Transcript Document

MM207 Statistics
Welcome to the Unit 7 Seminar
With Ms. Hannahs
Unit 7 Assignments and Final Exam
• Most (if not all) of your MSL StatCrunch
Homework and Quiz work in Unit 7 is about
looking at the data tables given and
interpreting them. See Help on Contingency
tables for computing probabilities.
• Expected value problems and Section 6.5
(and/or/conditional probability) problems will
be computed with pencil and paper and not
StatCrunch, best I can tell.
• Get started on the final as I will not take any
late final exams (due date is Tuesday, March
13). By the end of Unit 7, you should be able
to have problems 1-6 completed!! Show all
work on computations and paste StatCrunch
results when needed.
Statistical Significance
A set of measurements or observations in a
statistical study is said to be statistically
significant if it is unlikely to have occurred by
chance.
Is the probability that the observed difference occurred by chance less than or
equal to 0.05 (or 1 in 20)?
If the answer is yes (the probability is less than or equal to 0.05), then we
say that the difference is statistically significant at the 0.05 level.
If the answer is no, the observed difference is reasonably likely to have
occurred by chance, so we say that it is not statistically significant.
The choice of 0.05 is somewhat arbitrary, but it’s a figure that statisticians
frequently use. Nevertheless, other probabilities are often used, such as
0.1 or 0.01.
Example: An event is considered
“significant” if its probability is less
than or equal to 0.05.
Is it significant to be dealt an ace when you are
dealt one card from a standard 52-card deck?
(There are four aces in the deck.)
a.
Yes
b.
No
Example: An event is considered
“significant” if its probability is less
than or equal to 0.05.
Is it significant to be dealt an ace when you are
dealt one card from a standard 52-card deck?
(There are four aces in the deck.)
a.
Yes
b.
No
Note: this is an example of theoretical probability
Three Approaches to Finding Probability
• A theoretical probability is based on assuming that all
outcomes are equally likely. It is determined by dividing
the number of ways an event can occur by the total
number of possible outcomes.
• A relative frequency probability is based on observations
or experiments. It is the relative frequency of the event of
interest.
• A subjective probability is an estimate based on
experience or intuition.
Theoretical Probability
• Experiment: Rolling a single die
• Sample Space: All possible outcomes from experiment
S = {1, 2, 3, 4, 5, 6}
• Outcomes are the most basic possible results of observations or
experiments
• Event: a collection of one or more outcomes (denoted by capital
letter)
Event A = {3}
Event B = {even number}
• Probability = (number of favorable outcomes) / (total number of
outcomes)
• P(A) = 1/6
• P(B) = 3/6 = 1/2
Theoretical Probability – OR problems
Section 6.5 Page 272 green box ALWAYS works.
P(A or B) = P(A) + P(B) – P(A and B)
Do not use Page 271 green box. Let’s look at examples and see this
is better.
Recall: S = {1, 2, 3, 4, 5, 6}
Event A = {3}
Event B = {even number}
Event C = {4, 6}
• Find P(B or C)
P(B or C) = P(B) + P(C) – P(B and C) = 3/6 + 2/6 – 2/6 = 3/6 = 1/2
• Find P(A or B)
P(A or B) = P(A) + P (B) – P(A and B) = 1/6 + 3/6 – 0/6= 4/6 =2/3
Counting Possible Outcomes
• Suppose process A has a possible outcomes and
process B has b possible outcomes. Assuming the
outcomes of the processes do not affect each other, the
number of different outcomes for the two processes
combined is: a * b
• This idea extends to any number of processes. If a third
process C has c possible outcomes, the number of
possible outcomes for the three processes combined is:
a * b * c.
• Tree Diagrams are good for this. See page 241
Applying the Counting Rule
How many outcomes are
there if you roll a fair die
and toss a fair coin?
The first process, rolling a
fair die, has six outcomes
The second process,
tossing a fair coin, has
two outcomes (H, T).
