. From 3000 observations on flood tide levels, the USOEP summarized the data in
ID: 2923044 • Letter: #
Question
. From 3000 observations on flood tide levels, the USOEP summarized the data in a table Levels 0-0.5 0.5-1.0 1.0-1.5 1.5-2.0 2.0-2.5 2.5 Counts 1194 304 209 134 90 1069 We want to test whether or not these levels have an exponential distribution with -2 (ie. P X > -e-2t (a) Calculate the appropriate expected values and check to see whether the 2 approximation is appropriate for this set of bins (b) How many degrees of freedom will our 2 distribution have? (e) What does our test conclude? (d) Suggest how we might improve the test in this situation.Explanation / Answer
(a) Appropriate expected values will be calculated by calculating cumulative probability over the interval.
so Here distribution function with = 2
f(t) = e-2t
F(t) = 1 - e-2t
so for the interval (0 , 0.5)
F(0 < t < 0.5) = F(0.5) - F(0) = e-2*0 - e-2*0.5 = 1 - e-1 = 0.6321
P(0.5 < t < 1) = F(1) - F(0.5) = e-2*0.5 - e-2 *1 = e-1 - e-2 = 0.2325
P(1 < t <1.5) = F(1.5) - F(1) = e-2*1 - e-2 *1.5 = e-2 - e-3 = 0.0855
P(1.5 < t < 2) = F(2) - F(1.5) = e-2*1.5 - e-2 *2 = e-4 - e-3 = 0.0315
P(2 < t < 2.5) = F(2.5) - F(2) = e-2*2 - e-2 *2.5 = e-4 - e-5 = 0.0116
P( t > 2.5+) = F(2.5+) = e-2*2.5 = 3-5 = 0.0067
No the chi-square appoximation is not appropriate at all for this set of bins.
X2 = 55330.8
(b) Degrees of freedomm = 5
(c) Here we can conclusivedly conclude that the given distribution is not an exponential distribution.
(d) The only problem with this data set is the observed values for 2.5+ . That shows that is totally incorrect interval formation and wrong prediction of lembda
Levels Counts Probability Expected (O-E)^2/E 0-0.5 1194 0.6321 1896.3 260.0988 0.5-1.0 304 0.2325 697.5 221.9961 1.0-1.5 209 0.0855 256.5 8.796296 1.5-2.0 134 0.0315 94.5 16.51058 2.0-2.5 90 0.0116 34.8 87.55862 2.5+ 1069 0.0067 20.1 54735.88 3000 1.0 55330.8Related Questions
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