Formula’s From population Measurements of association
value to %: If linear relation is expected: Pearson’s
( Nx )×100
(affected by outliers → thus create first a scatterplot
Variance: -> SD
Percentage= Association
s= √ Var
Standard error: Not necessarily linear : Spearman’s (based on ranks)
s Less vulnerable for outliers, takes into account monotonic but
SE= From % to non-linear relationships
√n population value:
Standard error for
correlation coefficient:
x= (Percentage
100 )× N Cramer: 0-1
linear rel. = all symmetric (no IV and DP considered) %
difference = asymmetric
1/sqrt(n-3)
48# van population = F-test 0-1.
Confidence interval: 0.48 x population
Estimate ± z ( 2 ) × SE
Kendall tau-b =Colom’s and rows even Kendall tau-c= Colom’s and rows uneven
Steps Calculate In R Interpretation
Confidence interval
^p x #1 amount of people who answer yes , #2 = n Estimator: a quantity that you compute based on sample data
st:
1: Find
for proportion
2: Calculate SE ^p= prop.test (…, …)
3: Find Z (2) N binom.test(…,…)
(we use prop.test or binom.test because our dependent variable is Estimate: the actual value that you get when computing an estimate
4: Calculate CI 95% dichotomous)
√
Filter:
^p ( 1− ^p ) summary(datasett$variable==1)
SE= *Output is fractions -> % = x100
n #CI in dataset
table (dataset)
binom.test (amount of people who do it, whole population)
^p ± z ( 2 ) × SE table (dataset$variable) à proportion in this variable
T-test 1: Hypothesis
x−μ # T-value One-sample t-test of differences
2: Calculate your t-test t= T=(x-u)/(s/sqrt(n)) H0: 𝜇2 - 𝜇1 = 0 (no change); HA: 𝜇2 - 𝜇1 ≠ 0 (change)
3: Find p-value using R
4. interpret t-value
se #P value (transform t in negative)
pt (t,df) * 2
5. CI (by hand) Two-sample t-test: Tests if the mean of group 1 differs from the mean of group 2.
H0: μ(married) = μ(single) (no change) HA: μ(married) ≠ μ(single) (or μ(married) >
s When p-value = <0,05, we reject H0 μ(single) if one-sided)
SE= Which means that the average in this sample is significantly
√n
different from the population.
When p-value = > 0,05 we can’t reject H0 2-sided hypothesis: H0: 𝜇 = 100; HA: 𝜇 is not 100
1-sided hypothesis: H0: 𝜇 = 100; HA: 𝜇 < 100
df =n−1
t.test: compare mean from a single group to a known value. (one sample t-test and
paired t-test are similar) → paired is only measured overtime and the difference will be
margin of error=t−value∗SE the one sample t-test. Independent two-sample t-test whether two groups differ from
each other
mean ± z ( 2 ) × SE
value to %: If linear relation is expected: Pearson’s
( Nx )×100
(affected by outliers → thus create first a scatterplot
Variance: -> SD
Percentage= Association
s= √ Var
Standard error: Not necessarily linear : Spearman’s (based on ranks)
s Less vulnerable for outliers, takes into account monotonic but
SE= From % to non-linear relationships
√n population value:
Standard error for
correlation coefficient:
x= (Percentage
100 )× N Cramer: 0-1
linear rel. = all symmetric (no IV and DP considered) %
difference = asymmetric
1/sqrt(n-3)
48# van population = F-test 0-1.
Confidence interval: 0.48 x population
Estimate ± z ( 2 ) × SE
Kendall tau-b =Colom’s and rows even Kendall tau-c= Colom’s and rows uneven
Steps Calculate In R Interpretation
Confidence interval
^p x #1 amount of people who answer yes , #2 = n Estimator: a quantity that you compute based on sample data
st:
1: Find
for proportion
2: Calculate SE ^p= prop.test (…, …)
3: Find Z (2) N binom.test(…,…)
(we use prop.test or binom.test because our dependent variable is Estimate: the actual value that you get when computing an estimate
4: Calculate CI 95% dichotomous)
√
Filter:
^p ( 1− ^p ) summary(datasett$variable==1)
SE= *Output is fractions -> % = x100
n #CI in dataset
table (dataset)
binom.test (amount of people who do it, whole population)
^p ± z ( 2 ) × SE table (dataset$variable) à proportion in this variable
T-test 1: Hypothesis
x−μ # T-value One-sample t-test of differences
2: Calculate your t-test t= T=(x-u)/(s/sqrt(n)) H0: 𝜇2 - 𝜇1 = 0 (no change); HA: 𝜇2 - 𝜇1 ≠ 0 (change)
3: Find p-value using R
4. interpret t-value
se #P value (transform t in negative)
pt (t,df) * 2
5. CI (by hand) Two-sample t-test: Tests if the mean of group 1 differs from the mean of group 2.
H0: μ(married) = μ(single) (no change) HA: μ(married) ≠ μ(single) (or μ(married) >
s When p-value = <0,05, we reject H0 μ(single) if one-sided)
SE= Which means that the average in this sample is significantly
√n
different from the population.
When p-value = > 0,05 we can’t reject H0 2-sided hypothesis: H0: 𝜇 = 100; HA: 𝜇 is not 100
1-sided hypothesis: H0: 𝜇 = 100; HA: 𝜇 < 100
df =n−1
t.test: compare mean from a single group to a known value. (one sample t-test and
paired t-test are similar) → paired is only measured overtime and the difference will be
margin of error=t−value∗SE the one sample t-test. Independent two-sample t-test whether two groups differ from
each other
mean ± z ( 2 ) × SE