H
1
2
3
T
H
T
H
T
H
4
T
5
H
Sample
Space is
1H
1T
2H
2T
3H
3T
4H
4T
5H
5T
6H
6T
T
Therefore, there are 6 ×
6
H
2 = 12 outcomes for the
T
two processes together
(1H, 1T, 2H, 2T, . . . , 6H, Note: Probability of getting a 1
and then an H is 1/12
6T).
And Probability Problems
• And Probability for Independent Events
Two events are independent if the outcome of
one event does not affect the probability of the
other event. Consider two independent events A
and B with probabilities P(A) and P(B). The
probability that A and B occur together is
P(A and B) = P(A) * P(B)
The key to And Problems is interpreting when
there is “with replacement” (meaning
independent) as well as when there is “no
replacement” (meaning dependent)
Example “And” - Independent
P(A and B) = P(A) * P(B)
Setup: Two cards are to be selected with
replacement from a deck of cards.
Example 1: Find the probability that two red cards will
be selected.
P(Red and Red) = P(Red) * P(Red)
= (26/52) * (26/52) = 1/4
Example 2: Find the probability that a red card and
then a black card will be selected.
P(Red and Black = P(Red) * P(Black)
= (26/52) * (26/52) = 1/4
And Probability for Dependent Events
Two events are dependent if the outcome of one
event affects the probability of the other event. The
probability that dependent events A and B occur
together is
P(A and B) = P(A) * P(B given A)
where P(B given A) means the probability of event B
given the occurrence of event A.
This is often written as P(A) * P(B|A)  fancy notation
Example “And” - Dependent
P(A and B) = P(A) * P(B given A)
= P(A) * P(B|A)  fancy notation
Setup: Two cards are to be selected without
replacement from a deck of cards.
Example 1: Find the probability that two red cards
will be selected.
P(Red and Red) = P(Red) * P(Red)
= (26/52) * (25/51) = 25/102.
Example 2: Find the probability that a red card and
then a black card will be selected.
P(Red and Black) = P(Red) * P(Black)
= (26/52) * (26/51) = 26/102
Relative Frequency (Empirical)
Probability
•
Probability of an Event Not Occurring
If the probability of an event A is P(A), then the probability that
event A does not occur is P(not A).
Because the event must either occur or not occur, we can write:
P(A) + P(not A) = 1 or P(not A) = 1 – P(A)
The event not A is called the complement of the event A; the
“not” is often designated by a bar, so Ā means not A.
Example: A quarterback completes 67% of his passes, what is
the probability that he will NOT complete his next pass?
Answer: P(not A) = 1 – P(A) = 1 - .67 = .33. There is a 33%
chance he will not complete the next pass.
Probability Distributions
A probability distribution represents the
probabilities of all possible events. Do
the following to make a display of a
probability distribution:
1. List all possible outcomes. Use a
table or figure if it is helpful.
2. Identify outcomes that represent the
same event. Find the probability of
each event.
3. Make a table in which one column
lists each event and another column
lists each probability. The sum of all
the probabilities must be 1.
Here is the probability
distribution for our Sample
Space from earlier
1H
1/12
1T
1/12
2H
1/12
2T
1/12
3H
1/12
3T
1/12
4H
1/12
4T
1/12
5H
1/12
5T
1/12
6H
1/12
6T
1/12
Total = 12/12 = 1
Note: this is a uniform distribution
Creating a Probability Distribution Another example
If we toss three coins,
we have a total of 2 ×
2 × 2 = 8 possible
outcomes:
HHH, HHT, HTH,
HTT, THH, THT, TTH,
and TTT, as shown in
Figure 6.4 b.
Figure 6.4 b Tree diagram
showing the outcomes of tossing
three coins.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2- 18
Creating a Probability Distribution
What is the probability distribution for the number of heads
that occurs when three coins are tossed simultaneously?
The number of different outcomes when three coins are
tossed is 2 × 2 × 2 = 8.
Question. What is the
probability of getting
exactly 2 heads?
Answer 3/8.
Question. What is the
probability of getting
2 or more heads?
Answer 4/8 = 1/2.
