MATH1725 Introduction to Statistics: Worked examples

Question (lecture 14).
If Var[X] = 4 and Var[Y ] = 9 and corr(X,Y ) = 0.1, obtain cov(X + 2Y,X − Y ).
Answer: 26
Question (lecture 14).
If X ∼ N(1,9) and Y ∼ N(1,16) and X and Y are independent, what is pr{|X − Y | < 5}
Answer: 27
Question (lecture 14).
Suppose that X 1 ,X 2 ,... ,X n are independent and identically distributed random variables with
common mean E[X i ] = µ and common variance Var[X i ] = σ 2 . Let ¯X denote the mean of the X i
with mean µ and variance σ 2 /n. By writing (X i − ¯X) 2 = ({X i − µ} − { ¯X − µ}) 2 and expanding
the bracket, show that
S 2 = 1 n∑
(X i −
n − 1
¯X) 2
has mean E[S 2 ] = σ 2 .
Answer: 28
i=1
Question (lecture 15).
In January 2011 Durham police reported a “significant increase in road accidents during December
[2010] ...mainly due to severe weather”. During December 2010 there were 336 reported collisions,
up from 308 in the previous December. By fitting a suitable model to these data, test whether
there is indeed a significant difference in the number of accidents between December 2009 and
December 2010. Source: http://www.bbc.co.uk/news/uk-england-12261462
(This is harder than you would get in the examination – I have not done anything like this in the
module. Use the approximation that if X ∼ Poisson(µ) and µ is large, then X ≈ N(µ,σ 2 = µ).)
Answer: 29
Question (lecture 15).
Two independent samples gave values 3, 6, 5, 2 for sample 1 and 2, 2, 3, 3, 5 for sample 2.
Assuming that the samples come from independent normal distributions with known variances 4
and 1 respectively, test at the 5% level whether the difference in mean equals zero against the
alternative that it does not equal zero.
Answer: 30
26 cov(X, Y ) = corr(X, Y ) × p Var[X]Var[Y ] so cov(X, Y ) = 0.6 and cov(X +2Y, X −Y ) = Var[X]+cov(X, Y ) −
2Var[Y ] = −13.4.
27 X − Y ∼ N(0,25) so we want pr{−5 < X − Y ≤ +5}. pr{X − Y ≤ 5} = Φ(1) = 0.8413 so pr{X − Y > 5} =
0.1587 and answer is 0.6826.
28 Recall that Var[X i] = E[(X i − µ) 2 ] = σ 2 and Var[ ¯X] = E[( ¯X − µ) 2 ] = σ 2 /n. Also notice that ({X i − µ} −
{ ¯X − µ}) 2 = (X i − µ) 2 + ( ¯X − µ) 2 − 2(X i − µ)( ¯X − µ) and P i (Xi − µ) = n( ¯X − µ). Thus P P
i (Xi − ¯X) 2 =
i (Xi − µ)2 − n( ¯X − µ) 2 . Now take expectations.
29 A suitable model is to assume accidents occur randomly and independently in time. Assuming a constant level
of car usage we are using a Poisson process model. Thus the number X 1 of accidents in December 2010 satisfies
X 1 ∼ Poisson(µ 1). Similarly the number X 2 of accidents in December 2009 satisfies X 2 ∼ Poisson(µ 2). We want to
test whether µ 1 = µ 2. For µ i large, X i ≈ N(µ i, µ i) for i = 1,2 independently so X 1 − X 2 ≈ N(µ 1 − µ 2, µ 1 + µ 2).
Thus if H 0 is true, and µ 1 = µ 2 = µ,
X1 − X2
U = √ ≈ N(0,1).
2µ
Assuming the null hypothesis is true, we would estimate µ by ˆµ = 1 (336+308) = 322. Thus, replacing µ by ˆµ = 322
2
we obtain U = 1.103. Since |U| < 1.96, we accept the null hypothesis at the 5% level. The observed increase in
accidents was not significant!
30 n 1 = 4, ¯x 1 = 4, σ1 2 = 4, n 2 = 5, ¯x 2 = 3, σ2 2 = 1. Testing H 0: µ 1 − µ 2 = 0 vs. H 1: µ 1 − µ 2 ≠ 0. Test statistic is
15