Expert answer:Probability Analysis

Expert answer:A General Manger of Harley-Davidson has to decide on the size of a new facility. The GM has narrowed the choices to two: large facility or small facility. The company has collected information on the payoffs. It now has to decide which option is the best using probability analysis, the decision tree model, and expected monetary value.Options:

Facility
Demand Options
Probability
Actions
Expected Payoffs

Large
Low Demand
0.4
Do Nothing
($10)

Low Demand
0.4
Reduce Prices
$50

High Demand
0.6

$70

Small
Low Demand
0.4

$40

High Demand
0.6
Do Nothing
$40

High Demand
0.6
Overtime
$50

High Demand
0.6
Expand
$55

Determination of chance probability and respective payoffs:

Build Small:

Low Demand
0.4($40)=$16

High Demand
0.6($55)=$33

Build Large:

Low Demand
0.4($50)=$20

High Demand
0.6($70)=$42

Determination of Expected Value of each alternative Build Small: $16+$33=$49 Build Large: $20+$42=$62
au_bus499_statisticalterms.docx

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SAMPLING MEAN:
DEFINITION:
The term sampling mean is a statistical term used to describe the properties of statistical
distributions. In statistical terms, the sample mean
from a group of observations is an
estimate of the population mean . Given a sample of size n, consider n independent random
variables X1, X2… Xn, each corresponding to one randomly selected observation. Each of these
variables has the distribution of the population, with mean
and standard deviation
. The
sample mean is defined to be
WHAT IT IS USED FOR:
It is also used to measure central tendency of the numbers in a database. It can also be said that
it is nothing more than a balance point between the number and the low numbers.
HOW TO CALCULATE IT:
To calculate this, just add up all the numbers, then divide by how many numbers there are.
Example: what is the mean of 2, 7, and 9?
Add the numbers: 2 + 7 + 9 = 18
Divide by how many numbers (i.e., we added 3 numbers): 18 ÷ 3 = 6
So the Mean is 6
SAMPLE VARIANCE:
DEFINITION:
The sample variance, s2, is used to calculate how varied a sample is. A sample is a select number
of items taken from a population. For example, if you are measuring American people’s weights,
it wouldn’t be feasible (from either a time or a monetary standpoint) for you to measure the
weights of every person in the population. The solution is to take a sample of the population, say
1000 people, and use that sample size to estimate the actual weights of the whole population.
WHAT IT IS USED FOR:
The sample variance helps you to figure out the spread out in the data you have collected or are
going to analyze. In statistical terminology, it can be defined as the average of the squared
differences from the mean.
HOW TO CALCULATE IT:
Given below are steps of how a sample variance is calculated:
• Determine the mean
• Then for each number: subtract the Mean and square the result
• Then work out the mean of those squared differences.
To work out the mean, add up all the values then divide by the number of data points.
First add up all the values from the previous step.
But how do we say “add them all up” in mathematics? We use the Roman letter Sigma: Σ
The handy Sigma Notation says to sum up as many terms as we want.
•
•
Next we need to divide by the number of data points, which is simply done by
multiplying by “1/N”:
Statistically it can be stated by the following:
•
This value is the variance
EXAMPLE:
Sam has 20 Rose Bushes.
The number of flowers on each bush is
9, 2, 5, 4, 12, 7, 8, 11, 9, 3, 7, 4, 12, 5, 4, 10, 9, 6, 9, 4
Work out the sample variance
Step 1. Work out the mean
In the formula above, μ (the Greek letter “mu”) is the mean of all our values.
For this example, the data points are: 9, 2, 5, 4, 12, 7, 8, 11, 9, 3, 7, 4, 12, 5, 4, 10, 9, 6, 9, 4
The mean is:
(9+2+5+4+12+7+8+11+9+3+7+4+12+5+4+10+9+6+9+4) / 20 = 140/20 = 7
So:
μ=7
Step 2. Then for each number: subtract the Mean and square the result
This is the part of the formula that says:
So what is xi? They are the individual x values 9, 2, 5, 4, 12, 7, etc…
In other words x1 = 9, x2 = 2, x3 = 5, etc.
