FIT5197 Statistical Data Modelling

Question one

1. P(A U B U C) = P(A) + P(B) + P(C) – P(A ∩ C) – P(A ∩ B) – P(B ∩ C) + P(A ∩ B ∩ C)
2. e to eliminate double counting, we add the probabilities of the individual events then subtract the intersection of each two events. Some outcomes that cuts across all the events are also removed in the process hence the intersection of all the events is added back.

P(A U B U C U D) = P(A) + P(B) + P(C) + P(D) – P(A ∩ B) – P(A ∩ C) – P(A ∩ D) – P(B ∩ C) – P(B ∩ D) – P(C ∩ D) + P(A ∩ B ∩ C) + P(A ∩ B ∩ D) + P(A ∩ C ∩ D) + P(B ∩ C ∩ D) - P(A ∩ B ∩ C ∩ D).

i.e the reasoning is the same as in the previous case but we now subtract the intersection of all the four events as it is already covered in the intersections per 3 events.

Question two

P(x) = 1 -

Q(P) =  =  since we are only interested in the functional form

This is also the inverse CDF since it is not invertible.

10% quantile, Q(P) =  = 0.1

2x – x^2 = 0.2

x =

= 1.894 and 0.1056.

Question three

H(G) = 1/k i.e since the probability of the groups is the same.

H(I) = 1/nk

H(I/G) = H(G) * H(I)

H(G,I) = H(G) + H(I).

Question four

Mean (x) =

= q + 1-q

Variance(x) = E(x^2) –(E(x))^2

= q + 1-q –

Question five.

Check the required R code attached.

Question six

First we construct the Z score as:

Z =

    = 2.67

From the Z table, this corresponds to a probability of 0.0038 hence probability that IQ > 140.

Θ ~ bin(40, 0.0038)

Θ =

    = 0.00097146.

Hence the probability of more than two people having an IQ of more than 140 is 0.00097146

Question seven

∼N(0,)

∼N(0,1)

∴  ∼χ2 (1) = Γ(1/2,2)

∼Γ(1/2,2) = Γ(1/2,2)

E ~  = Γ(3/2,2)

Mean =( 3/22)

        = 3
Variance = 3/2

               = 6

Question eight

The samples are from normally distributed populations hence are also normally distributed.

We test the hypothesis for the difference in mean.

(A) =

         = 11.17

(B) =

         = 11.9875

n(A) = 10

n(B) = 8

H? :

H1: 

At 5% level of significance, the critical values of Z are -1.96 and +1.96 hence we reject the null hypothesis if  is not met.

Z =

Z =

   = - 28. 19.

We reject the null hypothesis since -28.19 is less than -1.96 hence we conclude that the level of contamination between the lakes are different.

Question nine

Since there more heart attacks from the placebo taking people than aspirin taking ones, it is clear that taking aspirin does not increase chances of heart attack.

Question ten

Computing the correlation coefficient,

= 0.4838( refer to the excel sheet attached as regards where the data is obtained.)

Height and salary are correlated as shown by the above result. To confirm whether the relationship is by chance or significant, we use t test as follows

t = r

         = 0.4838

         = 1.748

At t =1.75, p< 0.05 hence we conclude that a lawyers height is related to his salary

References

1. Craig, L. H. Mathematical Statistics.
2. Dennis, W., William, M., & Richard, L. (2007). Mathematical Statistics with Applications.W.S publishers.
3. Effron, B. (n.d.). The Efficiency of Logistic Regression compareed to normal discriminantanalysis. Vol 70.
4. Mood, & Crawbill. introduction to the theory of Statistical Inference.
5. R, V. H., J, W. M., & A, T. C. (2003). Introduction to Mathematical Statistics, 6th edition.Prentice Hall.
6. S, J. S., & E, H. (1995). Calculus: one and several variables, 7th ed.
7. (2004). Calculus concepts and contexts : Multivariate and Single variable.Brooks/Cole Pub.
8. Thomas, & Finney. (1988). Calculus and analytic geometry 7th ed.Addison-Wisley.