3 Coin Probability Distribution
Law of Large Numbers
• The law of large numbers (or law of averages) applies
to a process for which the probability of an event A is
P(A) and the results of repeated trials do not depend on
results of earlier trials (they are independent).
• It states: If the process is repeated through many trials,
the proportion of the trials in which event A occurs will be
close to the probability P(A). The larger the number of
trials, the closer the proportion should be to P(A).
Roulette Example
•
Expected Value
• The expected value of a variable is the weighted average of all its
possible events. Because it is an average, we should expect to find the
“expected value” only when there are a large number of events, so that
the law of large numbers comes into play.
Consider two events, each with its own value and probability. The expected value is:
expected value = (value of event 1) * (probability of event 1)
+ (value of event 2) * (probability of event 2)
This formula can be extended to any number of events by including more terms in the
sum.
Example: Suppose you pay $2
to roll a loaded die and win $6
if it comes up a 1 or 6, but
nothing otherwise. What is
your expected value?
ev = (value of event 1) * (probability of
event 1) + (value of event 2) * probability
of event 2)
= (NET win of $4)(prob of rolling 1 or 6) +
(NET loss of $6) (prob of rolling 2,3,4,5)
= ($4) (7/21) + (-$2) (14/21)
= (28/21) – (28/21)
= 0. Your expected value is 0.
Num
1
2
3
4
5
6
Prob
1/21
2/21
3/21
4/21
5/21
6/21
Expected Value - example
Question: You go to a social function and buy a $1
raffle ticket to win the one grand prize of $500. If
1200 tickets are sold what is your expected value?
Answer: ev =(value of event 1) * (probability of
event 1) + (value of event 2) * (probability of
event 2)
= (NET win of $499)(prob of winning) + (NET loss
of $1) (prob of losing)
= (499)(1/1200) + (-1)(1199/1200)
= (499/1200) – (1199/1200)
= -700/1200 = -7/12 = -.5833. If you did this 100
times, you should expect to lose $58.33.
Winning the Lottery
A $1 lottery tickets have the following probabilities: 1 in 5 to win
a free ticket (worth $1); 1 in 100 to win $5; 1 in 100,000 to win
$1,000; and 1 in 10 million to win $1 million.
What is the expected value of a lottery ticket?
Event
Purchase ticket
Win free ticket
Win $5
Win $1,000
Win
$1,000,000
Value
-$1
$1
$5
$1,000
$1,000,000
Probability
1
1/5
1/100
1/100,000
1/10,000,000
Value * Probability
-$1 x 1
$1 x 1/5
$5 x $100
$1,000 x 1/100,000
$1,000,000/10,000,000
Result
-$1.00
$0.20
$0.05
$0.01
$.10
Expected Value =
$-0.64
Thus, averaged over many tickets, you should expect to lose 64¢ for each
lottery ticket that you buy. If you buy, say, 1,000 tickets, you should expect
to lose about 1,000 × $0.64 = $640.
Accident Rates
Travel risk is often expressed in terms of an accident rate
or death rate.
For example, suppose an annual accident rate is 750
accidents per 100,000 people.
• This means that, within a group of 100,000 people, on
average 750 will have an accident over the period of a year.
•
The statement is in essence an expected value, which means
it also represents a probability: It tells us that the probability of
a person being involved in an accident (in one year) is 750 in
100,000, or 0.0075.
Death Rates
•
Vital Statistics
Computing Probabilities in StatCrunch
Create a Contingency table from data sets to compute answers
to And, Or, and Conditional Probabilities problems you may find in
The textbook, MSL homework or Quiz, and the final exam.
Computing Probabilities in StatCrunch
Example 1: Find the probability of a house
having exactly 3 bedrooms or 3 full bathrooms.
This is an OR problem.
n = 40
P(of 3 bedrooms or 3 bathrooms)
= P(3 bedrooms) + P(3 bathrooms) - P(3 bedrooms and 3 bathrooms)
= (19/40) + (6/40) - (2/40)
= 23 / 40 = .575 or .58 rounded to 2 decimal places
Questions?