So it says “for each value, subtract the mean and square the result”, like this
Example (continued):
(9 – 7)2 = (2)2 = 4
(2 – 7)2 = (-5)2 = 25
(5 – 7)2 = (-2)2 = 4
(4 – 7)2 = (-3)2 = 9
(12 – 7)2 = (5)2 = 25
(7 – 7)2 = (0)2 = 0
(8 – 7)2 = (1)2 = 1
We need to do this for all the numbers
Step 3. Then work out the mean of those squared differences.
To work out the mean, add up all the values then divide by how many.
First add up all the values from the previous step.
= 4+25+4+9+25+0+1+16+4+16+0+9+25+4+9+9+4+1+4+9 = 178
But that isn’t the mean yet, we need to divide by how many, which is simply done by multiplying
by “1/N”:
Mean of squared differences = (1/20) × 178 = 8.9
This value is called the variance.
STANDARD DEVIAITON:
DEFINITION:
This descriptor shows how much variation or dispersion from the average exists.
The symbol for Standard Deviation is σ (the Greek letter sigma).
It is calculated using:
In case of a sample the ‘N’ in this formula is replaced by n-1.
WHAT IT IS USED FOR:
It is used to determine the expected value. A low standard deviation indicates that the data points
tend to be very close to the mean (also called expected value); a high standard deviation indicates
that the data points are spread out over a large range of values.
HOW TO CALCULATE IT:
To determine the standard deviation, you need to take the square root of the variance.
EXAMPLE PROBLEM:
Let’s look at the previous problem and compute the standard deviation. The standard deviation
as mentioned earlier is nothing more than the measure of dispersion (spread). It can be calculated
by taking the square root of the variance. In case of the previous problem where the variance was
8.9, its corresponding standard deviation would be the square root of 8.9 which is 2.983
σ = √(8.9) = 2.983…
HYPOTHESES TESTING:
DEFINITION:
Hypothesis testing is a topic at the heart of statistics. This technique belongs to a realm known as
inferential statistics. Researchers from all sorts of different areas, such as psychology, marketing,
and medicine, formulate hypotheses or claims about a population being studied.
WHAT IT IS USED FOR:
Hypothesis testing is used to determine the validity of these claims. Carefully designed statistical
experiments obtain sample data from the population. The data is in turn used to test the accuracy
of a hypothesis concerning a population. Hypothesis tests are based upon the field of
mathematics known as probability. Probability gives us a way to quantify how likely it is for an
event to occur. The underlying assumption for all inferential statistics deals with rare events,
which is why probability is used so extensively. The rare event rule states that if an assumption is
made and the probability of a certain observed event is very small, then the assumption is most
likely incorrect.
The basic idea here is that we test a claim by distinguishing between two different things:
1. An event that easily occurs by chance
2. An event that is highly unlikely to occur by chance.
If a highly unlikely event occurs, then we explain this by stating that a rare event really did take
place, or that the assumption we started with was not true.
HOW TO USE THE TEST FOR DECISION MAKING PURPOSES:
1. Formulate the null hypothesis (commonly, that the observations are the result of pure
chance) and the alternative hypothesis (commonly, that the observations show a real effect
combined with a component of chance variation).
2. Identify a test statistic that can be used to assess the truth of the null hypothesis.
3. Compute the P-value, which is the probability that a test statistic at least as significant as the
one observed would be obtained assuming that the null hypothesis were true. The smaller the value, the stronger the evidence against the null hypothesis.
4. Compare the -value to an acceptable significance value (sometimes called an alpha value).
If
, that the observed effect is statistically significant, the null hypothesis is ruled out, and
the alternative hypothesis is valid.
EXAMPLE OF HYPOTHESIS TESTING (TWO-TAIL TEST)
If you are told that the mean weight of 3rd graders is 85 pounds with a standard deviation of 20
