1. SA3202, Solution for Tutorial 3
# of Boys Total
# of Families
Expected
Probability
H0: The number of boys follows a binomial distribution
The estimated probability that a born child is a boy =.5159
T=.6270, df=5-1-1=3 ( 1 free parameter). 95% table= Do not reject H0. That is, it seems that the number of boys follow a binomial distribution.
# of Goals Total
Frequency
Expected
Probability
The estimated lambda=total goals scored/ total number of matches=1.39
After combining the last three categories, T=.6088, G=.59845, df=5-1-1=3, 95% table=7.815
Don’t reject H0. It seems that the number of goals scored follows a Poisson distribution.
5/7/2019
SA3202, Solution for Tutorial 3

2. SA3202, Solution for Tutorial 3
Interval [0,.3) [.3, .6) [.6,.9) [.9,1.2) [1.2, 1.5] Total
Observed
Expected
For uniform distribution over [0,1.5], each interval has probability 1/5=.2 so that for each interval, the expected frequency is 25*.2=5. There is no any free parameter involved.
T=10.8, G= , df=5-1-0=4, 95% table= Reject H0. Therefore, it is unlikely that the rupture follows a uniform distribution over [0,1.5].
Interval [10,20) [20,30) [30,40) [40, 50] Total
Observed
Expected
Probability
Since the hypothetical distribution is normal distribution, which has a range over the whole
Real line, we should regard [10,20) as “20 and less”, [40,50] as “40 and over”, so that
p1=P(X<20)=P(Z<(20-30)/10= -1)=P(Z>1)=.1587
p2=P(20<X<30)=P(-1<Z<0)=P(Z<0)-P(Z<-1)= =.3413.
By Symmetry, p3=p2=.3413, p4=p1=.1587, no free parameter.
T=3.1169, G= , df=4-1=3, 95%table= Do not reject H0. It seems that the N(30,100) fits the data well. (Other interval division schemes are acceptable too).
5/7/2019
SA3202, Solution for Tutorial 3