Answer for lab 9; Question 3.10
Q1
The Geom() has expression f (y) = (1 )y1 , for independent Y1 , . . . , Yn , the likelihood
is
f (y1 , . . . , yn |) = n (1 )
yi n
.
Q2
From gure 2.5, we see that when a = b = 1, Beta(a, b) distribution is an Uniform one
Answers to Lab 5
Parts 17
R code to generate 10000 binomial samples from Bin (n = 40, = 0.5) and
for each sample calculate , se( ) and a condence interval for :
b = rbinom(10000, 40, 0.5)
pi.hat = b / 40
se.pi.hat = sqrt(pi.hat*(1-pi.hat)/40)
z = qnorm(0.
Answers to the R parts of Lab 3
Part 5
To generate 1000 samples of size n = 10 from the N (0, 22 ) distribution, we
rst create a 1000 10 matrix to hold the data:
y = matrix(0, nrow = 1000, ncol = 10)
where the argument 0 will be repeated to initialize the
Brief Answers to Lab 1
Q2
To make histogram, use R command hist(x, breaks=20). The second argument breaks
species the number of bars to draw on the histogram.
Figure 1: histogram for question 2
Q3
To obtain sample mean and variance, use mean() and var().
Name (Family, Given): Signature:
Student ID:
UBC-Vancouver STAT 305 Quiz #3 2011.03.18 Time: 45 minutes
Aids: Formula sheet (8% x 11, 2-sided); calculator
Instructor: W.J. Welch This examination has 4 pages.
1. An actuary models the distribution of th
2-58
CHAPTER 2. ESTIMATION
2.11 [Quiz #2, 2010-11, Term 1] Yu et al. (Science, 2006, pp. 16001603)
studied the statistical properties of gene expression. They argued that
for a particular gene in E. coli , the number of expression events (y) per
cell cycl
Answer for lab 12; Question 5.6
Q1
The null hypothesis is that the two factors are independent. So we have
H0 : ij = i. .j for all i, j Ha : ij = i. .j
Q2Q5 (not part of the lab)
The estimated row probabilities are 1. = 1634/2132 and 2. = 498/2132. The es
Answer for lab 6
Q1
In method of moment estimation, we use sample mean to estimate the population mean, so
Y = . Using the sample, moment = 370/298 1.24.
Q2
The Poisson distribution has expression f (y) = (y exp()/y!, for independent Y1 , . . . , Yn ,
the