pounds, and you find that the mean weight of a group of 22 students is 95 pounds, do you
question that that group of students is a group of third graders?
•
The z-score is ((x-bar) – µ)/(*sigma*/(n^.5)); the numerator is the difference between the
observed and hypothesized mean, the denominator rescales the unit of measurement to
standard deviation units. (95-85)/(20/(22^.5)) = 2.3452.
•
The z-score 2.35 corresponds to the probability .9906, which leaves .0094 in the tail
beyond. Since one could have been as far below 85, the probability of such a large or
larger z-score is .0188. This is the p-value. Note that for these two tailed tests we are
using the absolute value of the z-score.
•
Because .0188 < .05, we reject the hypothesis (which we shall call the null hypothesis) at the 5% significance level; if the null hypothesis were true, we would get such a large zscore less than 5% of the time. Because .0188 > .01, we fail to reject the null hypothesis
at the 1% level; if the null hypothesis were true, we would get such a large z-score more
than 1% of the time.
DECISION TREE:
DEFINITION:
A schematic tree-shaped diagram used to determine a course of action or show a statistical
probability. Each branch of the decision tree represents a possible decision or occurrence. The
tree structure shows how one choice leads to the next, and the use of branches indicates that each
option is mutually exclusive.
WHAT IT IS USED FOR:
A decision tree can be used to clarify and find an answer to a complex problem. The structure
allows users to take a problem with multiple possible solutions and display it in a simple, easyto-understand format that shows the relationship between different events or decisions. The
furthest branches on the tree represent possible end results.
HOW TO APPLY IT:
1. As a starting point for the decision tree, draw a small square around the center of the left side
of the paper. If the description is too large to fit the square, use legends by including a number
in the tree and referencing the number to the description either at the bottom of the page or in
another page.
2. Draw out lines (forks) to the right of the square box. Draw one line each for each possible
solution to the issue, and describe the solution along the line. Keep the lines as far apart as
possible to expand the tree later.
3. Illustrate the results or the outcomes of the solution at the end of each line. If the outcome is
uncertain, draw a circle (chance node). If the outcome leads to another issue, draw a square
(decision node). If the issue is resolved with the solution, draw a triangle (end node). Describe
the outcome above the square or circle, or use legends, as appropriate.
4. Repeat steps 2 through 4 for each new square at the end of the solution lines, and so on until
there are no more squares, and all lines have either a circle or blank ending.
5. The circles that represent uncertainty remain as they are. A good practice is to assign a
probability value, or the chance of such an outcome happening.
Since it is difficult to predict at onset the number of lines and sub-lines each solution generates,
the decision tree might require one or more redraws, owing to paucity of space to illustrate or
represent options and/or sub-options at certain spaces.
It is a good idea to challenge and review all squares and circles for possible overlooked solutions
before finalizing the draft.
EXAMPLE:
Your company is considering whether it should tender for two contracts (MS1 and MS2) on offer
from a government department for the supply of certain components. The company has three
options:
•
•
•
tender for MS1 only; or
tender for MS2 only; or
tender for both MS1 and MS2.
If tenders are to be submitted, the company will incur additional costs. These costs will have to
be entirely recouped from the contract price. The risk, of course, is that if a tender is
unsuccessful, the company will have made a loss.
The cost of tendering for contract MS1 only is $50,000. The component supply cost if the tender
is successful would be $18,000.
The cost of tendering for contract MS2 only is $14,000. The component supply cost if the tender
is successful would be $12,000.
