Download STAT 111 Recitation 6

STAT 111 Recitation 6
Xin Lu Tan
[email protected]
October 18, 2013
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Miscellaneous
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Please turn in homework 5.
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Please pick up homework 6 and the graded homework 4.
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Please check your grade and let me know during the next
recitation if there are any grade discrepancies (please show me
your graded homework as well).
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Keep a copy of your answers to question 3 in homework 6, as
these will be needed for one of the questions on homework 7.
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Midterm: October 21 (Monday), 6-8pm.
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Bring a PEN (not a pencil) and a pocket calculator that is
capable of doing powers, for example computing something
like 56 . NO cheat sheets are allowed.
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Please arrive at the exam at 5.50pm, and the exam will be
handed out then.
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Exam time is 1 hour 45 minutes, and it will start at 6.15pm.
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You should check for typos, ambiguities, errors, etc in the
exam and raise questions about these before 6.15pm.
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Homework 4: Problem 2
Consider the random variable T4 , the sum of the numbers turning
up on FOUR rolls of a fair die. From your answer to Homework 3,
Question 3, find the probability distribution of T4 .
Let T4 = Y1 + Y2 , where Y1 is the sum of the numbers turning up
on rolls 1 and 2 of the die, and Y2 is the sum of the numbers
turning up on rolls 3 and 4 of the die. The probability distribution
of Y1 is (after considering all combination of numbers in rolls 1
and 2, which is the same as the probability distribution of Y2 )
vk
P(vk )
2
3
4
5
6
7
8
9
10
11
12
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
So the probability distribution of T4 is
vk
P(vk )
4
5
6
7
8
1
1296
4
1296
10
1296
20
1296
35
1296
···
···
23
24
4
1296
1
1296
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Homework 4: Problem 6
Use the mean and variance of X1 , the random number to turn up
on one roll of a fair die, together with the relevant magic formulas
(given in class) to find quickly the mean and variance of the
average of the four numbers to turn up on four rolls of this die.
We have
35
mean of X1 = 3.5,
variance of X1 =
12
We know that if Xi , 1 ≤ i ≤ n are i.i.d. with mean µ and variance
σ 2 , then for
X1 + X2 + · · · + Xn
X̄ =
,
n
mean of X̄ = µ,
Here n = 4, µ = 3.5, σ 2 =
mean of X̄ = 3.5,
variance of X̄ =
35
12 ,
σ2
.
n
so we have
variance of X̄ =
35/12
35
= .
4
48
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Other comments on Homework 4
For Problem 8 and 9, when calculating z-scores, you should divide
by standard deviation σ, not variance σ 2 .
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Let X ∼ N(µ, σ 2 ), i.e. X is a random variable having a
normal distribution with mean µ and variance σ 2 .
Facts:
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I
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If X ∼ N(µ, σ 2 ), let Z = X −µ
σ , then Z ∼ N(0, 1).
P(Z ≤ −z) = P(Z ≥ z) = 1 − P(Z ≤ z).
To find P(a < X < b), do the following to translate
probability statement about X into probability statement
about Z :
X −µ
b−µ
a−µ
< √
< √
P(a < X < b) = P √
2
2
2
σ
σ
σ
a−µ
b−µ
=P
<Z <
σ
σ
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Other comments on Homework 5, Problem 3
Let T4 be the sum of the numbers turning up on four rolls of a fair
die, what is the probability that T4 ≤ 15 (please use normal
approximation)?
You can do
P(T4 ≤ 15)
or you can argue that since the smallest possible value of T4 is 4,
the required probability is
P(4 ≤ T4 ≤ 15).
Both answers will be accepted.
Note: Actually n = 4 is not big enough for us to use normal
approximation, but in any case if you were asked to compute
probabilities for T4 without being given the exact probability, and
if you were not asked to compute them explicitly, then you can just
use normal approximation.
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Estimation of a binomial parameter θ
If X ∼Binomial(n, θ), and we observe x, what can we say about θ?
We know that mean of X = nθ, variance of X = nθ(1 − θ). If we
let P = Xn , then mean of P = θ, variance of P = θ(1−θ)
.
n
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We estimate θ by p = xn .
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An approximate 95% confidence interval for θ is
r
r
p(1 − p)
p(1 − p)
to
p+2
,
p−2
n
n
or the conservative
1
p−√
n
to
1
p+√ .
n
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Estimation of a mean µ
If X1 , X2 , . . . , Xn are i.i.d. with mean µ and variance σ 2 , and we
observe x1 , x2 , . . . , xn , what can we say about µ?
