Elementary Statistics and
Inference
22S:025 or 7P:025
Lecture 21
1
Elementary Statistics and
Inference
22S:025 or 7P:025
Chapter 17
2
17 – Expected Value and Standard Error
A.
The Expected Value
In a chance process (probability experiment) the
outcomes of the chance process vary, but cluster
around the “expected value” or the average.
„
Example: Toss a coin 100 times
times.
1
P( H ) =
2 on each toss – you would expect about
50% of the outcomes to be heads – this is the
expected value.
Suppose in 100 tosses you obtained 57 heads, the
difference between 57 and 50 is called chance error –
(+7).
3
1
17 – Expected Value and Standard Error
(cont.)
Example: Suppose you selected 100 draws from the box,
with replacement:
1
1
1
5
„
You would expect about ¾ of the 100 draws to be a “1”,
and about ¼ of the draws to be a “5”.
„
The expected sum would be (25 x 5) + (75 x 1) = 125 +
75 = 200.
4
17 – Expected Value and Standard Error
(cont.)
The formula for the Expected Sum is:
E ( sum) = (average of the number in the box) ∗ (number of draws)
⎛1+1+1+ 5 ⎞
E ( sum) = ⎜
⎟ × 100 = 200
4
⎝
⎠
E ( sum) = n ⋅ (avg of box) = n ⋅ X
Note: see description on page 289
5
17 – Expected Value and Standard Error
(cont.)
Example (page 289):
Example 1: Suppose you are going to Las Vegas to play
Keno. Your favorite bet is a dollar on a single number.
When you win, they give you the dollar back and two
dollars more. When you lose, they keep the dollar.
There is a chance in 4 to win. About how much should
you expect to win (or lose) in 100 plays, if you make this
bet on each play?
$2.00
-$1.00
-$1.00
-$1.00
P(W ) =
1
4
n=100 plays
6
2
17 – Expected Value and Standard Error
(cont.)
average of the box =
2 + (−3)
= $ − .25
4
E (sum) = 100 · avg of box = 100 (-$
( $.25)
25) = $
$-25
25.00
00
You would expect to lose $25 in 100 plays.
7
17 – Expected Value and Standard Error
(cont.)
Exercise Set A – (p. 290) #1, 2, 3, 4, 5
1.
Find the expected value for the sum of 100 draws at
random with replacement from the box –
(a)
0
1
1
6
(b)
-2
-1
0
(c)
-2
-1
3
(d)
0
1
1
2
8
17 – Expected Value and Standard Error
(cont.)
3.
Someone is going to play roulette 100 times, betting a
dollar on the number 17 each time. Find the expected
value for the net gain. (See pp. 283-284.)
9
3
17 – Expected Value and Standard Error
(cont.)
6.
A game is fair if the expected value for the net gain
equals 0: on the average, players neither win nor lose.
A generous casino would offer a bit more than $1 in
winnings if a player staked $1 on red-and-black in
pay
y to make
roulette and won. How much should theyy p
it a fair game? (Hint: Let X stand for what they should
pay. The box has 18 tickets X and 20 tickets -$1 .
Write down the formula for the expected value in terms
of X and set it equal to 0.)
10
17 – Expected Value and Standard Error
(cont.)
18 ⋅ X 20
−
=0
38
38
18 X − 20 = 0
18 X = 20
Expected gain =
X = $1.11
11
17 – Expected Value and Standard Error
(cont.)
B.
The Standard Error of the Sum
Given the box:
0
2
3
4
6
Avg of box = 15 / 5 = 3
12
4
17 – Expected Value and Standard Error
(cont.)
„
In 25 draws with replacement,
Expected Sum = E(sum)=n·avg of box
E(sum)=25 · 3=75
„
The actual sum will be
Sum=expected value + chance error
„
The chance error is a function of the standard deviation
of the box.
„
The chance error, called Standard Error (SE) is:
SE = n ⋅ (SD of box)
See Note p. 291
13
17 – Expected Value and Standard Error
(cont.)