The cost of tendering for both contracts MS1 and MS2 is $55,000. The component supply cost if
the tender is successful would be $24,000.
For each contract, possible tender prices have been determined. In addition, subjective
assessments have been made of the probability of getting the contract with a particular tender
price as shown below. Note here that the company can only submit one tender and cannot, for
example, submit two tenders (at different prices) for the same contract.
Option
Possible Probability
tender
of getting
prices ($) contract
MS1 only
130,000
115,000
0.20
0.85
MS2 only
70,000
65,000
60,000
0.15
0.80
0.95
MS1 and MS2 190,000
140,000
0.05
0.65
In the event that the company tenders for both MS1 and MS2 it will either win both contracts (at
the price shown above) or no contract at all.
•
•
•
What do you suggest the company should do and why?
What are the downside and the upside of your suggested course of action?
A consultant has approached your company with an offer that in return for $20,000 in
cash, she will ensure that if you tender $60,000 for contract MS2, only your tender is
guaranteed to be successful. Should you accept her offer or not and why?
Solution
The decision tree for the problem is shown below.
Below we carry out step 1 of the decision tree solution procedure which (for this example)
involves working out the total profit for each of the paths from the initial node to the terminal
node (all figures in $’000).
Step 1
•
•
•
•
•
•
path to terminal node 12, we tender for MS1 only (cost 50), at a price of 130, and win the
contract, so incurring component supply costs of 18, total profit 130-50-18 = 62
path to terminal node 13, we tender for MS1 only (cost 50), at a price of 130, and lose the
contract, total profit -50
path to terminal node 14, we tender for MS1 only (cost 50), at a price of 115, and win the
contract, so incurring component supply costs of 18, total profit 115-50-18 = 47
path to terminal node 15, we tender for MS1 only (cost 50), at a price of 115, and lose the
contract, total profit -50
path to terminal node 16, we tender for MS2 only (cost 14), at a price of 70, and win the
contract, so incurring component supply costs of 12, total profit 70-14-12 = 44
path to terminal node 17, we tender for MS2 only (cost 14), at a price of 70, and lose the
contract, total profit -14
•
•
•
•
•
•
•
•
path to terminal node 18, we tender for MS2 only (cost 14), at a price of 65, and win the
contract, so incurring component supply costs of 12, total profit 65-14-12 = 39
path to terminal node 19, we tender for MS2 only (cost 14), at a price of 65, and lose the
contract, total profit -14
path to terminal node 20, we tender for MS2 only (cost 14), at a price of 60, and win the
contract, so incurring component supply costs of 12, total profit 60-14-12 = 34
path to terminal node 21, we tender for MS2 only (cost 14), at a price of 60, and lose the
contract, total profit -14
path to terminal node 22, we tender for MS1 and MS2 (cost 55), at a price of 190, and
win the contract, so incurring component supply costs of 24, total profit 190-55- 24=111
path to terminal node 23, we tender for MS1 and MS2 (cost 55), at a price of 190, and
lose the contract, total profit -55
path to terminal node 24, we tender for MS1 and MS2 (cost 55), at a price of 140, and
win the contract, so incurring component supply costs of 24, total profit 140-55- 24=61
path to terminal node 25, we tender for MS1 and MS2 (cost 55), at a price of 140, and
lose the contract, total profit -55
Hence we can arrive at the table below indicating for each branch the total profit involved in that
branch from the initial node to the terminal node.
Terminal node Total profit $’000
12
62
13
-50
14
47
15
-50
16
44
17
-14
18
39
19
-14
20
34
21
-14
22
111
23
-55
24
61
25
-55
We can now carry out the second step of the decision tree solution procedure where we work
from the right-hand side of the diagram back to the left-hand side.
Step 2
•
•
For chance node 5 the EMV is 0.2(62) + 0.8(-50) = -27.6
For chance node 6 the EMV is 0.85(47) + 0.15(-50) = 32.45
Hence the best decision at decision node 2 is to tender at a price of 115 (EMV=32.45).
•
•
•
