1.1 Average

\[\bar x = \frac{\sum x_i}{n}\]

1.2 Standard deviation

\[\eqalign{\displaystyle\sigma &=& \sqrt{\frac{\sum (x - \bar x)^2}{n-1}}\\ &=&\sqrt{\frac{\sum x^2 - \left(\sum x\right)^2/n}{n-1}}\\}\]

1.3 Method 1: Group A: Sums of deviations

1.4 Method 1: Group B: Sums of deviations

1.5 Method 2: Group A: Sum of squares

1.6 Method 2: Group B: Sum of squares

1.7 By computer function

a = c(4,4,4,5,5,5,5,6)
cat("a =", mean(a),"+/-", sd(a))

a = 4.75 +/- 0.7071068

b = c(4,5,5,6,6,7,7,8)
cat("b =", mean(b),"+/-", sd(b))

b = 6 +/- 1.309307

1.8 Data points vs Population modelling

1.9 Normal distribution

\[f\left(x\right) = \frac{1}{\sqrt{2\pi \sigma}}e^{-\frac{\left(x-\mu\right)^2}{2\sigma^2}}\]

2.1 SE vs Std Dev

2.2 Standard deviations

2.3 Sample size

3.1 t test - One sample

\[{\displaystyle t={\frac {{\bar {x}}-\mu _{0}}{ \frac{s}{\sqrt {n}}}} = \frac{\Delta x}{\frac{\sigma}{\sqrt{n}}}}\]

\[df = n - 1\]

3.2 2 Sample of equal size equal variance and size

\[{\displaystyle t={\frac {{\bar {X}}_{1}-{\bar {X}}_{2}}{s_{p}{\sqrt {\frac {2}{n}}}}}}={\frac {{\bar {X}}_{1}-{\bar {X}}_{2}}{\sqrt {\frac {s_{1}^{2}+s_{2}^{2}}{n}}}}\]

\[{\displaystyle s_{p}={\sqrt {\frac {s_{1}^{2}+s_{2}^{2}}{2}}}}\] \[df= 2n - 2\]

3.3 2 Sample t-test

\[t = \frac{\bar x_1 - \bar x_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\]

\[s_p = \sqrt{\frac{(n_1-1)s_1^2 +(n_2-1)s_2^2}{n_1+n_2-2}}\]

\[df = n_1 + n_2 -2\]

3.4 Welches t-test

\[t = \frac{\quad\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

\[{\displaystyle \mathrm {d.f.} ={\frac {\left({\frac {s_{1}^{2}}{n_{1}}}+{\frac {s_{2}^{2}}{n_{2}}}\right)^{2}}{{\frac {\left(s_{1}^{2}/n_{1}\right)^{2}}{n_{1}-1}}+{\frac {\left(s_{2}^{2}/n_{2}\right)^{2}}{n_{2}-1}}}}} \]

4.1 Simple experimental design

To have 95% confidence on experiments with only 0.5 \(\sigma\) separation:

\[n = t^2\left(\frac{\sigma}{\Delta x}\right)^2=1.96^2\left(\frac{\sigma}{0.5\sigma}\right)^2=15.3664=16\]

4.2 Example

120 people work at Company Q, 85 of which drink coffee daily, find the 99% confidence interval of the true proportion of people who drink coffee at Company Q on a daily basis.

\[\eqalign{CI &=& \hat p \pm z \times \sqrt{\frac{p(1-p)}{n}}\\ &=&\frac{85}{120} \pm 2.58 \times \sqrt{\frac{\frac{85}{120}\left(1 - \frac{85}{120}\right)}{120}}= 92.83\\}\]

4.3 Unlimited population

Sample size necessary to estimate the proportion of people in a supermarket that identify as vegan with 95% confidence, and a margin of error of 5%. Assume a population proportion of 0.5, and unlimited population size.

\[n=\frac{z^2 \times \hat p(1- \hat p)}{\epsilon^2}=1.96^2 \times \frac{0.5(1-0.5)}{0.05^2}=384.16\]

