week 4.pdf - 8/09/2020 Section B (Weeks 4 – 7) One variable...

8/09/2020
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Section B
(Weeks 4 – 7)
One variable analysis
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Map of course
Section Weeks Title …contains
A 1 - 3 Informal inference with 1 and 2
variables; probability
Critical thinking; variables; data display – bar charts, boxplots,
histograms, scatterplots; summary statistics; probability
definitions and rules
B 4 - 7 Formal inference: hypothesis
testing with 1 variable
Binomial and Normal probability distributions; Sign Test; t-test
for a mean; Z-test for a mean; Central Limit Theorem
C 8 - 11 Formal inference: hypothesis
testing using 2 variables
Chi-square test; ANOVA; t-test for 2 independent means;
linear regression
D 12 Consolidation Revision, linking the topics together
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Binomial distribution
Week 4
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• We consider the possible outcomes of a random variable, and the
probabilities of each of those outcomes
• This gives us our probability distribution
• So probability distributions are also either discrete or continuous
• The discrete probability distribution we will use is the Binomial Distribution
(The continuous one we’ll use shortly is the Normal Distribution)
• Probability distributions are mathematical models of real life – they never
perfectly fit reality!
(If they do, we should probably be suspicious)
Probability distributions
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Coin toss exercise: the random variable ‘X’ counts the number of
heads obtained (out of 3 trials)
Experimental (observed) results Theoretical (expected) results
• ܲ ܺ = 0 =
• ܲ ܺ = 1 =
• ܲ ܺ = 2 =
• ܲ ܺ = 3 =
5
• ܺ = 0 :
• ܺ = 1 :
• ܺ = 2 :
• ܺ = 3 :
read as ‘the probability that X equals 3 is …’
Probability distribution for coin toss exercise
࢞ 0 1 2 3
ࡼ(ࢄ = ࢞) 0.125 0.375 0.375 0.125
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• With this categorical
variable, the proportions
of people within all
categories add to 1
• In the same way, the
probabilities within a
probability distribution
will always add to 1
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Binomial distribution
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Example of binomial probability distribution
• Let’s say that Lionel Messi (brilliant Argentinian football player) is practising his
penalty shots (against an equally talented goalkeeper)
• Every time Messi takes a shot, we’ll say his chance of scoring (‘success’) is 0.8
(we know this from past experience). Each shot is a Bernoulli trial as he will either
score or not score
• We want to know the probability of the event that he scores 5 or fewer times out
of 9 attempts
(what do you guess that this probability might be?)
• So we want to find ܲ(ܺ ≤ 5) , where X is the random variable representing the
number of successes (goals) out of 9 shots
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‘is less than or equal to’

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The necessary conditions for a binomial
distribution
• If (and only if) the following conditions are met, we can use the
binomial distribution to find this probability
1) Fixed number of trials [ ݊ = ]
2) Fixed probability of success [݌ = ]
3) Two possible outcomes [ or ]
4) Trials are independent* [he doesn’t get better or worse depending on
previous results]
*Discuss!