Question

A particular area in a town suffers a high burglary rate. A sample of 100 streets is taken,
and in each of the sampled streets, a sample of six similar houses is taken. The table below
shows the number of sampled houses, which have had burglaries during the last six months.

No. of houses burgled x	0	1	2	3	4	5	6
No. of streets f	39	38	18	4	0	1	0

(i) (a) State any assumptions needed to justify the use of a binomial model for the number of
houses per street which have been burgled during the last six months.
(b) Derive the maximum likelihood estimator of X bar, the probability that a house of the
type sampled has been burgled during the last six months.
(c) Determine the probabilities for the binomial model using your estimate of X bar, and,
without doing a formal test, comment on the fit.

(ii) An insurance company works on the basis that the probability of a house being burgled
over a six-month period is 0.18. Carry out a test to investigate whether the binomial
model with this value of p provides a good fit for the data.

NOTE:THERE WAS A MISTAKE ON (b) IT IS ESTIMATOR OF X BAR NOT ESTIMATOR OF P but i have corrected it

Answer 1

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RGui (32-bit) Windows Help View Misc Packages Edit File STOP t-c (39 , 38,18,4,0,1, 0) > > prob-NULL > tcalculation of expected frequencies > for (i in 1:7) + (7-i): prob [i]=choose (6, i-1)* (.18(i-1 ) )* ( 1-.18) + +} > #creating data frame for the given data >mat data.frame (n =0:6, obsFreq f, expFreq 100*prob) > #expected frequencies are less than 5..So the classes are merged > mat$groups <- cut (mat$n, breaks c(0:3,7), right FALSE, labels c ("0","1", "2","3 or more")) > mat expFreq 39 30.400667142 n obsFreq groups 1 0 38 40.039903066 18 1 1E6352 3 or more 5 4 1.058787936 0.092966746 3 or mo 6 5 re 7 6 0.003401222 3 or more > obsFreqGroup <-1 > expFregGroup <- tapply (matsexpFreq, mat$groups, aum) > data.frame (obs obsFreqGroup, exp expFreqGroup) pply (matsobsFreq, matsgroups, sum) obs exp C 39 30.400667 38 40.039903 2 18 21.973118 5 7.586312 3 or more > #chisq test for goodness of fit with p.18 > saveChiTest <-chisq.test (obsFreqGroup, p expFreqGroup/00) > saveChiTest Chi-squared test for given probabilities data: obsFreqGroup X-squared = 4.1365, df = 3, p-value 0.2471