Download Elementary Statistics and Inference Elementary Statistics and

Elementary Statistics and
Inference
22S:025 or 7P:025
Lecture 24
1
Elementary Statistics and
Inference
22S:025 or 7P:025
Chapter 18 (cont.)
2
Chapter 18 (cont.)
E.
The Scope of the Normal Approximation
„
The probability sums for tickets drawn from a box
model also approaches a Normal Distribution as the
number of draws with replacement increases
increases.
„
If the box differs from 50% of tickets with one value,
and 50% with a different value – we need more draws.
(See page 319)
3
1
Chapter 18 (cont.)
„
Suppose
1
10
9 1
3
SD =
⋅ =
10 10 10
avg =
9
0
1
1
4
Chapter 18 (cont.)
„
Note the approximation of sums as the number of draws
increases – the approximation to the Normal Distribution
gets better as the number of draws increases.
5
Chapter 18 (cont.)
6
2
Chapter 18 (cont.)
„
The normal approximation works for sums from any box
model – note example on page 321.
7
Chapter 18 (cont.)
8
Chapter 18 (cont.)
„
Average of the box=
SD of the box =
1+ 2 + 3
=2
3
(1 - 2) 2 + ( 2 − 2) 2 + (3 − 2) 2
1+1
2
=
=
3
3
3
SD of the box = .67
E ( sum) = n ⋅ avg of the box = (25)(2) = 50
SE ( sum ) = n ⋅ SD of the box = 25 .67 = 4.09
9
3
Chapter 18 (cont.)
„
The sum is approximated by the Normal Distribution
even if the box with the tickets is unusual.
10
Chapter 18 (cont.)
„
Note the approximation to the Normal Distribution as the
number of draws increases.
11
Chapter 18 (cont.)
12
4
Chapter 18 (cont.)
Exercise Set C – (pp. 324-325)
2 – (p. 324)
A biased coin is tossed 10 times and has one chance in
10 off landing
l di heads.
h d It iis ttossed
d 400 times.
ti
E
Estimate
ti t
the chance of obtaining 40 heads.
„
„
Because the number of tosses is large, even though the
chances of obtaining a head is small – the Normal
Approximation applies.
13
Chapter 18 (cont.)
1
1
9
0
(1×1 + 9 × 0) =
1
10
10
1 9
3
SD of the box = (1 - 0)
×
=
= .30
10 10 10
avg of the box =
⎛1⎞
E (number of Heads) = n ⋅ avg of the box = (400)⎜ ⎟ = 40
⎝ 10 ⎠
SE (number of Heads) = n ⋅ SD of the box = 400 (.3) = 6
14
Chapter 18 (cont.)
SE=6
39.5
40.5
M=40
-.083
ƒ
.083
Number of
Heads
X − Mean
Z=
SE
Area between ±Z=.083 ~6%
15
5
Chapter 18 (cont.)
F.
Central Limit Theorem
Review Exercises – (pp. 327-329) #1, 2, 3, 4, 5, 8, 9, 10, 11
16
17
18
6