Exam II Solutions,Ma442 Spr04

# Exam II Solutions,Ma442 Spr04 - 1(15 pts The following...

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Unformatted text preview: 1. (15 pts.) The following histogram contains 62 data values -— namely 62 intervals, in days, between “serious” earthquakes, across the globe, from December 16, 1902 through March 4, 1977‘. 20 10 . 0 II. - -_ o 200 400 600 800 1000 12001400 1600 1800 2000 Interval a. The data are best modeled by a(n) (choose one): (E ”mm JP ”10'“ch 5m- ﬂzi Frequency . ii. Gamma _ ,L. A I iii. Normal ( ' e . __ .. iv. Poisson , 'A '- HO 9 AMA " —— Sty-o b. The average of the 62 values is approximmly 440 days. Suppose a “serious” earthquake has just occurred. Estimate the chance that the next earthquake occurs within 44 days. ' WEWFHIVA \$91M 5 +1») 3: ++ -‘/ 1< 1 7““ an ‘f‘l‘D ° . +4 - /f-1.p 9C . g: -6 D ~ 5 2: .. Z I 'f' I ‘ The data are from Hand, DJ. (1994), A Handbook of Small Datasets, Chapman & Hall, p. 203 where a “serious" earthquake is carefully deﬁned. .— - ‘7’ A " I " d 4.. . 0?; L 2. (25 pts.) On the evening of 4/20/04 I [email protected] numbers between 0 and l on my TI- 83 calculator. As a partial test on the randomness of my calculator, test the hypothesis that numbers are equally likely to fall in any one of the 5 intervals [0.00,0.20), [0.20,0.40), [0.40, 0.60), [0.60,0.80),[0.80, l .00) by answering the questions below. . Data: A _:mm— Number of Data Poin @waa —' _-—_i_ 5» {a —_—__ ;. __ a. Compute (Pearson’s) 12 statistic. b. Give a rough sketch of the 12 curve that the 12 statistic follows if the null hypothesis of equally likely intervals is true. c. What is the degrees of freedom number for the 12 curve? d. On your graph (part b), label the numeric value along the horizontal axis for which there is area 0.05 to the right. e. On your graph (parts b,d), indicate the region of 2'2 statistic values for which you accept the null hypothesis. Likewise, indicate the region of 12 statistic values for which you reject the null hypothesis. f. Come to a conclusion about the null hypothesis. That is, do you accept or reject the null hypothesis (use a = 0.05) ? @ (6—5» ‘ cw. ‘ (we? 6%. ‘ 64-40‘ ?<°"54‘2"’H‘ - 5p )+ 5'0 )4. 7’ + f) 4. 5—D - 14o : 3. (10 pts.) A random variable is a Beta random variable if it has density 11“ + ﬂ) (1-! _ -| f(x) = —I"(a)F(ﬁ) x (l x)’ 0 < x < l 0 otherwise for some parameters a and ,6. Random variables having such a density have the following mean and variance: a 0,1 ___ “)6 a+,a’ (a+ﬂ+1)(a+ﬂ)’ ﬂ: Subpose you have a random variable which you believe to follow some Beta (e.g. by checking the histogram). If I = 0.60 is the average of the data values and s1 = 0.28 is the sample variance of the data values state, but do n_ot try to solve, the equation(s) you would need to solve to ﬁnd the method of moments estimates for a and [3. far A; if a“ 5 774m” (5/ 55f {\ (X = 0.52 00-? ﬁ 4. (15 pts.) Here is a histogram of the waist circumferences of 507 physically active (several hours of exercise per week) individualszz W 30 i i” ' u 10 0 5 5 75 5 ﬁ 105 116 WalstClrcurfetmee . a. Waist girth is best modeled by a(n) (circle one): VIE-WWW IF ”A“ 6"“ ‘ L K 36’, '- ‘ I 2 e i. E n .nential A 1 X t ’ ,_ 1 2 «'6’ = x iii. Normal 5" 2-75 2. iv. Poisson 0‘ F1: 5 b. Suppose that the average of the 507 data values is about 93.3 cm and the standard variance, s2, is about 100.6 cmz. Write down, but don’t evaluate, an integral which estimates the proportion of such physically active people who have waist circumference of at least 95 cm. " - 5L* 7078' PM a ' 27. gm = 2&- 3’ 94.530 52. _ ’0 I 0(—l ”W? Jo: F00 2 75) - S (6"; m; X a ?5 at“ o( is rep/“‘4 by ﬁrm: 34.9% F “ ' (’93., = L078 2 Collected by Dr. Grete Heinz, Dr. Louis J. Peterson, San Jose State University, and people trained by them. For further infomiation, see the article by Heinz, G., Peterson, L., :lohnson, R. and Kerk, C. (2003), “Exploring relationships in body dimensions”, Journal of Statistics Education, vol. 11, no. 2, jse/vl ln2/datasets.heinz.html. 5. (15 pts.) Suppose3 that cars am've mid-afternoon to the McDonalds on Mt. Rushmore at a rate of roughly 1 per minute. Also suppose that cars arrive independently. a. Estimate the chance that no cars arrive in 2 minutes.( A: 1) b. Estimate the chance that at most 3 cars arrive in 5 minutes. C ?e 7.5) @ X: #éwtwls/ Vitﬁn—Q 3:1 V(1€=?>)=2,57 e L = 4" a .354— @ XsLarHVJS/ VWSBV. 3‘5 FOX 53) : VKSC>D)+ Pl2>0+ P(z‘='z) +ﬂx=3) I - , 3 Zog'}+. 5 e 54— f? < 5+ ,5; Q a! It 2: 3,! (hp-5+ 375+ L23) e”: a 0.2650 3 Based upon data collected from 2:13-2:58 pm on October 26, 2000. 6. (10 pts.) Let X1,X2,X3 be i.i.d. random variables having mean u and variance 02. Compute the mean squared error of [I (an estimate of ,u) where 2X1 —)(2 + 3X3 4 . 2 , @ 7262..» \$5620 + {— 552:) ,0. ,2: mm) '3- szxo 41:1.) wwﬁé) mm) 2 “L = V9442) + [Eb-’1) ‘M31 = go- 7. (10 pts.) Multiple Choice: / a. If ms.e.(i¢,) < ms.e.(ﬁz) where [1, and it, are estimate of ,u, then which estimate is to be preferred for estimating p (circle one): b. If we decide to restrict our attention to unbiased estimates/of a parameter using .- A.“ .e. measure, then this is equivalent to choosing the unbiased estimate having the smallest varian w 1' False (circle one). . 7" c. To check whether a random quantity can reasonably be modeled by some exponential, especially in the small sample case, we can check an exponential probability plot. For us to reasonably trust an exponential model, the points in this plot should follow what kind of pattern (circle one)? il ii - onential ‘j m. ‘ormal iv. None of the above ...
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