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We know that mean of X̄ = µ, variance of X̄ = σn .
x1 +x2 +···+xn
.
n
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We estimate µ by x̄ =
I
An approximate 95% confidence interval for µ is
s
x̄ − 2 √
n
where
s2 =
to
s
x̄ + 2 √ ,
n
x12 + x22 + . . . + xn2 − nx̄ 2
.
n−1
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Estimation of the difference of two binomial parameters
θ1 − θ2
If X1 has a binomial distribution with index n1 and parameter θ1 ,
and X2 has a binomial distribution with index n2 and parameter θ2 ,
and we observe X1 = x1 , X2 = x2 , what can we say about θ1 − θ2 ?
We know that if Pi = Xnii , then mean of Pi = θi , variance of Pi =
pi (1−pi )
≤ 4n1 i , for i = 1, 2.
ni
I We estimate θ1 − θ2 by
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p1 − p2 , where pi =
xi
ni ,
for i = 1, 2.
An (conservative) approximate 95% confidence interval for
θ1 − θ2 is
r
r
1
1
1
1
p1 − p2 −
+
to
p1 − p2 +
+ .
n1 n2
n1 n2
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Estimation of the difference between two means µ1 − µ2
If X11 , X12 , . . . , X1n1 are i.i.d. with mean µ1 and variance σ12 , and
X21 , X22 , . . . , X2n2 are i.i.d. with mean µ2 and variance σ22 , and we
observe x11 , x12 , . . . , x1n1 , x21 , x22 , . . . , x2n2 ,, what can we say about
µ1 − µ2 ?
σ2
We know that mean of X̄i = µi , variance of X̄i = nii , for i = 1, 2.
xi1 +xi2 +···+xini
ni
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We estimate µ1 − µ2 by x̄1 − x̄2 , where x̄i =
i = 1, 2.
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An approximate 95% confidence interval for µ1 − µ2 is
s
s
s12
s22
s12
s2
+
to
x̄1 − x̄2 + 2
+ 2,
x̄1 − x̄2 − 2
n1 n2
n1 n2
where
si2 =
, for
2 + x 2 + . . . + x 2 − n x̄ 2
xi1
i i
i2
ini
.
ni − 1
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Practice Problem:
Let X be a normal random variable with mean 5 and variance 9.
What is the number a such that P(X < a) = 0.9332?
The following “Z ” chart gives “less than” probabilities for positive values
of z, i.e. P(Z ≤ z) for Z ∼ N(0, 1).
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Practice Problem:
Let X1 ∼ N(3, 9), X2 ∼ N(4, 16), and X1 and X2 are independent,
what is the probability distribution of 2X1 − X2 ?
Facts:
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If X ∼ N(µ, σ 2 ), then aX ∼ N(aµ, a2 σ 2 ).
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If X1 ∼ N(µ1 , σ12 ), X2 ∼ N(µ2 , σ22 ) and X1 , X2 are
independent, then X1 + X2 ∼ N(µ1 + µ2 , σ12 + σ22 ).
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Review: Variance Formula for Discrete Random Variable
If X is a random variable that takes on possible values v1 , v2 . . . , vk
with respective probabilities P(v1 ), P(v2 ), . . . , P(vk ), then
µ = v1 P(v1 ) + v2 P(v2 ) + · · · + vk P(vk )
=
k
X
vi P(vi )
i=1
σ 2 = (v1 − µ)2 P(v1 ) + (v2 − µ)2 P(v2 ) + · · · + (vk − µ)2 P(vk )
k
X
=
(vi − µ)2 P(vi )
σ2 =
=
i=1
v12 P(v1 )
k
X
OR
+ v22 P(v2 ) + · · · + vk2 P(vk ) − µ2
vi2 P(vi ) − µ2
i=1
The two formulae for σ 2 are equivalent (give same answer)!
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My advice on preparing for the midterm
1. Know all the formulas in the “What you should know” notes
given by Professor Ewens by heart.
2. Know how to use Z-chart and binomial chart.
3. Know everything about the binomial random variable and
normal random variable that has been discussed in class so far.
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My advice on preparing for the midterm
1. Know all the formulas in the “What you should know” notes
given by Professor Ewens by heart.
2. Know how to use Z-chart and binomial chart.
3. Know everything about the binomial random variable and
normal random variable that has been discussed in class so far.
Good luck on the exam!!!!
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