„
In the box above, the SD of box = 2.
(0 − 3) 2 + (2 − 3) 2 + (3 − 3) 2 + (4 − 3) 2 + (6 − 3) 2
5
9 + 1 + 0 + 1 + 9 20
2
=
=4
S =
5
5
S =2
S2 =
ƒ SE = 25 ⋅ ( 2) = 10
14
17 – Expected Value and Standard Error
(cont.)
For the box example
0
2
3
4
6
n = 25 draws with
replacement
The E ( sum) = n ⋅ avg box = 25(3) = 75
SE ( sum) = n ⋅ SD of box = 25 (2) = 10
15
5
17 – Expected Value and Standard Error
(cont.)
„
In 25 draws with replacement, we would expect the sum
of the draws to be 75 give or take 10.
Note: The sum of draws is likely to be around the expected
value give or take the standard error.
value,
error
Note: Observed values in a chance process are rarely
more than 2 or 3 standard errors away from the
Expected Value.
16
17 – Expected Value and Standard Error (cont.)
Exercise Set B – (p. 293) #1, 2, 4
17
18
6
19
17 – Expected Value and Standard Error
(cont.)
C.
ƒ
The Normal Curve
For a large number of draws from a box, with
replacement, the “sum of the draws” is approximately
normally distributed.
distributed
20
17 – Expected Value and Standard Error (cont.)
Example: Suppose 25 draws with replacement are made
from the box, with tickets as shown:
0
2
3
4
6
X = avg of box = 3,
S = SD of box = 2
E(sum) = n ⋅ avg of box = 25 (3) = 75
SE (sum) = n ⋅ SD of box = 25 (2) = 10
Now the sums are approximately normally distributed –
with mean = 75, and S = 10.
ƒ Find probability (chance) that the sum for any 25 draws
will be between 50 and 100.
21
7
17 – Expected Value and Standard Error
(cont.)
SE(sum)=10
98.76%
sum
50
E(sum)=75 100
-2.50
0
Z=
2.50
sum − mean
SE
ƒ
Using Normal Curve table – we find
98.76% of the
scores between ±2.50
Standard Deviations from the
mean.
22
17 – Expected Value and Standard Error
(cont.)
Example 2. In a month, there are 10,000 independent
plays on a roulette wheel in a certain casino. To
keep things simple, suppose the gamblers only
stake $1 on red at each play. Estimate the chance
$250 from these
that the house will win more than $
plays. (Red-or-black pays even money, and the
house has 20 chances in 38 to win.
Solution:
What is probability that casino will gain
$250.00 or more from 10,000 plays of roulette.
23
17 – Expected Value and Standard Error
(cont.)
20
$1.00
18
-$1.00
avg of box = expected gain on one play =
20 18
−
= $.
$ 05
38 35
20(1 − .05) 2 + 18( −1 − .05) 2
38
20(.95) 2 + 18( −1.05) 2
2
S =
38
S 2 = .9972
S2 =
24
8
17 – Expected Value and Standard Error
(cont.)
SD of box = $.998 ≈ $1.00
E ( sum) = n ⋅ avg of box
E ( sum) = 10,000(.05) = $500.00
SE ( sum) = n ⋅ SD of box = 10,000 (1) = 100
25
17 – Expected Value and Standard Error
(cont.)
Find Probability (Chance) that sum is greater than $250.
SE(sum)=100
X
250 E(sum)=500
Z
0
26
17 – Expected Value and Standard Error
(cont.)
250 − 500 − 250
=
= −2.50
100
100
Area between ± 2.50 = 98.76%
Z=
Area less than - 2.50 =
ƒ
100 - 98.76
98 76
= .62%
2
So about 99.32% for casino to win more than $250 in
10,000 roulette plays.
Exercise Set C – (pp. 296-297) #1, 2, 3, 4, 5
27
9
28
29
30
10