Chapter 17. The Expected Value and Standard Error
Using a Box Model we want to answer the questions:
- What should be the EXPECTED SUM of the DRAWS? This is the EXPECTED VALUE for the SUM.
- When the game is actually played, how much will my ACTUAL sum of draws differ from the EXPECTED sum of draws? This is the STANDARD ERROR for the Sum.
Key Words
Sum of the Draws – Is found by drawing tickets at random from a box and then adding up the numbers on the tickets.
Expected Value – the value expected for the sum of draws made at random with replacement from a box.
Standard Error – The amount the numbers drawn at random differ from the expected value.
EXPECTED VALUE
The expected value from the sum of draws made at random with replacement from a box equals
(number of draws)(average of box)
Example 3
Find the expected value for the sum of 100 draws with replacement from the box:
-2, 4, 4, 7, 7
Solution
The average of the box is
= 4
There are 100 draws. Therefore, the expected value from the sum of draws with replacement is
(number of draws)(average of box)
(100)(4) = 400
STANDARD ERROR
The likely size of the chance error in a sum of draws process is given by the standard error or SE. For instance, the sum of the draws from a box will be around the expected value, give or take a standard error or so.
There is a formula to use in computing the SE for a sum of draws made at random with replacement from a box. It is called the square root law.
Standard Error for sum = SquareRoot(Number of Draws)SDofBOX
Example 4, Exercise B, Page 293, #1
Solution
(a)Expected value = (number of draws)(average of box)
Number of draws = 100
Average of box = = 4.
Expected value = (100)(4) = 400
To find the standard error, apply the following equation:.
Standard error =
The deviations from the average are (-3), (-2), (-1), 0, 1, 2, 3. The SD of the box is
= 2
The standard error is = 20.
(b)The sum of the draws will be around 400, give or take 20 or so.
(c)You should guess that the sum is 400 and that the sum would be off by 20 or so.
USING THE NORMAL CURVE
The purpose of this section is to find out what is the chance that the sum of the draws will be in a given range when the number of draws from a box is large and is made with replacement. Provided the number of draws is sufficiently large, the normal curve can be used to figure chances from the sum of the draws. To find the area underneath the normal curve, find the expected value and standard error and then covert to standard units.
If we take the sum of 25 draws made at random with replacement from a box, what is the chance that the sum will fall between 50 and 100? The numbers in the box are: 0, 2, 3, 4 and 6. The sum would fall within the range of 0 and 150 because (25 draws)0 = 0 and (25 draws)6 = 150. To find the chance, convert to standard units and use the normal curve. Standard units say how many SE’s a number is away from the expected value.
Expected value = (number of draws)(average of box)
Number of draws = 25
Average of box = =3
Expected value = (25)(3) = 75
Standard Error = (SD of box)
SD of box = 2
Standard Error = (2) = 10
The expected value for the sum is 75 and the SE is 10, so 100 is 2.5 SE’s above the expected value
= +2.5
and 50 is -2.5 SE’s below the expected value
= -2.5.
The interval from 50 to 100 is the interval within 2.5 SE’s of the expected value, so the area underneath the curve according to the z-table is 99%.
WHAT IS THE DIFFERENCE BETWEEN THE SE AND THE SD?
The SD measures the spread between the numbers on the tickets in the box. The SE measures the spread between the SUMs of the numbers on the tickets in the box.
SUM of DRAWS FOLLOWS the NORMAL CURVE!!!!
The SUMs of the DRAWS will follow the NORMAL CURVE as the histogram EVEN when the numbers on the tickets in the box do NOT. That is what the central limit theorm in probability theory says. Look at the spreadsheet Central Limit Theorem Spreadsheet to see that this is true.
In the spreadsheet we draw 50 times from a box with 18 1’s and 20 0’s, and we take the sum of the 50 draws. We do this 500 times and draw the histogram for the sum of draws. You can see in the picture that it looks like the normal curve.
A SHORT-CUT For THE SD OF A BOX
There is a short-cut to finding the SD for boxes with only two kinds of tickets. This shortcut applies ONLY to boxes with only two different numbers on the tickets. There can be many tickets in the box but they can only have some tickets with one of the numbers and the rest of the tickets have the other number. The numbers do NOT have to be ZERO and ONE. When the tickets in the box show only two different numbers there is a larger number (“big”) and a smaller number (“small”), the SD of the box equals
For example, take the box 1, 1, 1, 5. Since there are only two different numbers in the box, use the short-cut.
1.73
Example 5, Page 304, # 1
Solution
(a)The draws can be as small as 1001=100 or as large as 10010=1,000.
(b)Expected value = (number of draws)(average of box)
Number of draws = 100
Average of box = = 7
Expected value = (100)(7) = 700
Standard error =
SD of box = = 3
Standard error = =30
Converting to Standard Units:
=+1.67
= --1.67
Referring to the z-table, the area underneath the curve when the z-values are –1.67 and 1.67 is approximately 90%.
Example 6, Page 304, # 3
Solution
Using the short-cut equation:
(a) (iii) because[1-(-2)]= 3
(b) (i) because (16-15)=
(c)(v) because [1-(-1)]= 2
(d)(iv) because (1-0)=
(e)(ii) because (2-0)= 2
CLASSIFYING AND COUNTING
Some chance processes involve counting. The box model needs to be setup correctly and then the square root law can be used. In a counting problem we are interested in COUNTING the number of times something of interest happens. We COUNT that as drawing a ONE out of the box, and when it doesn’t happen, that is like drawing a ZERO out of the box. So the box in such problems is a ZERO-ONE box. The steps for this are
STEP 1) CLASSIFY the event of interest as a ONE. When it doesn’t happen that is a ZERO.
STEP 2) The number of ONEs in the box should be set up so that the chances of drawing a ONE should be the same as the chance the event of interest occurs.
The following example shows how this is done.
Example 7, Pg. 303, #2
Solution
The expected value for the number of fives is
The box model will be applied to figure out the deviation from the expected value of 20 to 8 and 32 respectively.
The box model for this problem
becomes
because we are concerned with what is the chance of getting 5’s. Notice that on each draw, the number of 5’s either goes up by 1 or stays the same:
- 1 is added to the count if the draw is 5
- 0 is added to the count if the draw is 1, 2, 3, or 4
The short-cut method for calculating the SD of the box involves two numbers, in this case the two numbers are 0 and 1. Using the short-cut method, the SD is
SD =
The SE is
To find out the chance of getting between 8 and 32 tickets marked “5”, use the following logic. The expected number of 5s is 20. So we need the area under the normal curve centered at 20 from 8 to 32.
FIRST Convert 8 and 32 to standard units to get two z values.
z1=(8-20)/4=-3, z2=(32-20)/4=3.
NOW go to the z-table and find the area under the Standard normal curve from –3 to +3. This is about 99%, and that is the chances of getting between 8 and 32 5s.
.