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# 9 - MIT OpenCourseWare http/ocw.mit.edu 14.30 Introduction...

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MIT OpenCourseWare http://ocw.mit.edu 14.30 Introduction to Statistical Methods in Economics Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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Problem Set #9 14.30 - Intro. to Statistical Methods in Economics Instructor: Konrad Menzel Due: Friday, May 8, 2009 Question One: Confidence Intervals (Adapted from BainIEngelhardt p. 384) Consider a random sample of size n from a normal distribution, Xi N N(p, a2). 1. If it is known that a2 = 15, find a 90% confidence interval for p based on the estimate Z = 25.3 with n = 16. Solution to 1: We just plug in the components to the formula for the first of the "Important Cases" from the lecture notes: A 2 - u2 - 15 which gives, for 0 = Z = 25.3, a = .lo, and ag - - with a 90% probability, or the confidence interval covers the truth with a 90% probability (since p isn't random-the CI is random). 2. Based on the information in (I), find a one-sided lower 90% confidence limit for p. Also, find a one-sided upper 90% confidence limit for p. Solution to 2: A one-sided lower 90% confidence limit for p and one-sided upper 90% confidence limit for p correspond to a lower/upper bound on p. We just adjust the probabilities for the upper and lower limit so that the 10% error is on the lower or upper end:
and for the upper end Notice that the one-sided confidence intervals are actually shorter from f than the two-sided version, because we've allocated all of the error to just one side. 3. For a confidence interval of the form given by (f - V1(l - B) 5, f + W1 (1 - 2) 31, derive a formula for the sample size required to obtain an interval of specified length, A. If a2 = 9, then what sample size is needed to achieve a 90% confidence interval of length 2? Solution to 3: The length of a confidence interval is just the difference between the two bounds: If for a2 = 9, a = .lo, and X = 2, we will need a sample size as follows: So, we'll need a sample of size 25 in order to achieve a confidence interval of length 2 (or slightly less). 4. Suppose now that a2 is unknown. Find a 90% confidence interval for p if 3 = 25.3 and s2 = 15.21 with n = 16. Solution to 4: We now just need to use the formula which using the t distribution's quantiles: where the n - 1 subscript denotes the degrees of freedom parameter of the t distribution. This gives, for 0 = f = 25.3, a! = .lo, and s2 = 15.21

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Note that this 90% confidence interval is slightly wider due to two factors: 15.21>15 and the t distribution's critical values (quantiles) are wider than the Normal's. Most of the difference here is due to the t distribution's wider critical values. 5. Based on the data in (4), find a 99% confidence interval for a2. Also show for n = 14. (Hint: What is the distribution of \$? It's a Xi-1, which happens to approach the Normal distribution, but use the Xip1 for this part.) Solution to 5: To construct this confidence interval, we're going to use case 4 from the "Important Cases" in the lecture notes. We want to find constants a and b such that In other words, we're going to focus on a symmetric confidence interval. We just need to find a and b such that Fs2(b) = 0.995 and FS2 (a) = 0.005.
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9 - MIT OpenCourseWare http/ocw.mit.edu 14.30 Introduction...

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