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UW-Madison ECON 310 - EconStats310 - October 17

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1 Econ 310 Professor Wallace October 17, 2013 Lecture: - The first midterm exam - Estimation (Chapter 10, 12) o Point Estimation o Interval Estimation Midterm #1 Descriptives - Each TA graded a different question for consistency. If you have concerns about how a question was graded you should contact the TA that did the graded via email. o Zack graded #1 ([email protected]) o Xianwei graded #2 ([email protected]) o Pedro graded #3 ([email protected]) o Zhewen graded #4 ([email protected]) o Leo (Chenxin) graded in #5 ([email protected]) - The mean was 52 points (out of 120) o Question 1 mean was 21 o Question 2 mean was 5 (ouch~) o Question 3 mean was 9 o Question 4 mean was 10 o Question 5 mean was 7 - The mean was also 52 points. - The high score was 101.5. There were 3 scores at 100 or above - The low score was below 15.2 - Distribution of scores - Midterm #1 Curve - Considerations o The exam was too long to complete, but I’ve given a lot of thought to it and I don’t think the questions were too difficult. o There is a tradeoff because it is difficult to write a exam that spans the scope of topics that we covered that is short that it can be completed in 75 minutes. o Because of the length the exam was basically a 100 point exam rather than a 120 point exam. A 80+ AB 65-79.5 B 55-64.5 BC 45-54.5 C 35-44.5 D 25-34.5 F <25 05101520Percent0 10 20 30 40 50 60 70 80 90 100 110 120Midterm 1 Scores (out of 120)3 - Distribution of Grades 05101520PercentF D C BC B AB AGrade4 Point Estimation Point estimator – draws inference about a population by estimating the value of an unknown parameter using a single point value. Sampling error – the difference between the sample statistic and the population parameter Illustration: When using the sample mean to estimate the population mean the sampling error is X Standard error - The standard error is simply the standard deviation of the sampling error. Illustration: When using the sample mean to estimate the standard error is  XXSVar Xnn       or in the special case where the population is binomial and X is a sample proportion P   11PPppppnn  5 Properties of point estimators - Unbiased – an unbiased estimator of a population parameter is an estimator with an expected value equal to the parameter. Illustration: Assuming that iX is independently distributed with mean  and variance 2 the sample mean X is an unbiased estimator for  because EX. Thus, the expected value of the sampling error is equal to zero. - Consistency – an unbiased estimator is said to be consistent if the difference between the estimator and the parameter grows smaller as the sample size grows larger. Illustration: The sample mean X is a consistent estimator for  because  lim 0nPX   This result should make sense because X is unbiased and    lim lim0nnVar X Var X    or, in other words, the variance difference between estimator and the population goes to zero as n gets large. - Relative efficiency – if there are two unbiased estimators of a parameter, the one whose variance is smaller is said to have relative efficiency6 Interval Estimation Interval estimator – draws inference about a population parameter by estimating the value using an interval Example 1: Assume that we have taken a sample of size 10,000 and compute 30,000x  and 20,000. Find an interval such that we are 95% certain that  will lie in the interval. For any standard normal random variable we know that  1.96 1.96 0.95PZ    If this is true for any standard normal random variable then it should be true for Xn. Thus, 1.96 1.96 0.95XPn    What we want to do is solve the inequality inside for 7 Multiply through by n 1.96 1.96 0.95PXnn                 Multiply through by -1 and switch inequality signs. 1.96 1.96 0.95PXnn                 Rearrange so that smaller bound is on the left 1.96 1.96 0.95PXnn                 Add X to both sides 1.96 1.96 0.95P X Xnn                What the above statement says is that 95% percent of the time we will draw a sample mean that is within 1.96 standard deviations of .8 1.96 1.96 0.95P X Xnn                Implication: The interval 1.96xn will contain  for 95% of the values of X - The 95% of the time that we draw a value of X within one 1.96 standard deviations of the  (its mean) the interval will include  - The 5% of the time that we draw a value of X farther than 1.96 standard deviations from (its mean) our confidence the interval will not include  - We can never be completely certain whether the interval contains  or not9 For our specific example we can be 95% confident that that the population mean is in the interval 20,00030,000 1.9610,000 or between 29,608 and 30,392 In this example: - 95% is the confidence level - The confidence coefficient is 0.95 - The significance level (  ) = 1 – Confidence Coefficient General formula for  1 100% confidence interval: /2SxZn where: - S is the sample standard deviation (use the population standard deviation if you have it); - 1 is the confidence coefficient; and - /2Z is the Z value providing an area /2in the upper tail of a standard normal probability distribution. Example 2: Using the same numbers calculate 90% confidence interval. We just use different Z values in your calculations. Here we want 0.05Z instead of 0.025Z 90% confidence interval for : 20,00030,000 1.64510,000   30,000 1.645 200 30,000 329    We can be 90% confident that the population mean is between 29,671 and


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UW-Madison ECON 310 - EconStats310 - October 17

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