# 9781305251809 Chapter 14 Problem 14E.png - Chapter 14 ...

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Term
Fall
Professor
TamreCardoso
Tags
Null hypothesis, Probability theory, probability density function, Discrete probability distribution, Chi square distribution

Unformatted text preview: Chapter 14, Problem 14E Problem The article "Feeding Ecology of the Red-Eyed Vireo and Associated Foliage-Gleaning Birds" (Ecological Monographs, 1971: 129-152) presents the accompanying data on the variable X = the number of hops before the first flight and preceded by a flight. The author then proposed and fit a geometric probability distribution [ p(x) = P(X = x) = p . q for x = 1, 2, . .., where q = 1 - p] to the data. The total sample size was n = 130. 2 3 4 5 6 7 8 9 10 11 12 Number of Times x 48 31 20 9 6 5 4 2 1 1 Observed a. The likelihood is (p- . q) . ... . (p- . q) = p2-" . q". Show that the mle of p is given by p = (Ex, - n)/Ex,, and compute p for the given data. b. Estimate the expected cell counts using p of part (a) [expected cell counts = n . (p)"- . q for x = 1, 2, . . . 1, and test the fit of the model using a x2 test by combining the counts for x = 7, 8, . .., and 12 into one cell (x 2 7). . The likelihood is (pi- . q) . ... .(pl . q) = pEx- - q". Show that the mle of p is given by p = (Ex; - n)/Ex; and compute p for the given data. b. Estimate the expected cell counts using p of part (a) [expected cell counts = n . (p)' ' . q for x = 1, 2, . . . ], and test the fit of the model using a y' test by combining the counts for x = 7, 8, . .., and 12 into one cell (x 2 7). Step-by-step solution Step 1 of 4 It is given that the data on variable X =the number of hops before the first flight and proceeded by a flight. a) Find the value of p for the given data. The Geometric probability mass function for the given data as shown below. P(X = x) = p" q, for x = 1,2, ...., Here, q=1-p The likelihood function for p obtained below. L(P) = [1p q Apply logarithm on both sides log [L(p) ] = log( P2 -q) (Ex-n) logp+nlog(1-p) Apply derivation with respect to p and equating them to 0. alog (L) Ex-n n =0 op P I-P 2x-n(1-p)-mp = 0 P(1-P) 2x-n-pyx+np-"p= 0 P(1-P Ex-n Ex Total of the x values is, 48(1)+31(2)+20(3)+9(4)+6(5)+5(6) 1+7(4)+8(2)+9(1)+10(1)+11(2)+12(4) =363 Thus, Lx-n 363-130 363 = 0.642 Therefore, the value of p is [0.642] Step 2 of 4 b) The null and alternative hypothesis can be defined as follows: Ho : The data not fit the g ric distribution. H, : The data fit the geometric distribution. Consider the calculations shown in the below table: Expected Observed (O-E) (0-E) (0-E) E=n(P)" q 48.00 130(0.642) (1-0.642) 1.46 2.1316 0.04580 = 46.54 2 31.00 29.87 .13 1.2769 0.04275 3 20.00 19.18 0.82 0.6724 0.03506 9.00 12.31 -3.31 10.9561 0.89002 6.00 9 -1.90 3.61 0.45696 6 5.0 5.07 -0.07 0.0049 0.00097 11.00 325 7.75 60.0625 18.48077 130.00 124.12 5.88 34.5744 0.27856 Step 3 of 4 x' = ) (observed - estimated expected)" estimated expected S (0-E) (0.04580+0.04275 +0.03506 +0.89002 +0.45696 + 0.00097 +18.48077 +0.27856] x' =20.231 Step 4 of 4 The degrees of freedom is, k-1-m=7-1-1 From the chi-square distribution table, the critical value at 0.05 level of the significant for 5 degrees of freedom is 11.07 Here, it can be observed that the calculated value is greater than the critical value. Hence, the null hypothesis should be rejected. Therefore, it can be concluded that there is sufficient evidence to support that the data fit the geometric distribution....
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