psych 218 - M2 Flashcards
1
Q
Types of probability
A
- ‘a priori’: before data / observation
- based on expectations
- p(A) = # of A-possible events / N-possible events
- i.e. dice landing on an odd roll: p(A) = 3/6 = 0.5
- ‘a posteriori’: after data observation
- based on what we observe
- would not equal a priori, but should be relatively similar > will be skeptical if very different
- p(A) = # of A-observed / N-observed
- p(A) = 58 odd rolls / 100 rolls = 0.58
- have to decide if difference (0.50 and 0.58) is systematic or random factors
- dice is fair / not enough evidence its unfair: maintain probability at 0.5
- dice is unfair: predict probability at 0.58 for the future
2
Q
Probability terms
A
- mutually exclusive events (P & Q): events that cannot happen at the same time
- i.e. dice cannot land on even and odd
- P and Q will be complementary: P = 1 - Q
- P and Q will never happen together: p(P & Q) = 0
- independent events: events that have no influence on each other
- i.e. first dice roll does not influence second dice roll
- correlation between independent events: p = 0.000
- exhaustive set of events: describe all possible events
- i.e. even and not even dice roll
- we know whether dice roll is “A” or “not A”
3
Q
Probability Rules
A
- addition rule: used when talking about one of several possible outcomes
- p(A or B) = p(A) + p(B) - p(A and B)
- if events are mutually exclusive, p(A and B) = 0
- multiplication rule: quantifies probability of successive events
- independent successive events: i.e. probability of rolling sum of 11?
- p(5) x p(6|5) = 1/6 x 1/6 = 1/36
- p(6) x p(5|6) = 1/6 x 1/6 = 1/36
- add up: 1/36 + 1/36 = 2/36
- dependent successive events: i.e. probability of being dealt 2 Kings?
- p(king) = 4/52
- p(king | king) = 3/52
- p(2 kings) = 4/52 x 3/52
- independent successive events: i.e. probability of rolling sum of 11?
4
Q
Goal of using inferential statistics?
A
- if we can assume population is normally distributed, we can predict where they fall
- can calculate the probability of sampling a certain z-score
5
Q
5 rules for binomial distribution
A
- series of N trials
- multiple choice test with 20 questions
- only 2 outcomes
- right v wrong answer
- outcomes are mutually exclusive
- if you’re right, you cannot be wrong
- outcomes are independent
- probability of solving Q1 is independent from Q2
- probability of P is consistent
- always 25% because there are 4 options
6
Q
z-distribution (= normal distribution)
A
- shows how likely observations are when many outcomes are possible
- expectation = a priori probability
- n x p (can be quantified before data collection)
- there will be deviations from expectations (a priori ≠ a posteriori)
- as N gets bigger, distribution will look more and more normal
7
Q
Binomial v Normal Distribution
A
- as N increases, binomial distribution will look more and more like a normal distribution
- can solve problems using z-scores and normal curves
- as P and Q deviate from 0.50, binomial distribution looks less normal
- can only use normal distribution if:
- N x P ≥ 10
- N x Q ≥ 10
8
Q
Parameters of the normal distribution
A
- mean of distribution
- μ = N x P
- standard deviation of distribution
- σ = √ N x P x Q
9
Q
Example: Mosquito Preference
A
- N = 20, X = 15 bites
- a priori: P(bite) = 0.5
- a posteriori: P(bite) = 0.75 (15 out of 20 bites)
- assuming a priori probability, determine how strange the result is:
- p(≥ 15) = 0.0148 + 0.0046 + 0.0011 + 0.0002 + 0.000 + 0.000 = 0.02
- must add up all values more extreme to get area under the curve of a priori
- there is a 2% chance of X getting 15+ bites if null is true (= mosquito has no preference)
- as 2% is very low, can conclude that the mosquito has preference
10
Q
How to Solve Binomial
A
- [1] Figure out what you are solving for
- the probability you want to obtain (a priori)
- evaluating the outcome we observe (a posteriori)
- [2] Consider outcomes that would be even more extreme
- [3] Look at the table
- how many observations do we have (N)
- probability of an event (a priori probability)
- translate score into number of questions
- i.e. if I need 80% to pass, need to solve at last 40 questions
11
Q
Normal Approximation
A
- converting the information we have into a z-score (can use normal curve)
- 50 questions, need to get 80% score
- p(correct) = 0.7 (based on student’s knowledge)
- z = (X - μ) / σ = 1.54
- X: number of questions needed to get right
- 50 x 0.8 = 40
- μ = N x P
- 50 x 0.7 = 35
- σ = √ N x P x Q
- √50 x 0.7 x 0.3 = 3.24
- X: number of questions needed to get right
12
Q
Hypothesis Terms
A
- hypothesis: proposed explanation for an observation / phenomenon
- hypothesis testing: construct 2 hypothesis and have them compete with each other
- must be mutually exclusive
- must be exhaustive
- null hypothesis (Ho): any observed change is due to chance or unexplained factors
- IV does not effect DV / no relationship / groups do not differ from each other
- always start out assuming null hypothesis is true (because we can quantify it mathematically)
- the extent we deviate from expectation: can decide if its due to randomness or something systematic
- alternative hypothesis (H1): observed change is not due to chance or unexplained factors
- if non-directional, null will be “IV has no effect on DV”
- if directional, null will be “IV does not have effect on DV, or effect is opposite to alternate”
13
Q
Hypothesis Testing
A
- reject null if
- p < α
- z-obt > z-crit
- fail to reject null if
- p > α
- z-obt < z-crit
14
Q
Statistical Significance
A
- we can claim statistical significance when we can reject null hypothesis
- can rule out chance as the only reason why treatment group did better
- chance is still a factor, but chance on its own is not able to account for the size of discrepancy – difference can be attributed to the treatment
- note: statistically significant does not indicate how meaningful treatment is
- treatment group has lower fear levels, but we do not know strength or magnitude of the effect
- thus, must calculate effect size to see extent
- better to say “statistically reliable”
15
Q
Sign Test
A
- ignores magnitude, only considers sign (+ / -)
- requires repeated measures design
- steps:
- [1] Check textbook for N number of events
- [2] Check column for a priori probability
- [3] Add up all extreme scores to see how likely it is to occur
- if two-tailed, all extreme scores include scores on the other side of the distribution