y*) and SD(y*) Rules for y*, s* 2, and s*

Expected Value and Standard Deviation of Linear Transformation a + by Let y=number of repairs a new computer needs each year. Suppose E(y)= 0.20 and SD(y)=0.55 The service contract for the computer offers unlimited repairs for $100 per year plus a $25 service charge for each repair. What are the mean and standard deviation of the yearly cost of the service contract? Cost = $100 + $25y E(cost) = E($100+$25y)=$100+$25E(y)=$100+$25*0.20= = $100+$5=$105 SD(cost)=SD($100+$25y)=SD($25y)=$25*SD(y)=$25*0.55= =$13.75

Addition and Subtraction Rules for Random Variables n E(X+Y) = E(X) + E(Y); n E(X-Y) = E(X) - E(Y) n When X and Y are independent random variables: 1. Var(X+Y)=Var(X)+Var(Y) 2. SD(X+Y)= SD’s do not add: SD(X+Y)≠ SD(X)+SD(Y) 3. Var(X−Y)=Var(X)+Var(Y) 4. SD(X −Y)= SD’s do not subtract: SD(X−Y)≠ SD(X)−SD(Y) SD(X−Y)≠ SD(X)+SD(Y)

Example: rv’s NOT independent n X=number of hours a randomly selected student from our class slept between noon yesterday and noon today. n Y=number of hours the same randomly selected student from our class was awake between noon yesterday and noon today. Y = 24 – X. n What are the expected value and variance of the total hours that a student is asleep and awake between noon yesterday and noon today? n Total hours that a student is asleep and awake between noon yesterday and noon today = X+Y n E(X+Y) = E(X+24-X) = E(24) = 24 n Var(X+Y) = Var(X+24-X) = Var(24) = 0. n We don't add Var(X) and Var(Y) since X and Y are not independent.

a2a2 c 2 =a 2 +b 2 b2b2 Pythagorean Theorem of Statistics for Independent X and Y a b c a 2 + b 2 = c 2 Var(X) Var(Y) Var(X+Y) SD(X) SD(Y) SD(X+Y) Var(X)+Var(Y)=Var(X+Y) a + b ≠ c SD(X)+SD(Y) ≠SD(X+Y)

9 25= Pythagorean Theorem of Statistics for Independent X and Y = 5 2 Var(X) Var(Y) Var(X+Y) SD(X) SD(Y) SD(X+Y) Var(X)+Var(Y)=Var(X+Y) ≠ 5 SD(X)+SD(Y) ≠SD(X+Y)

Example: meal plans n Regular plan: X = daily amount spent n E(X) = $13.50, SD(X) = $7 n Expected value and stan. dev. of total spent in 2 consecutive days? (assume independent) n E(X 1 +X 2 )=E(X 1 )+E(X 2 )=$13.50+$13.50=$27 SD(X 1 + X 2 ) ≠ SD(X 1 )+SD(X 2 ) = $7+$7=$14

Example: meal plans (cont.) n Jumbo plan for football players Y=daily amount spent n E(Y) = $24.75, SD(Y) = $9.50 n Amount by which football player’s spending exceeds regular student spending is Y-X n E(Y-X)=E(Y)–E(X)=$24.75-$13.50=$11.25 SD(Y ̶ X) ≠ SD(Y) ̶ SD(X) = $9.50 ̶ $7=$2.50

For random variables, X+X≠2X n Let X be the annual payout on a life insurance policy. From mortality tables E(X)=$200 and SD(X)=$3,867. 1) If the payout amounts are doubled, what are the new expected value and standard deviation?  Double payout is 2X. E(2X)=2E(X)=2*$200=$400  SD(2X)=2SD(X)=2*$3,867=$7,734 2) Suppose insurance policies are sold to 2 people. The annual payouts are X 1 and X 2. Assume the 2 people behave independently. What are the expected value and standard deviation of the total payout?  E(X 1 + X 2 )=E(X 1 ) + E(X 2 ) = $200 + $200 = $400 The risk to the insurance co. when doubling the payout (2X) is not the same as the risk when selling policies to 2 people.

Estimator of population mean  n y will vary from sample to sample n What are the characteristics of this sample-to- sample behavior?

Numerical Summary of Sampling Distribution of y  Unbiased: a statistic is unbiased if it has expected value equal to the population parameter.

Numerical Summary of Sampling Distribution of y

Standard Error  Standard error - square root of the estimated variance of a statistic  important building block for statistical inference

Shape? n We have numerical summaries of the sampling distribution of y n What about the shape of the sampling distribution of y ?

THE CENTRAL LIMIT THEOREM The World is Normal Theorem

The Central Limit Theorem (for the sample mean y) n If a random sample of n observations is selected from a population (any population), then when n is sufficiently large, the sampling distribution of y will be approximately normal. (The larger the sample size, the better will be the normal approximation to the sampling distribution of y.)

The Importance of the Central Limit Theorem n When we select simple random samples of size n, the sample means we find will vary from sample to sample. We can model the distribution of these sample means with a probability model that is Shape of population is irrelevant

Estimating the population total 

 Expected value

Estimating the population total   Variance, standard deviation, standard error

Finite population case  Example: sampling w/ replacement to estimate 

Finite population case  Example: sampling w/ replacement to estimate  SampleProb of Sample  V(  ) {1, 2} {1, 3}.0835/ {1, 4} {2, 3}.0855/ {2, 4} {3, 4}.3235/ {1, 1} {2, 2} {3, 3}.1615/20 {4, 4}.16100

Finite population case  Example: sampling w/ replacement to estimate   From the table:

Finite population case  Example: sampling w/ replacement to estimate 

Finite population case  Example: sampling w/ replacement to estimate   Example Summary

Finite population case  Sampling w/ replacement to estimate pop. total   In general

Finite population case  Sampling w/ replacement to estimate pop. total 

Finite population case  Sampling w/ replacement to estimate pop. total  In reality, do not know value of y i for every item in the population. BUT can choose  i proportional to a known measurement highly correlated with y i.

Finite population case  Sampling w/ replacement to estimate pop. total  Example: want to estimate total number of job openings in a city by sampling industrial firms. Many small firms – employ few workers; A few large firms – employ many workers; Large firms influence number of job openings; Large firms should have greater chance of being in sample to improve estimate of total openings. Firms can be sampled with probabilities proportional to the firm’s total work force, which should be correlated to the firm’s job openings.

Finite population case  Sampling without replacement to estimate pop. total  Thus far we have assumed a population that does not change when the first item is selected, that is, we sampled with replacement. Example: population {1, 2, 3, 4}; n=2, suppose equally likely. Prob. of selecting 3 on first draw is ¼. Prob. of selecting 3 on second draw depends on first draw (probability is 0 or 1/3) When sampling without replacement this is not true

Finite population case  Sampling without replacement to estimate pop. total  Worksheet

End of Chapter 3