Stats 2 (AQA) Formula
(everything in blue in formula book)
Expectation (mean) & Variance of Random Variable from Probability Distribution
πππ(π) = π 2 = ∑ π₯π 2 ππ − π 2
πΈ(π) = π = ∑ π₯π ππ
= πΈ(π 2 ) − π 2
( ο οΉ x . ο is theoretical mean, x is sample mean.)
πππ(ππ + π) = πππ(ππ) = π2 πππ(π)
πΈ(ππ + π) = πΈ(ππ) + π = ππΈ(π) + π
Poisson distribution (a discrete distribution)
π(π = π₯) = π −π
π~ππ(π)
ππ₯ =
ππ₯
π₯!
π = π, π 2 = π
π
π
π₯ π₯−1
If π~ππ(π1 ) and π~ππ(π2 ), then the distribution of (π + π)~ππ(π1 + π2 ).
Continuous Random Variables
Probability Density Function (pdf) = f
Cumulative Distribution Function (cdf) = F
π
π(π₯) = πΉ(π₯)
ππ₯
πΉ(π₯) = ∫ π(π‘) ππ‘
π₯
−∞
(replace lower limit with lower limit of function when calculating)
πΉ(−∞) = 0
πΉ(+∞) = 1
Three methods to find specific probabilities of X:
π
1. π(π < π < π) = ∫π π(π₯) ππ₯ (ie integrate to find area between two values)
2. Having integrated, evaluate between limits (similar to finding normal distribution
probabilities).
π(π < π < π) = πΉ(π) − πΉ(π)
πΉ(π > π) = 1 − πΉ(π)
3. Use geometry of graph of π(π₯) (where possible, splitting it into triangles, trapezia etc).
To find median value, m, solve for m either:
π
or
0.5 = ∫−∞ π(π₯)ππ₯
0.5 = πΉ(π)
To find, for example, the value at 95th percentile, solve for d;
π
or
0.95 = ∫−∞ π(π₯)ππ₯
Mean and variance…
0.95 = πΉ(π)
∞
πΈ(π) = π = ∫ π₯π(π₯) ππ₯
−∞
∞
2
πΈ(π 2 ) = ∫ π₯ π(π₯) ππ₯
−∞
∞
πΈ(π(π)) = ∫ π(π₯)π(π₯) ππ₯
−∞
If distribution is symmetrical then πΈ(π) = ππππ‘πππ π£πππ’π.
∞
πππ(π) = πΈ(π 2 ) − π 2 = ∫ π₯ 2 π(π₯) ππ₯ − π 2
−∞
Rectangular (Uniform) Distribution
π(π₯) =
1
πΈ(π) = (π + π)
2
1
π−π
πππ(π) =
πΉ(π₯) =
1
(π − π)2
12
π₯−π
π−π
Estimation
The t distribution (two random variables, πΜ
and π, i.e. unknown population variance).
π=
πΜ
− π
π
√π
π~π‘π or π~π‘π−1
π = ππ’ππππ ππ πππππππ ππ πππππππ = π − 1
(parameter π pronounced ‘nu’)
Confidence intervals given by
π₯Μ
± "c"
π
√π
π 2
π₯Μ
± "c"√
π
Hypothesis Testing
Null Hypothesis
π»0 : π = 435
Alternative Hypothesis
π»1 : π < 435
One tailed
π»1 : π ≠ 435
Two tailed
Acceptance of null hypothesis does not mean it is true, rather that the data provides no
evidence to prefer the alternative.
Type 1 error – to reject H0 (and accept H1) when H0 is actually true.
Type 2 error – to reject H1 (and accept H0) when H1 is actually true.
π(ππ¦ππ 1 πππππ) = π πππππππππππ πππ£ππ
Test Procedure:
1. Write down the two hypotheses.
2. Identify an appropriate test statistic and the distribution of the corresponding random
variable.
3. Identify the significance level (usually given). This is also π(ππ¦ππ 1 πππππ).
4. Determine the critical region (should be done before collecting data).
5. Calculate the value of the test statistic.
6. Determine and clarify in context the outcome of the test.
ππ Contingency Tables Tests
Part A – The ππ Distribution
π 2 ~ππ2
π 2 = 2π
(where π = π − 1)
π=π
π
(ππ − πΈπ )2
π =∑
πΈπ
2
π=1
(where π = number of different possible outcomes)
All expected frequencies must be greater than 5
Part B – Contingency Tables
Associated vs independent
To calculate expected frequencies from observed frequencies use
πππ€ π‘ππ‘ππ × ππππ’ππ π‘ππ‘ππ
πππππ π‘ππ‘ππ
π = (π − 1)(π − 1)
Yates’s correction for a 2x2 table (π = 1 ):
4
ππΆ2
=∑
1
(|ππΌ − πΈπ | − 0.5)2
πΈπ
π
π 2
=
(|ππ − ππ| − )
ππππ
2
Where…
a
c
r
b
d
s
m
n
N