For chance node 7 the EMV is 0.15(44) + 0.85(-14) = -5.3
For chance node 8 the EMV is 0.80(39) + 0.20(-14) = 28.4
For chance node 9 the EMV is 0.95(34) + 0.05(-14) = 31.6
Hence the best decision at decision node 3 is to tender at a price of 60 (EMV=31.6).
•
•
For chance node 10 the EMV is 0.05(111) + 0.95(-55) = -46.7
For chance node 11 the EMV is 0.65(61) + 0.35(-55) = 20.4
Hence the best decision at decision node 4 is to tender at a price of 140 (EMV=20.4).
Hence at decision node 1 we have three alternatives:
•
•
•
tender for MS1 only EMV=32.45
tender for MS2 only EMV=31.6
tender for both MS1 and MS2 EMV = 20.4
Hence the best decision is to tender for MS1 only (at a price of 115) as it has the highest
expected monetary value of 32.45 ($’000).
INFLUENCE OF SAMPLE SIZE:
DEFINITION:
Sample size is one of the four interrelated features of a study design that can influence the
detection of significant differences, relationships, or interactions. Generally, these survey designs
try to minimize both alpha error (finding a difference that does not actually exist in the
population) and beta error (failing to find a difference that actually exists in the population).
WHAT IT IS USED FOR:
The sample size used in a study is determined based on the expense of data collection and the
need to have sufficient statistical power.
HOW TO USE IT:
We already know that the margin of error is 1.96 times the standard error and that the standard
error is sq.rt ^p(1 ^p)/n. In general, the formula is ME = z sq.rt ^p(1-^p)/n
where
*ME is the desired margin of error
*z is the z-score, e.g., 1.645 for a 90% confidence interval, 1.96 for a 90% confidence interval,
2.58 for a 99% confidence interval
_ ^p is our prior judgment of the correct value of p.
_ n is the sample size (to be found)
EXAMPLE:
If ^p =0.3 and Z=1.96 and ME =0.025 then the necessary sample size is:
ME= Z sq.rt (^p*1-^p)/n
0.025 = 1:96 sq.rt (0.3*0.7)/n
n=1291 or 1300 students
POPULATION MEAN:
DEFINITION:
The population mean is the mean of a numerical set that includes all the numbers within the
entire group.
WHAT IT IS USED FOR:
In most cases, the population mean is unknown and the sample mean is used for validation
purposes. However, if we want to calculate the population mean, we will have to construct the
confidence interval. This can be achieved by the following steps:
HOW TO USE IT:
•
•
•
•
The sample statistic is the sample mean x¯
The standard error of the mean is s/sq.rt n where s is the standard deviation of individual
data values.
The multiplier, denoted by t*, is found using the t-table in the appendix of the book. It’s a
simple table. There are columns for .90, .95, .98, and .99 confidence. Use the row for df =
n − 1.
Thus the formula for a confidence interval for the mean is x¯±t∗ (s/sq.rt n)
EXAMPLE:
In a class survey, students are asked if they are sleep deprived or not and also are asked how
much they sleep per night. Summary statistics for the n = 22 students who said they are sleep
deprived are:
•
Thus n = 22, x¯ = 5.77, s = 1.572, and standard error of the mean = 1.572/sq.rt 22=0.335
•
•
•
•
A confidence interval for the mean amount of sleep per night is 5.77 ± t* (0.335) for the
population that feels sleep deprived.
Go to the t-table in the appendix of the book and use the df = 22 – 1 = 21 row. For 95%
confidence the value of t* = 2.08.
A 95% confidence interval for μ is 5.77 ± (2.08) (0.335), which is 5.77 ± 0.70, or 5.07 to
6.7
Interpretation: With 95% confidence we estimate the population mean to be between
5.07 and 6.47 hours per night.
RANDOM SAMPLING
Random sampling is a sampling technique where we select a group of subjects (a sample) for
study from a larger group (a population). Each individual is chosen entirely by chance and each
member of the population has a known, but possibly non-equal, chance of being included in the
sample.
By using random sampling, the likelihood of bias is reduced.
WHEN RANDOM SAMPLING IS USED:
Random sampling is used when the researcher knows little about the population.
THE STEPS ASSOCIATED WITH RANDOM SAMPLING:
1.
2.
3.
4.
5.
6.
Define the population
Choose y …
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