Summer 2020, Statistics Midterm, Solutions

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Question 1

###
1.1

.056
###
1.2

> .216+.604
[1] 0.82

###
1.3

> .604/(.604+.056)
[1] 0.9151515

###
1.4

No.
Previous two answers would have to be the same.

###
1.5

> .056 + .604
[1] 0.66

X ~ Bern(.66)

E(X) = p = .66

###
1.6

p(1-p)  
> .66*(1-.66)
[1] 0.2244

###
1.7
> sqrt(.2244)
[1] 0.4737088


###
1.8

> .216/(.216+.124)
[1] 0.6352941


P(Y=1 | X=0) = .635

Y is more likely to be big when X is big so the move up and down together.

corr =  0.3450683

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Question 2

###
2.1

> .1*.12 + .9*.14
[1] 0.138

###
2.2

> .1^2 * 5.25 + .9^2 * 9.76 + 2*.1*.9*3.063
[1] 8.50944

###
2.3

> sqrt(8.50944)
[1] 2.917094

###
2.4

> .138 + 2*2.9*c(-1,1)
[1] -5.662  5.938

###
2.5
> pnorm(0,0.138,2.917094)
[1] 0.4811341

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Question 3

###
3.1

3.5

###
3.2

12.25

###
3.3

> sqrt(12.25)
[1] 3.5

###
3.4

.95

###
3.5

.68

###
3.6

.84

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Question 4

###
4.1

Basically yes.
Histogram has the bell shape and no obvious pattern in sequence plot.
Maybe variance goes down at the end?
Maybe right tail a bit heavier.


###
4.2

> .629/sqrt(107)
[1] 0.06080773

###
4.3

> .168 + 2*0.06080773*c(-1,1)
[1] 0.04638454 0.28961546

###
4.4

Huge

###
4.5

yes, 4.6 percent is still a lot different from 0.

###
4.6

> .168 + 0.06080773*c(-1,1)
[1] 0.1071923 0.2288077


###
4.7

for 95:
> qnorm(.05/2)
[1] -1.959964

for 68:
> qnorm(.32/2)
[1] -0.9944579
> 

for 99:
> qnorm(.01/2)
[1] -2.575829

so you use 2.6.

> .168 + 2.6*0.06080773*c(-1,1)
[1] 0.009899902 0.326100098


###
4.8
> .168 + 2*.629*c(-1,1)
[1] -1.090  1.426


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Question 5

###
5.1

> 520/1000
[1] 0.52

###
5.2

> sqrt(.52*(1-.52)/1000)
[1] 0.01579873

###
5.3

> .52  + 2*.0158*c(-1,1)
[1] 0.4884 0.5516


###
5.4

It is big in that we don't know what side of .5 we are on.
It is small in that we know we are pretty close to .5.

###
5.5

> sqrt(.52*(1-.52)/9000)*2
[1] 0.01053249

To get it 3 time smaller you need the sample size 9 times bigger.

##########
Question 6

###
6.1

Yes.
Pretty noisy but looks like a bit of a linear trend.


###
6.2

> 3.71340 + 0.27148*1
[1] 3.98488

###
6.3

sigmat ~ .5 , +/1 is 1.

4 +/- 1

pretty big, which makes sense given the scatter plot.

###
6.4

> 0.27148 + 2*0.02837*c(-1,1)
[1] 0.21474 0.32822

pretty small!!

###
6.5

p-val is 0 => reject

###
6.6

> 3.71340/165.119
[1] 0.02248924


###
6.7

dy = bhat xy

> 0.27148 * 2
[1] 0.54296

which  is quite a bit.

###
6.8


Seems like there is strong evidence for a weak relationship given the 
small confidence interval for beta1 and the big sigmahat.


#########
Question 7

###
7.1

Only way to get this is if the price went up P(U)= .6

###
7.2

would need 2 ups in a row P(UU) = .6*.6 = .36

###
7.3

The price would have to go up by 2:

P(U) = .6

###
7.4

p:  99,102
pr: .4, .6

###
7.5

p(UU) = .36, P2 = 104
p(DD) = .4^2 = .16, P2=98
p(DU) = .4*.6 = .24, P2 = 101
p(UD) = .4*.6 = .24, P2 = 101

p: 98, 101, 104
pr: .16, .48, .36

###
7.6

8 possibilites for the price paths!!

P(UUU) = .6^3 = .216, P3 = 106
3 ways to have 2 U and 1 D : 3*.6*.6*.4 = .432, P3 = 103
3 ways to have 2D and 1 U: 3*.4*.4*.6 = .288, P# = 100
P(DDD) = .4^3 = .064, P3 = 97

p: 97, 100, 103, 106
pr: .064, .288, .432, .216


###
7.7

If P3 > 101 C is P3-101, otherwise it is 0

c: 0, 2, 5
p(c): .352 (.288+.064), .432, .216
p