231hw7solutionsF11

231hw7solutionsF11 - Will Landau September 22, 2011 STAT...

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Will Landau September 22, 2011 STAT 231 Problem Set 7 Solutions Note: for most of the book problems, I borrowed the solution from the instructor’s manual. Let me know if you would like me to elaborate on anything. Exercise 7.1 (Prof. Vardeman’s Ex. 1) . I start by making a new data table with 1000 rows, and column labeled X1, . .., X10, Mean(n=2), Mean(n=5), and Mean(n=10): For each of the Xi’s, I ﬁll the column with a formula: I select ”Random” from the drop-down menu indicated below: 1

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I select Random Uniform as my formula or just write ”Random Uniform()” in the formula box: Once I’ve done that for all the Xi’s, I calculate the means for n=2, 5, and 10 with the same formula utility: For n=2, for example, I write ”Mean(X1,X2)” as my formula: 2
Now, I generate the output by going to Analyze ¿ Distribution: I click and drag X1, Mean(n=2), Mean(n=5), and Mean(n=10) from the list on the left to the white box next to the ”Y, Columns” button and click ”OK”: 3

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I click the red triangle next to the ”Distributions” heading and select ”Stack” to stack the histograms. To make sure all the plots are on the same scale, I click that red triangle from before and select ”Uniform Scaling”: 4
Then, I show cdf plots like so: Now, for n=1, 2, 5, and 10, I want to approximate the probability that X falls within 1 / 12 n of 1/2. That is, if F is the cdf of X , I want to estimate F (1 / 2 + 1 / 12 n ) - F (1 / 2 - / 12 n . For each of the following values of n, you should get the corresponding values for 1 / 2 ± 1 / 12 n : 5

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Value 1 / 2 - 1 / 12 n 1 / 2 + 1 / 12 n n = 1 0.2113 0.7887 n = 2 0.2959 0.7041 n = 5 0.3709 0.6291 n = 10 0.4087 0.5913 Use the crosshair tool to calculate the corresponding cdf values: Notice that in the above screenshot, the crosshair tool is selected (highlighted blue) at the top of the output window. I got the following results. Yours should be similar: Value F (1 / 2 - 1 / 12 n ) F (1 / 2 + 1 / 12 n ) P (1 / 2 - 1 / 12 n X 1 / 2 + 1 / 12 n ) n = 1 0.2333 0.7833 0.55 n = 2 0.2056 0.8222 0. 6166 n = 5 0.1778 0.8389 0.6611 n = 10 0.1611 0.8444 0.6833 Now, the far right of the above column estimates P ( - 1 Z 1) = 0 . 6826. Our estimates get better as n increases. Furthermore, the estimates increase as n increases. This is because as we increase n, the tails of the empirical pdfs thin out. The tails are thick for small n because X is an average of uniformly distributed random variables. Exercise 7.2
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231hw7solutionsF11 - Will Landau September 22, 2011 STAT...

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