Solution to Assignment 1, winter 2015
Introduction to Statistics/MAT2375
1. Textbook problems.
5.4-22 : (a) We have
E(exp(t(X1 + X2 ))) = (1 − 2t)−r/2
and
E(exp(tX1 )) = (1 − 2t)−r1 /2 .
Since X1 and X2 are independent
E(exp(t(X1 + X2 ))) = E(exp(tX1 ))E(exp(tX2 )).
Therefore
(1 − 2t)−r/2 = (1 − 2t)−r1 /2 E(exp(tX2 )).
This gives
E(exp(tX2 )) = (1 − 2t)(r−r1 )/2 .
Therefore
(b)
X2 ∼ χ2 (r − r1 ).
5.5-16 : (a) Since d.f. = 8, from the t table we have
t0.025 = 2.306.
(b)
√
√
P (X̄ − 2.306S/ 8 < µ < X̄ + 2.306S/ 8) = 0.95.
5.6-14 : We need to have
20
X
P(
Xi > a) = 0.2
i=1
Since from C.L.
20
X
Xi
approximate
∼
N (200, 80).
i=1
Approximately
P20
P
a − 200
i=1 Xi − 200
√
> √
80
80
This gives
!
= 0.2.
a − 200
√
= 0.842.
80
Solve for a to get
a = 207.53
6.2-10 :
> x
[1] 9.5 10.7 8.3 9.8
[16] 9.9 10.9 12.3 9.2
[31] 2.9 9.8 5.7 8.2
[46] 9.3 8.2 9.9 11.6
[61] 6.6 7.3 16.7 11.0
>hist(x)
> summary(x)
Min. 1st Qu. Median
2.900
8.675
9.400
>mean(x)
[1] 9.421875
9.1
9.3
8.1
8.7
9.4 9.6 11.9
9.3 10.5 9.4
8.8 9.7 8.1
5.0 9.9 6.3
Mean 3rd Qu.
9.422 10.220
2
9.5 12.6 10.5 8.9 11.4 12.0 12.4
9.4 8.2 10.4 9.3 8.7 9.8 9.1
8.8 10.3 8.6 10.2 9.4 14.8 9.9
6.5 10.2 8.8 8.0 8.7 8.9 6.8
Max.
16.700
>var(x)
[1] 4.32872
>sd(x)
[1] 2.080558
>mean(x)-sd(x)
[1] 7.341317
> sum(x<mean(x)+sd(x)&x>mean(x)-sd(x))/length(x)
[1] 0.75
> sum(x<mean(x)+2*sd(x)&x>mean(x)-2*sd(x))/length(x)
[1] 0.9375
> stem(x)
The decimal point is at the |
2
4
6
8
10
12
14
16
|
|
|
|
|
|
|
|
9
07
35683
01122236777888991123333444455678889999
223455790469
0346
8
7
See the histogram in page 4. In a normal population these probabailities
are 68% and 95% (75% and 93.75% based on the data in this question).
6.3-6. We have the p.d.f. for Wr
g(w) =
n!
n!
(F (w))r−1 f (w)(1−F (w))n−r =
wr−1 (1−w)n−r
(r − 1)!(n − r)!
(r − 1)!(n − r)!
This shows Wr ∼ β(r, n − r + 1) distribution. Therefore
E(Wr ) =
r
.
n+1
(see the cover pages of your textbook for details). For part (a) take r = 1
and r = n as special cases.
bf Question 2.
3
20
0
10
Frequency
30
Histogram of x
5
10
15
x
>x=rbinom(20000,1,0.46)
>X=matrix(x,ncol=200)
>m=apply(X, 1,mean)
> m
[1] 0.545 0.475 0.505
[13] 0.490 0.475 0.455
[25] 0.460 0.480 0.405
[37] 0.500 0.445 0.465
[49] 0.460 0.505 0.475
[61] 0.445 0.480 0.485
[73] 0.460 0.465 0.490
[85] 0.445 0.505 0.495
[97] 0.455 0.470 0.455
0.425
0.415
0.455
0.435
0.430
0.460
0.460
0.490
0.455
0.505
0.415
0.485
0.400
0.490
0.435
0.515
0.465
0.460
0.430
0.480
0.490
0.395
0.400
0.455
0.480
0.415
0.515
0.440
0.490
0.445
0.465
0.435
0.425
0.450
0.460
0.480
0.475
0.495
0.385
0.470
0.480
0.470
0.450
0.465
0.485
0.425
0.525
0.475
0.440
0.515
0.415
0.445
0.420
0.390
0.485
0.465
0.485
>hist(m)
>plot(qqnorm(m))
As we can see the histogram is symmetric and the qqplot is alomst linear
and it confirms normality. This is in fact confirming central limit theorem.
4
0.485
0.450
0.490
0.505
0.465
0.455
0.470
0.470
0.470
0.445
0.485
0.440
0.465
0.445
0.435
0.410
Normal Q−Q Plot
Sample Quantiles
Frequency
10
0
0.40
5
0.45
15
0.50
20
25
0.55
Histogram of m
0.40
0.45
0.50
0.55
−2
m
−1
0
1
2
Theoretical Quantiles
Question 3. As you can notice the qqplot and histogram confirms the observations are not coming from a normal distribution.
> r1=rnorm(500,10,5)
> r2=rnorm(500,20,5)
> r=r1*r2/(r1+r2)
> par(mfrow=c(2,1))
> sum(r<8&r>6)/500
[1] 0.348
> hist(r)
> plot(qqnorm(r))
5
100
50
0
Frequency
150
Histogram of r
−10
−5
0
5
10
r
10
5
0
−5
−10
Sample Quantiles
Normal Q−Q Plot
−3
−2
−1
0
1
Theoretical Quantiles
6
2
3