4.4 Finite population

Sample size needed to estimate the proportion of German speakers in a college with 95% confidence, and a margin of error of 5%. Assume a population proportion of 0.2, and population of 1,200 students.

\[\eqalign{n &=& \frac{\frac{z^2 \times \hat p(1- \hat p)}{\epsilon^2}}{1+\frac{z^2 \times \hat p(1- \hat p)}{\epsilon^2 N}} = \frac{\frac{1.96^2 \times 0.2(1- 0.2)}{0.05^2}}{1+\frac{1.96^2 \times 0.2(1- 0.2)}{0.05^2 \times 1200}}\\ &=&\frac{\frac{3.84\times 0.16}{0.0025}}{1+\frac{3.84\times 0.16}{0.0025\times 1200}} = \frac{245.76}{3.6144} = 67.995 \\ }\]

4.5 Variables

• z is the z score
• ε is the margin of error
• N is the population size
• p̂ is the population proportion

4.6 Yamane FORMULA

The variables in this formula are:

\[\eqalign{n &=& sample\ size\\ N &=& population\ of\ the\ study\\ e &=& margin\ of\ error\ in\ the\ calculation\\ }\]

\[n=\frac{N}{(1 + N e^2)}\]

5.1 One dimension

• Observed: Heads: 3, Tails: 7
• Expected: Heads: 5, Tails: 5

\[\chi^2 = \frac{(3-5)^2}{5} + \frac{(7-5)^2}{5} = \frac{4}{5} + \frac{4}{5} = 1.6\]

df = 1 ; p = 0.7940968

5.2 Contingency Table with rows and columns

\[\chi ^{2}=\sum _{{i=1}}^{{r}}\sum _{{j=1}}^{{c}}{(O_{{i,j}}-E_{{i,j}})^{2} \over E_{{i,j}}}\]

\[df = r + c - 2\]

6.1 Shoesize

male: 41 42 42 44 42 42 42 46 

female: 36 37 40 38 38.5 39 37 40 38 40 

    Welch Two Sample t-test

data:  ms and fs
t = 5.9314, df = 14.194, p-value = 3.461e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 2.731133 5.818867
sample estimates:
mean of x mean of y 
   42.625    38.350 

6.2 Dating Behaviour

       Phone EMail F2F
Male      27    52  95
Female    35    48  65

    Pearson's Chi-squared test

data:  dta
X-squared = 4.7488, df = 2, p-value = 0.09307

6.3 Raw counts

Observed

       Phone EMail F2F
Male      27    52  95
Female    35    48  65

Expected

          Phone    EMail      F2F
Male   33.50311 54.03727 86.45963
Female 28.49689 45.96273 73.54037

6.4 Relative as Percentages

Observed

           Phone    EMail      F2F
Male    8.385093 16.14907 29.50311
Female 10.869565 14.90683 20.18634

Expected

           Phone    EMail      F2F          
Male   10.404691 16.78176 26.85082  54.03727
Female  8.849967 14.27414 22.83863  45.96273
       19.254658 31.05590 49.68944 100.00000

6.5 Percentage (by row)

           V1    Phone    EMail      F2F
1   Male: Obs 15.51724 29.88506  54.5977
2 Female: Obs 23.64865 32.43243 43.91892
3    Expected 19.25466  31.0559 49.68944

6.6

\[ sd = \sqrt{\frac{\sum x^2 -\frac{(\sum{x})^2}{n}}{n-1}} =\sqrt{\frac{75-\frac{19^2}{5}}{4}} =0.84 \]

\[sd = \sqrt{\frac{\sum x^2 -\frac{(\sum{x})^2}{n}}{n-1}} = \sqrt{\frac{142 - \frac{24^2}{5}}{4}}=2.59\]

\[t = \frac{4.8-3.8}{\sqrt{\frac{0.84^2+2.59^2}{2}}}\]

7.1 IT Club Calendar

7.2 Post Prison therapy