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Topic 1: Sampling and Sampling Distributions

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Topic 1: Sampling and Sampling Distributions Topic 1 --- page 45 The _____________ of the Sample Mean(V( X ) ¾We also need to know the variance of the sampling distribution of ___for a given sample size n. Notation: The variance of the values of X is denoted by either: 2 VX() or σ X ¾The ________ is the average of the squared deviations of the variable X about its mean μ X . σμμ22==−=−VX()()()( EX X2 PX) XX∑ X X Continuation of the previous example: ()X − μ ()X − μ 2 (X − μ )2 P( X ) X P( X ) X X X 4 1 (4-7)=-3 9 16 16 5 2 1 (5-7)=-2 8 16 = 8 16 6 3 (6-7)=-1 3 16 16 4 1 (7-7)=0 0 7 16 = 4 8 3 (8-7)=1 3 16 16 9 2 = 1 (9-7)=2 8 16 8 16 10 1 (10-7)=3 9 16 16 Topic 1 --- page 46 2 XXPX−=()40 =25 . ∑ ()i 16 Mean = ___ Variance = ____ Recall the population variance = ____ 2 VX()==σ 2 σ Notice that: X n i.e. 5/2= 2.5 ¾This is not a coincidence either!! 2 Recall: The Var(X) = Var(Xi ∀ i) =σ . ‘n’ is the sample size. Sampling distribution is for all possible samples of size n. Proof: ⎛ 1 ⎞ VX()= V⎜ ∑ Xi ⎟ ⎝ n i ⎠ 2 ⎛ 1⎞ = ⎜ ⎟ VX ⎝ n⎠ ()∑ i 2 ⎛ 1⎞ = ⎜ ⎟ ()VX()i ⎝ n⎠ ∑ and Since X': s are independent under random sampling 1 =++++()σσσ222 σ 2 n2 []∑ L 1 = nσ 2 n2 1 = σ 2 . n Topic 1 --- page 47 Although we calculated the value of _____directly in this 4 element population of Xi’s, in problems where there are many values of X , direct calculation is impractical. As long as we ______the variance of the population σ 2 , we can calculate the VX(). This is because the variance of the random variable X is related 2 to σ , the population variance, and to the sample size by the formula: 2 VX()= σ __ The variance of X is always ____ than or equal to the population variance. The variance of the mean of a sample of n independent 1 observations is n times the ________ of the parent population (see footnote p. 256). VX()==σ221 σ X ( n) (Equation 7.6) When n=1, the samples contain only one observation and distribution of X and X are the _____. 2 ¾As n increases, σ X becomes _______ because the sample means will tend to be closer to the value of the population mean μ X . Topic 1 --- page 48 When n = N (in a finite population) all sample means will _____ the population mean and the VX()’s will equal ___. 2 With our example, the population variance (σ ) is known (= 5) and n=2: So the variance of X (VX()) is: VX()==σσ2211 =()52 =.5 X ( n) ( 2) What Happens toVX()as n _________? ¾Because each sample contains more information or more elements of the population as the sample size _________, the sample will be closer to the population, so expect ____ variability. Example: Suppose X ~N(0, 100) Randomly draw samples of size: (i) 10 (ii) 100 (iii) 1000 from this population. Calculate X 10 for all possible samples of size 10. Calculate X 100 for all possible samples of size 100. Topic 1 --- page 49 Calculate X 1000 for all possible samples of size 1000. Then we can show: Sampling Distributions for Xbar: Various Sample Sizes: 0 For n=10: dispersion if Xbar is quite wide around the mean of 0. For n=1000: less variation around the mean of zero. ¾When n approaches ________, there is no dispersion and variance of Xbar =0. Topic 1 --- page 50 Standard _____________of the Mean Notation: We usually denote the standard deviation of X ’s, σ X , the standard _____ of the mean. ¾The error refers to sampling _____. ¾σ X is a measure of the standard expected _____ when the sample mean is used to obtain information or draw conclusions about the unknown population mean. Standard Error of the mean: σσ2 σ X == n n (Equation 7.7) σ 2 5 2236. σ ====15811. In our example: X n 2 2 Notes: (i) μ X and σ X are parameters of the population of sample averages for all conceivable samples of size n. ¾ These parameters are usually _______. (ii) The population parameters (μ, σ2) are also usually _______. Topic 1 --- page 51 (iii) This means that we cannot use the relationships: μμ==and σ σ XXn to solve for values of one of these statistics. But these relationships allow us to test hypotheses about the population parameters on the basis of sample results. More on this later........ Next: We now have derived the mean and the variance of the sampling distribution, but have not said anything about the _____ of the sampling distribution of X . Recall that distributions with the same mean and variance can have very different ______. ¾We must now specify an assumption about the entire distribution of X ’s: Topic 1 --- page 52 Section 7.5 Sampling distribution of X , ______ Parent Population ¾It is typically not possible to specify the shape of the X ’s when the parent population is discrete and the sample ____ is small. ¾ However, the shape of a sample taken from a normally distributed parent population (X) can be specified. In this case, the X ’s are distributed normally. “ The sampling distribution of X ’s drawn from a normal parent population is a ______ distribution.” Recall: The mean of the X s is μ X = μ and the variance of X s 2 σ 2 = σ is X n . ¾Hence the sampling distribution of X is: 2 NN(,)μσ2 = (, μσ ) X ~ XX n when ever the parent population is ______. X~N(μ,σ2). ¾Meaning, regardless of the _____ of the parent population, the 2 μ σ 2 = σ mean and variance of X equal: X and X n . Topic 1 --- page 53 From the last example: X~N(0, 100). Hence, 100 XN10 ~,()0 10 = N (,)010 100 XN100 ~,()0 100 = N (,)01 100 XN1000 ~,()0 1000 = N (,.001). Remember: The normal distribution is a continuous distribution. (I.e. infinite number of different samples could be drawn.) (Error in text on page 259?) Example: Suppose all the possible samples of size 10 are drawn from a ______ distribution that has a mean of 25 and a variance of 50. That is, X is normally distributed with a mean μ=25 and variance σ2=50 : X~N(25,50). ¾Since the population mean μ=25, the mean of X s equal μ X =25. ¾Since the population variance σ2=50, the variance of the X ’s 2 σ 2 ===σ 50 5. equals X n 10 ¾Since X is ______, X is ________ distributed X ~N(25,5). Topic 1 --- page 54 What this means is: 68.3% of the sample means will fall within ± one _______ error of the mean: σ X ==5224.. μ +=±1σ X 25()(. 1 224 )= 2276 .to 2724 . _____% of the sample means will fall within ± two standard errors of the mean: μ +=±2σ X 25()(. 2 2 24 )= 25± 4 . 48⇒ 2052 .to 29 . 48 . 99.7% of the sample means will fall within ± three standard errors of the ____: μ +=±3σ X 25()(. 3 2 24 )= 25± 6 . 72⇒ 18 . 28to 3172 . Topic 1 --- page 55 The Standardized Form of the Random Variable X and σ _____ In Economics 245, we saw from Chapter 6 that it is easier to work with the standard normal form of a variable than it is to leave it in its original units. The same type of _______________ made on a random variable X, can be made on the random variable X . Recall, to _________ the random variable X to its standard normal form (Z), we subtract the mean from each value and divide by the standard deviation: ()X − μ Z = σ ←Z has a mean = 0 & variance = 1. Z~N(0,1). The standardization of X is ___________ the same way: ()X − μ X − μ Z = X = σ σ (Equation 7.9) X n ¾The random variable Z has a mean of zero and a variance of ___. Topic 1 --- page 56 Thus: When sampling from a normal parent population, the X − μ Z = distribution of σ will be ______ with mean zero and n variance equal to ___. (See figure 7.4.) Example: Suppose X is the height (in inches) of basketball players on all university teams in Canada during summer term. Suppose X~N(__,36). A random sample of nine players is drawn from this population. ¾ What is the probability that the sample average team player height is less than 80 inches? (What is P ( X ≤ 80)?) Solution: If X ~N(__,36), then X ~N(75,36/9=4). ¾Standardize the variable X : X − μ 80−__ 5 5 Z = = ===25. σ 6 6 2 n 9 3 Looking at the Cumulative Standardized Normal Distribution Table F(Z), on page 891, the P(Z ≤2.5) = 0.____. The probability that the average height of basketball players in our sample of size 9 is less than 80 inches is 99.38%. Topic 1 --- page 57 0.9938=CDF Z 0 2.5 Example: Let X be the amount of money customers owe on home mortgages at the Bank of Nova Scotia (in thousands of $). Suppose X~N(150,____). Draw a random sample of 25 from the population. What is the probability that the average amount owing is greater that $200? PX()≥ 200 ? Solution: X~N(150,____), so X ~N(150,____/25=324)=N(150, 324); X − μ 200− 150 50 50 Z = = ===278.
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